# ATAR Notes: Forum

## HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: jakesilove on January 28, 2016, 06:51:01 pm

Post by: jakesilove on January 28, 2016, 06:51:01 pm

If you have general questions about the HSC Mathematics course or how to improve in certain areas, this is the place to ask! 👌

Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.

To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

OTHER MATHEMATICS RESOURCES

Original post.
Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)

Hey everyone!

A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!

This is a community, so we want you to feel like you can post any type of 2U Mathematics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Maths is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.

I got an ATAR of 99.80, and a mark of 96 in the Mathematics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: liiz on February 03, 2016, 05:03:10 pm
Hey Jake - I'm struggling with this question from my maths textbook: "A rectangular box is to have a square base and no top. If it's volume is 500cm3, find the least area of sheet metal of which it can be made." I'm not understanding how all this maximum and minimum stuff is applied to these kind of questions, I thought it just had to do with graphs? Thanks!! :)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 03, 2016, 06:30:16 pm
Hey Jake - I'm struggling with this question from my maths textbook: "A rectangular box is to have a square base and no top. If it's volume is 500cm3, find the least area of sheet metal of which it can be made." I'm not understanding how all this maximum and minimum stuff is applied to these kind of questions, I thought it just had to do with graphs? Thanks!! :)

Hey Liiz:

HEY Im jake (obviously not haha). Im a year 12 student who completed my 2u HSC mathematics course in year 11 and lm just happy to help out here. Now, this type of question is amongst one of the most difficult ones in geometrical applications of calculus, and unfortunately in HSC exams there WILL be harder ones. But don't worry, once you have practised enough, you will begin to seize some tricks to approach these questions.

BEFORE YOU DO ANYTHING, DRAW A DIAGRAM WITH LABELS
1. Highlight all USEFUL INFORMATIONS (in this case, highlight rectangular box, square base, no top, 500cm^3 and least area)
2. Appoint two variables to the unknown sides (in this case, I named the side length of the square base as x, and the height of the rectangular box as L)
3. There will be at least one number quantity in every one of these questions in 2u mathematics, so the first equation you should construct, using your name variables to construct an equation that uses the numbers provided by the question
4. Draw out the relationship between the two variables through this equation that you have constructed
5. Construct another equation using your variables and the subject that is asked for in the question (In this case, for example, we constructed an Area equation which directly relates to what we are asked to find)
6. Substitute in the equivalent expression of a variable (In this case, for example, L = 500/x^2, so we substitute any L we see with 500/x^2) to reduce the total number of values down to one, so that we can construct an equation entire out of only one variable, which then allows us to perform differentiation
7. Clean up the equation after the substitution to make life easier
8. Differentiate the equation
9. Let this derivative = 0 to find any stationary points (In an Exam, YOU MUST STATE "LET dy/dx = 0 TO FIND ANY STATIONARY POINTS, OTHERWISE MARKS MAYBE DEDUCTED!!!)
10. Solve the derivative equation and find a value for your variable which will be your stationary point
11. Test both sides to show that a local minimum/maximum occurs at your stationary points
12. Substitute your minimum value back into the area equation (or maximum value if the question asks for maximum area) to find the minimum area (or the maximum area if you substitute in the maximum value)

So here is my solution:
(http://i.imgur.com/t6wtIk1.jpg)

Hope you find my solutions clear and useful! If you are confused with anything, dont hesitate to ask!!! :D

Best Regards
Jacky He
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 03, 2016, 06:40:09 pm
Hey Jake - I'm struggling with this question from my maths textbook: "A rectangular box is to have a square base and no top. If it's volume is 500cm3, find the least area of sheet metal of which it can be made." I'm not understanding how all this maximum and minimum stuff is applied to these kind of questions, I thought it just had to do with graphs? Thanks!! :)

Hey Liiz! I see that HPL has already posted an answer, but I'll throw mine in as well in case any of the methodology is different. Hope it helps, and thanks for the questions!

(http://i.imgur.com/NubpZ8v.png?1)
(http://i.imgur.com/0beu0z5.png?1)

If anyone else wants to ask or answer questions, I would really appreciate the contribution! Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 03, 2016, 07:25:29 pm
Hey Liiz! I see that HPL has already posted an answer, but I'll throw mine in as well in case any of the methodology is different. Hope it helps, and thanks for the questions!

(http://i.imgur.com/NubpZ8v.png?1)
(http://i.imgur.com/0beu0z5.png?1)

If anyone else wants to ask or answer questions, I would really appreciate the contribution! Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)

Great work Jake, I think it was a really good idea for you to first state the equation of how to calculate the volume of the rectangular box (cant believe l forgot to include that) and to state that x does not equal to 0 which is the condition which must be met despite obvious in this case. Great combined effort and a really good question Liiz!!!
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: brenden on February 03, 2016, 08:21:06 pm
Great work Jake, I think it was a really good idea for you to first state the equation of how to calculate the volume of the rectangular box (cant believe l forgot to include that) and to state that x does not equal to 0 which is the condition which must be met despite obvious in this case. Great combined effort and a really good question Liiz!!!
You have the best handwriting I've ever seen.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 03, 2016, 09:03:38 pm
You have the best handwriting I've ever seen.

This is the best forum I have ever seen! :)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: liiz on February 08, 2016, 08:50:06 pm
Thanks everyone for the help with the question!! It was really useful to see both solutions :)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Jacqui105 on February 09, 2016, 12:56:43 am
Hi,

Okay, so I've got a (possibly silly) question to ask, but I actually - GENUINELY - do not have any idea and need help:

How does one study for Maths?

I seem to have a really consistent system for each of my subjects as to approaching study in the most efficient way possible, but I can not seem to find an effective way of studying for Maths. I've tried everything I can think of. Please help.

Thank you
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jamonwindeyer on February 09, 2016, 08:40:05 am
Hi,

Okay, so I've got a (possibly silly) question to ask, but I actually - GENUINELY - do not have any idea and need help:

How does one study for Maths?

I seem to have a really consistent system for each of my subjects as to approaching study in the most efficient way possible, but I can not seem to find an effective way of studying for Maths. I've tried everything I can think of. Please help.

Thank you

Hey Jacqui105!

Firstly, absolutely not a silly question. I guarantee a large portion of the students reading this would be wanting to ask something similar.

This is a huge question that demands a huge response. We are planning a resource specifically answering this exact question for you, so stay tuned!  ;D

In the meantime, I'll direct you to this guide I wrote a little while ago that has helped quite a few people. It contains a whole bunch of tips for everything Math related, including study tips!

Good luck  :D

Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: caninesandy on February 14, 2016, 02:39:43 pm
Hello HSC Math geniuses :D
I have a math differentiation and integration assessment soon and was just wondering if you could help me on how to set answers out, especially with lengthy questions. Here is one question I have recently, recived the correct answers but was not sure on how to structure my answer exactly:
A magazine advertisement is to contain 50cm^2 of lettering with clear margins of 4cm each at the top and bottom and 2cm at each side. Find the overall dimensions if the total area of the advertisement is to be a minimum.
Thank you so much for your help!!!!!!
;D ;D ;) :D :)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 14, 2016, 04:24:24 pm
Hello HSC Math geniuses :D
I have a math differentiation and integration assessment soon and was just wondering if you could help me on how to set answers out, especially with lengthy questions. Here is one question I have recently, recived the correct answers but was not sure on how to structure my answer exactly:
A magazine advertisement is to contain 50cm^2 of lettering with clear margins of 4cm each at the top and bottom and 2cm at each side. Find the overall dimensions if the total area of the advertisement is to be a minimum.
Thank you so much for your help!!!!!!
;D ;D ;) :D :)

Hey Caninesandy!

This was actually quite a difficult question, which required me to draw the picture in order to figure out what was going on! My general tips for these questions are

1. Figure out the equations that you have
2. Sub in the equations, so that you have a differentiatable one
3. Differentiate, and find the required minimum
4. Sub the found value into the original, to find the solution.

Here is the solution I came up with. Hope it helps! This is exactly how I would set out any question similar to this :)

(http://i.imgur.com/eyyUD6Q.jpg?1)
(http://i.imgur.com/FbVA77a.jpg?1)

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 15, 2016, 08:59:38 am
Hey Caninesandy!

This was actually quite a difficult question, which required me to draw the picture in order to figure out what was going on! My general tips for these questions are

1. Figure out the equations that you have
2. Sub in the equations, so that you have a differentiatable one
3. Differentiate, and find the required minimum
4. Sub the found value into the original, to find the solution.

Here is the solution I came up with. Hope it helps! This is exactly how I would set out any question similar to this :)

(http://i.imgur.com/hY8GLxv.png?1)
(http://i.imgur.com/hQTKRDX.png?1)

Jake

Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jamonwindeyer on February 15, 2016, 10:06:53 am
Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)

Definitely a transcription error HPL  :D
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 15, 2016, 11:38:18 am
Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)

Yep, definitely an error there, thanks for pointing that out! I'll fix up the proof and update the answer in a second.

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: aqsarana_ on February 20, 2016, 03:51:20 pm
Hi!
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jamonwindeyer on February 20, 2016, 05:35:10 pm
Hi!
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.

Hey aqsarana! Awesome question. Your calculator should be in radian measure by default, since you must use radians for trigonometric calculus. The only time you switch to Degree measure is if degrees are used in the question, that is, values are given in degrees. This normally happens with geometry questions. For these questions, switch to degree measure, then switch back immediately when you are finished.

But yes, by default, leave your calculator in Radian measure. Only switch to Degrees if you see degree measurements in the question.

Hope this helps  ;D
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on February 20, 2016, 05:44:51 pm
To make what Jamon said more explicit.

You will find that in questions that ask you to be in degrees, there will be a degree symbol (°) present. As soon as that goes missing, naturally assume radians.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 20, 2016, 05:48:51 pm
And by personal experience there will definitely be cases where you would forget to change from radian mode to degrees modes and this is gonna be very disastrous. So some of my friends decided to bring two calculators, one labelled degrees and another labelled radians. If you are struggling to remember to change to radian mode then you might like to consider that method.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jamonwindeyer on February 20, 2016, 06:42:41 pm
And by personal experience there will definitely be cases where you would forget to change from radian mode to degrees modes and this is gonna be very disastrous. So some of my friends decided to bring two calculators, one labelled degrees and another labelled radians. If you are struggling to remember to change to radian mode then you might like to consider that method.

I also heard of students doing this! One person in my year did it, it's not a bad idea if you are so inclined!
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 21, 2016, 12:44:24 pm
Hi!
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.

Definitely agree with all of the above: If the question mentions degrees, use degrees. If the question doesn't mention degrees, use Radians. Simple as that! If you sometimes get confused, maybe it is worth getting into the habit of highlighting every time you see a degrees symbol in a question, so you know you need to put your calculator into degrees mode.

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Phillorsm on February 21, 2016, 09:57:47 pm
Hey jake, having trouble integrating this.
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 21, 2016, 10:01:45 pm
Hey jake, having trouble integrating this.
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )

Hey Phillorsm:

Is your question a definite or indefinite integral?

Assuming that your question is an indefinite integral, the answer should not be zero because after all, area cannot be zero right? I like your approach of drawing a semi-circle and you have correctly identified the domain of the semi-circle. So all you really had to do was apply pi(r^2)/2 to find the area of the semi circle, and that will give you the integral (l believe you have identified that r=2 because 4 = 2^2). Considering that this is an indefinite integral question, you will need to add +C at the end.

(sorry just ignore the sintheta stuff in the background, they are irrelevant)

(http://i.imgur.com/azgnjfQ.jpg)

If you have any further questions dont hesitate to ask!

Best Regards
Happy Physics Land
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 21, 2016, 10:07:51 pm
Hey jake, having trouble integrating this.
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )

Hey Phillorsm!

If you are looking to find the area underneath that curve, you've identified one great way to do it: by understanding that the graph is a semicircle, use the formula (pi*r^2)/2 !

However, you should also be able to integrate as normal and find the required result. I'm not at home at the moment, so can't write out a full solution, but hopefully someone else on the forums can!

Hope this helps! Let me know if there's anything you're unsure of.

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Loki98 on February 23, 2016, 08:56:39 pm
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 24, 2016, 12:09:02 am

Hey Loki!

This is one of the more difficult Finance question I've seen. Once you learn the general method, however, you will have no problem! Below is my solution :)

(http://i.imgur.com/MpSJxr4.png?1)
(http://i.imgur.com/IctwnG7.png?1)
(http://i.imgur.com/0GBauHq.png?1)

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Phillorsm on February 26, 2016, 09:23:55 pm
Hey Jake, I just got my results back for my first 2u exam for year 12. It was on sequence and series, applications of calculus, integration and logs and exponentials. I got a 55/78 because I didn't finish the last section (logs and expos) of the exam. My teacher did say I smashed the first three parts though.
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 26, 2016, 10:15:25 pm
Hey Jake, I just got my results back for my first 2u exam for year 12. It was on sequence and series, applications of calculus, integration and logs and exponentials. I got a 55/78 because I didn't finish the last section (logs and expos) of the exam. My teacher did say I smashed the first three parts though.
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?

Hey Phillorsm:

Im clearly not Jake (obviously) but I just wanna to be here to tell you that getting around a 70 for your first 2u assessment is really nothing to worry about. Im not just saying this to boost you with blind confidence, Im saying this because it's too early a stage to give up. If you've ever been to one of Jake's lectures, he told us his experience not being able to finish TWO PAGES of his trial exam for maths ext 1 upon his return from the international science forum. Yet he still got a remarkable 98. So there is absolutely nothing for you to worry about.

Another real life example from me: one of my classmates for accelerated maths only got 60 percent in his first exam, and throughout the entire course his mark fluctuated around 70 percent. Yet in the end he scored a mark of 85 for his 2u mathematics. So if you look at yourself now, you are in a much better situation. You know majority of your stuff in the first three sections and you are in a better starting point than my friend! As a person who's gone through 2u hsc mathematics before, I can tell you that in the end, after the tonnes of practise papers you have done, you will begin to realise how much you have improved and how surprisingly better you will be end up doing in your hsc exam. So dont get defeated by this one exam, because to be honest, I have failed 2 of my first assessments this year as well and I found out that in the end, they only contribute to 5% (sometimes not even) of your final mark.

Ok hmmm regarding exam techniques, I think you might have been stuck on a question for too long and that may have caused your inefficiency in time management. Of course you will hear people saying to you "skip the question if its taking too long", but I do empathise with you because sometimes we are fairly certain we have seen the particular question before and it shouldnt take us much time to complete it. Unfortunately sometimes we can make a mistake that we wouldnt realise because its too trivial or the teacher could have added something extra in the question to make the question one step harder. As a general rule of thumb for me, if there is a time-consuming question that you know you definitely can solve in a multiple choice section, SKIP IT AND COME BACK LATER. Surely, if we solve the question it boosts our ego, but its not worth spending 5 minutes on a question that merely awards you with 1 mark. If a question is in the free-response section and you have looked at it for 3 minutes not knowing where you should be starting, then I would also suggest to skip it and come back later.

Logarithmic function and exponentials definitely can be hard, perhaps one of the hardest concepts in 2u. I would recommend, ofc, to do past trial papers (you can find links in atarnotes to access trial papers) and practise those questions on the topic. Also practise some past HSC questions.

Tips I would give on logarithmic functions:
- (This is especially common with difficult integration questions involving logs). KNOW YOUR YEAR 8 GEOMETRY WELL! If you look up question 16b) on 2015 paper you will realise that you have to draw a rectangle and use integration about y-axis to figure out the area encompassed by the y-axis, the scope of the rectangle and the curve first, then use your area of rectangle to subtract that area to find the area you were asked to find. This is a VERY USEFUL TECHNIQUE.
- I often struggled with the idea of changing the bases, simply because its not frequently tested. But if the teacher wants to make a hard question, you would need to change the base as a technique to approach your final answer, so be familiar with the process
- I remember in one of my earlier exams, I was asked to find the area under a log curve and I accidentally wrote 1/x as an integral of ln x. Sometimes when the exam pressure kicks in, you really can be making those sorts of mistakes. So just remember that at your level, you definitely wouldnt be required to integrate ln x, and if you did end up having to integrate ln x, you know there must be an easier way.
- There is also that kind of typical question that's involved in both integration and the integration of logarithmic functions that involve several parts. The first part would most likely ask you to differentiate and then in the second step ask you to integrate. Keep in mind if you see this sort of question in integration/integration of logs and exponentials, you know that you can use part I in part II. You can somehow manipulate what you are required to integrate in part II into the outcome of differentiation in part I and then just integrate it saying "using part I, we know that _____ is a primitive function to _____, hence the integral is ______"

I will say this much for now, but yes definitely do not give up. The game has just started and sometimes its good to be on a lower end of a ladder because it avoids arrogance and it allows you to put in more effort to do better in the next exams.

Good luck!

Best Regards
Happy Physics Land
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 26, 2016, 11:15:57 pm
Hey Jake, I just got my results back for my first 2u exam for year 12. It was on sequence and series, applications of calculus, integration and logs and exponentials. I got a 55/78 because I didn't finish the last section (logs and expos) of the exam. My teacher did say I smashed the first three parts though.
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?

Hi Phillorsm!

Firstly, I just wanted to congratulate you on your fantastic 2u result! 71% is nothing to be worried about: in fact, I would say that the fact that you were able to do so well in the sections you completed puts you in good stead for the  HSC (more on that later).

Next, I want to echo the words of HPL. I absolutely agree with his advice, and won't repeat anything he has said. Specifically, the idea of skipping questions and coming back and knowing specific techniques to answer difficult questions.

The only point that I would add, when discussing ways to speed up exam completion, is DO PAST PAPERS! Do them again, and again, and again. This is the best way to speed up answering questions. The reason is that, if you can spend one minute less per question figuring out what you actually need to do (ie. by recognizing that an answer should be identical to a practice response you did the day before), you will gain so much extra time. This is extremely valuable in finishing assessment tasks!

So, my main tip for you is to just do hundreds of past papers, in timed conditions, again and again. There are never too many papers you can complete!

The fact that you know your stuff is a really good indicator of how you are going in the year. Timing can always be improved, and the more past papers you do the faster you will get in completing individual questions. Knowing the content in-depth, however, is a far more difficult task. So, congratulations for where you are at! Just keep doing past papers, and keep in mind the comments made by HPL.

Jake :)

Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: amandali on February 27, 2016, 08:10:47 am
cos(2x+pie/6)=1/root2     -pie<=x<=pie

how do i find the angles in a simple way without getting the angles confused?
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on February 27, 2016, 08:43:13 am
cos(2x+pie/6)=1/root2     -pie<=x<=pie

how do i find the angles in a simple way without getting the angles confused?
$\text{For starters, since your subject is not }x\text{ but rather }2x+\frac{\pi}{6}, \text{ you must change your boundaries:}\\ -\pi\le x\le \pi \\ \Leftrightarrow -2\pi\le 2x\le 2\pi \\ \Leftrightarrow -\frac{11\pi}{6}\le 2x+\frac{\pi}{6}\le \frac{13\pi}{6}$
$\text{Now that the boundaries have been changed, we can consider the quadrants using "ASTC".} \\ \text{In the positive direction, the cosine ratio is positive in 1st and 4th quadrants.} \\ \therefore \cos{\left(2x+\frac{\pi}{6}\right)}=\frac{1}{\sqrt{2}}\\ \Rightarrow 2x+\frac{\pi}{6}= \frac{\pi}{4}, \, 2\pi-\frac{\pi}{4}\\2x+\frac{\pi}{6}=\frac{\pi}{4}, \,\frac{7\pi}{4}\\ 2x=\frac{\pi}{12}, \frac{19\pi}{12} \\ x=\frac{\pi}{24}, \frac{19\pi}{24}$
$\text{Now, keep in mind that when we go in the negative quadrants we are going in REVERSE. The negative 1st quadrant corresponds to the positive 4th quadrant.} \\ \text{The cosine ratio is positive in both the negative 1st and negative 4th quadrants.} \\ \text{We have: }\\ \cos{\left( 2x+\frac{\pi}{6}\right) }=\frac{1}{\sqrt{2}}\\ \Rightarrow 2x+\frac{\pi}{6}=-\frac{\pi}{4}, \, -2\pi+\frac{\pi}{4}\\2x+\frac{\pi}{6}=-\frac{\pi}{4}, \, -\frac{7\pi}{4}\\ 2x=-\frac{5\pi}{12},-\frac{23\pi}{12}\\ x=-\frac{5\pi}{24},-\frac{23\pi}{24}$

Edit: Holy crap the mistakes.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Phillorsm on February 27, 2016, 05:03:16 pm
Thank you Jake and HPL, I really value all your help and time you spend here sharing your wisdom. HPL, what you said about not coming first giving you the motivation to work harder for the following exams was especially helpful :) Thanks again guys, now I need to get stuck into hardcore study mode for an ext1 exam on wednesday. Binomial, inverse, induction and harder 2u... Wish me luck!  ;D
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: amandali on February 27, 2016, 07:55:06 pm
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on February 27, 2016, 09:09:39 pm
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Weakly lacking information. There are three cases:
1. We want to minimise the time of the entire journey (this sounds extremely preferable to the others)
2. We want to minimise the distance of the entire journey (sounds very unfeasible as there is actually a very obvious answer)
3. We want to minimise only a part of the journey (e.g. reduce some walking

If the earlier answer didn't make sense, all the hints were given.
a) Rewrite boundaries because otherwise you don't know what you're trying to get at
b) Always use ASTC
c) Negative angles work in reverse (clockwise around ASTC)

Exact values are different. Use the formula sheet for exact values.
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 27, 2016, 09:12:34 pm
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Hey!

This was actually quite a difficult question! First, you need to construct the relevant equation. Then, the differentiation part is quite tricky for a 2U question. Finally, even solving the equation is a bit of work! Below is my answer, I hope you find it helpful! In the case of questions like this, just think through what is actually happening in the example. Then, always create some sort of equation that you can differentiate!

(http://i.imgur.com/sujfvuQ.png?1)
(http://i.imgur.com/FxQw2xR.jpg?1)

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on February 27, 2016, 09:20:05 pm
Assumed we wanted to minimise time:

P.S. Woah Jake what are you doing with the trig!
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: Happy Physics Land on February 27, 2016, 09:22:05 pm
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Hey Amanda:

Another great question from you amanda! Ok, so when I first looked at the question, my instinct tells me I will have to draw a diagram, because it involves distance and speed and even worse theres also time. And I will have to admit, despite my distaste towards drawing diagrams, it is essential for you to draw one because graphics help you to visualise stuff. So next time when you see these types of questions, definitely draw a diagram, its worth the time.

Ok and then I manipulated pythagorus theorem, made an algebraic expression for  the time to travel distance MS, made another algebraic expression for the time to travel distance BS. The time taken to travel the total distance now becomes T(MS) + T(BS), where T =time. So after establishing such a relationship, the rest is straight forward: simply repeat the conventional process of simplifying, differentiating, make derivative = 0, and then find stationary point. Afterwards you test a value on both sides of stationary point to determine its nature (I.e. maximum or minimum) and then make a concluding statement for your answer. Actually it is quite crucial to include a concluding statement, because that makes it clearer for the marker what your final result is and you would less likely to be deducted a mark on not stating the result clearly.

Anyways, my solution as below:

(http://i.imgur.com/M88wbvs.jpg)
(http://i.imgur.com/zIqIQMu.jpg)

Sorry for the messy working btw, if you have any further questions dont hesitate to ask! :)

Best Regards
Happy Physics Land
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jakesilove on February 27, 2016, 09:26:50 pm
Assumed we wanted to minimise time:

P.S. Woah Jake what are you doing with the trig!

So, this is a wayyyy easier solution. There are so many ways to answer questions like these, so whichever makes more sense to you! Still, thanks everyone for participating and answering questions on the forums!

My method is definitely more tricky, because I have a tendency to over-use trig. If you can use algebra, that's usually a better and simpler solution.

Jake
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: 16ebond on March 07, 2016, 06:09:18 pm
Hey Jake
So we have been doing e in class and I am a bit stuck on how to answer this question.
I have attached the question, so hopefully the file will open.
Thanks heaps
Em  :)
Title: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on March 16, 2016, 04:50:39 pm
I'm not sure if this was meant to be an unanswered question but here's a solution

Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.

(http://i.imgur.com/fzE9TxN.png)

So the breadth and length of the triangle are:

$B=e^{-x^2}$
$L= 2x \quad (x - (-x) = 2x)$

Hence we can combine these to give an expression for area:

$A=-2xe^{-x^2} \\ \frac{dA}{dx}=e^{-x^2}\times(-2x+4x^2)$

Set dA/dx=0 to maximise:

$e^(-x^2)\neq 0$
$\therefore -2x+4x^2=0\\ x=0, x=\pm\frac{1}{\sqrt{2}}$

Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).

So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.

I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex  ;D)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: jamonwindeyer on March 17, 2016, 01:05:32 pm
I'm not sure if this was meant to be an unanswered question but here's a solution

Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.

(http://i.imgur.com/fzE9TxN.png)

So the breadth and length of the triangle are:

$B=e^{-x^2}$
$L= 2x \quad (x - (-x) = 2x)$

Hence we can combine these to give an expression for area:

$A=-2xe^{-x^2} \\ \frac{dA}{dx}=e^{-x^2}\times(-2x+4x^2)$

Set dA/dx=0 to maximise:

$e^(-x^2)\neq 0$
$\therefore -2x+4x^2=0\\ x=0, x=\pm\frac{1}{\sqrt{2}}$

Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).

So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.

I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex  ;D)

LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion  ;D
Post by: kal.123 on March 19, 2016, 11:43:17 am
I was wondering how to draw the x-intercepts and vertex??? For Y= 3-2xsquared
Post by: jamonwindeyer on March 19, 2016, 12:00:28 pm
I was wondering how to draw the x-intercepts and vertex??? For Y= 3-2xsquared

Sure! So we have:

$y=3-2x^2$

The x-intercepts occur when y=0, so we just substitute and solve. This is a quadratic, so we can expect anywhere from o to 2 answers:

$3-2x^2=0 \\ 2x^2=3 \\ x^2=\frac{3}{2} \\ x=\pm\sqrt{\frac{3}{2}}$

Now, for the vertex, we are actually taught that the x-coordinate of the vertex is given by:

$x_\text{vertex}=\frac{-b}{2a}$

Now for your quadratic, a=2, b=0 and c=-3 (when rearranged), so the x-coordinate of the vertex will actually be zero. The y-coordinate will then be y=3 by substitution of x=0.

So, we have intercepts and a vertex, we can now draw a diagram. Hope this helps!

(http://i.imgur.com/jem7eQy.png?1)
Title: Re: 96 in 2U Maths: Ask me Anything!
Post by: RuiAce on March 19, 2016, 05:53:28 pm
LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion  ;D

Ah sweet. Thanks for the aid too Jamon!
Post by: imtrying on March 21, 2016, 04:26:50 pm
Hey :)
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful :)
Post by: jamonwindeyer on March 21, 2016, 09:39:11 pm
Hey :)
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful :)

Sure thing!!

$y = \frac{5-x}{(4x^2+1)^3}$

The quotient rule says:

$y'=\frac{u'v-v'u}{v^2}$

$u = 5-x \quad\quad u'=-1 \quad\quad v=(4x^2+1)^3 \quad\quad v'=3(4x^2+1)^2\times8x = 24x(4x^2+1)^2$

Note that our differentiation of that last term is done with the chain rule!

So the first derivative by substitution is:

$y'=\frac{(-1)(4x^2+1)^3-24x(4x^2+1)^2(5-x)}{(4x^2+1)^6}$

Through cancellation and then expansion, we obtain:

$y'=\frac{(-1)(4x^2+1)-24x(5-x)}{(4x^2+1)^4} \\ =\frac{20x^2-120x-1}{(4x^2+1)^4}$

Now for the second derivative, it works the exact same way!

$y'=\frac{(20x^2-120x-1)}{(4x^2+1)^{-4}}$

$y''= \frac{a'b-b'a}{b^2}$

$a=(20x^2-120x-1) \quad\quad a'=(40x-120) \quad\quad b=(4x^2+1)^{4} \quad\quad b'=32x(4x^2+1)^{3}$

$\therefore y''=\frac{(40x-120)(4x^2+1)^{4}-32x(20x^2-120x-1)(4x^2+1)^{3}}{(4x^2+1)^8} \\ =\frac{(40x-120)(4x^2+1)-32x(20x^2-120x-1)}{(4x^2+1)^5}$

Which through expansion and factorisation will turn into your solution for Part B!

I hope this helps!

Post by: Marlo365 on March 28, 2016, 06:57:44 pm
Hey Jake, so i got my 2u half yearly tomrrow and not sure what i'm supposed to be doing this very moment wether its past papers of relaxing haha, any advice would  be great
Post by: jakesilove on March 28, 2016, 07:37:43 pm
Hey Jake, so i got my 2u half yearly tomrrow and not sure what i'm supposed to be doing this very moment wether its past papers of relaxing haha, any advice would  be great

Hey Marlo!

I think that it's totally up to you! Generally, I would recommend that you just keep doing past papers until a set time. Say you want to get 8 hours of sleep: figure out when you need to stop, and GO TO BED!

If you're feeling nervous, do a few past papers. If you find that this is making you more nervous, call it quits for the night! I really don't have any particular suggestion as to what you should do: just whatever makes you feel the most comfortable. If you're totally indifferent, maybe have a crack at one more paper (who know if it will help you tomorrow!) but definitely don't stay up all night.

Absolute best of luck, I'm sure you'll do great!

Jake
Post by: imtrying on March 29, 2016, 06:11:02 pm
Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car. Answer: (a)$592.00     (b) $39 319.89 Title: Re: Mathematics Question Thread Post by: liiz on March 29, 2016, 06:16:09 pm Hello, needing some help to integrate this: x / 2x2 - 3 Thankyou! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on March 29, 2016, 06:37:49 pm Hello, needing some help to integrate this: x / 2x2 - 3 Thankyou! Hey Liiz! Sure thing! Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm. Remember, the derivative of the log of some function is: $\ln{f(x)} = \frac{f'(x)}{f(x)}$ For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4: $\int\frac{x}{2x^2-3}dx = \frac{1}{4}\int\frac{4x}{2x^2-3}dx \\\\ =\frac{1}{4} \ln{(2x^2-3)} + C \\ \text{where C is a constant of integration}$ This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps ;D Title: Re: Mathematics Question Thread Post by: jamonwindeyer on March 29, 2016, 06:44:01 pm Loan Repayments Question The smith family buys a car for$38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
(a) $592.00 (b)$39 319.89

I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack  ;D
Post by: imtrying on March 29, 2016, 06:58:30 pm
I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack  ;D

All good :)
Post by: RuiAce on March 30, 2016, 11:18:21 am
Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car. Answer: (a)$592.00     (b) 39 319.89 $\text{Let the monthly repayment be equal to }M\text{ and observe that we are compounding interest monthly.}\\\text{This means we consider 60 periods.}\\\text{From experimentation, I have also determined that the interest rate they gave was per annum.}\\\text{Monthly interest rate}=0.015/12=0.00125$ \begin{align*}{A}_{0}&=34200\\{A}_{1}&={A}_{0}\left(1.00125\right)-M\\&=34200\left(1.00125\right)-M\\{A}_{2}&={A}_{1}\left(1.00125\right)-M\\&=34200{\left(1.00125\right)}^{2}-M\left(1.00125\right)-M\\{A}_{3}&={A}_{2}\left(1.00125\right)-M\\&=34200{\left(1.00125\right)}^{3}-M{\left(1.00125\right)}^{2}-M\left(1.00125\right)-M\\&\vdots\\{A}_{n}&=34200{\left(1.00125\right)}^{n}-M{\left(1.00125\right)}^{n-1}-M{\left(1.00125\right)}^{n-2}-\dots-M\\&=34200{\left(1.00125\right)}^{n}-M\left(1+1.00125+{1.00125}^{2}+\dots+{1.00125}^{n-1}\right)\\&=34200{\left(1.00125\right)}^{n}-M\left(\frac{{1.00125}^{n}-1}{0.00125}\right)\end{align*} \text{We want }{A}_{60}=0\\\begin{align*}\Rightarrow 0&=34200{\left(1.00125\right)}^{60}-M\left(\frac{{1.00125}^{60}-1}{0.00125}\right)\\\Leftrightarrow M&=\frac{34200{\left(1.00125\right)}^{60}}{\left(\frac{{1.00125}^{60}-1}{0.00125}\right)}\\\Leftrightarrow M&=591.9981715...\end{align*}\\\text{So the monthly repayment to the nearest cent is }\592 $\text{The total amount of payments is simply}\\\text{Amount paid in the deposit + Amount paid every month for ALL 60 months}\\=3800 + 60*591.9981715...\\=39319.89029...\\\text{Therefore, to the nearest cent, }\39319.89$ Title: Re: Mathematics Question Thread Post by: liiz on March 30, 2016, 04:59:15 pm Hey Liiz! Sure thing! Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm. Remember, the derivative of the log of some function is: $\ln{f(x)} = \frac{f'(x)}{f(x)}$ For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4: $\int\frac{x}{2x^2-3}dx = \frac{1}{4}\int\frac{4x}{2x^2-3}dx \\\\ =\frac{1}{4} \ln{(2x^2-3)} + C \\ \text{where C is a constant of integration}$ This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps ;D Ah okay, yep I understand! Thanks so much Title: Re: Mathematics Question Thread Post by: SimonaB on March 30, 2016, 09:54:08 pm I've got a question that was on my exam and I got it wrong after literally 10 attempts. The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6 Thanks! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on March 30, 2016, 10:50:49 pm I've got a question that was on my exam and I got it wrong after literally 10 attempts. The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6 Thanks! Hey Simona! Welcome to the forums!! ;D Let's have a look. The first step for an area question of this nature is to draw a quick sketch, just to get a better picture of what we are dealing with. Wolfram Alpha helped me out here, but you don't need any great level of accuracy, we just want the position of the curves with relation to each other. Note also that this sketch from Wolfram Alpha has the axes centred at the first point of intersection, totally irrelevant to this question, just ignore the coordinates! ;D (http://i.imgur.com/nTy2OY1.png) Okay! So it is clear from a quick sketch that the sinx curve is above the cos2x curve in this domain. There is a really neat trick here. Even when the curves go above or below the x axis, any weird combination, we can always find the area by finding the area under the upper curve and subtracting the area under the lower curve, in the given domain. It's nice, and eliminates the need for weird sums of absolute values of integrals. Literally, the integral of the upper curve subtract the integral of the lower curve, works every time ;D So, the working would be: $A = \int^\frac{5\pi}{6}_\frac{\pi}{6}\sin{x}-\cos{2x}dx \\ = [-\cos{x}-\frac{1}{2}\sin{2x}]^\frac{5\pi}{6}_\frac{\pi}{6} \\ = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} \\ =\frac{3\sqrt{3}}{2}u^2$ I hope this helps!! Title: Re: Mathematics Question Thread Post by: SimonaB on March 30, 2016, 11:12:37 pm Thanks heaps! I see my mistake now.. ::) Title: Re: Mathematics Question Thread Post by: Maz on April 03, 2016, 04:47:10 pm hey can u please help me with how to differentiate this...i really don't know how to... i'd really appreciate it :) Title: Re: Mathematics Question Thread Post by: RuiAce on April 03, 2016, 04:56:02 pm hey can u please help me with how to differentiate this...i really don't know how to... i'd really appreciate it :) Can't see your image. If your integral was something along the lines of this then in NSW it belongs under MX2. $\int_{1}^{1+{x}^{2}}{\frac{1}{\sqrt{1+{e}^{t}}}\,dt}$ Also, you said differentiate? Did you mean you wanted to put d/dx in front of that integral? (or d/dt inside it?) Title: Re: Mathematics Question Thread Post by: Maz on April 03, 2016, 04:59:49 pm Can't see your image. If your integral was something along the lines of this then in NSW it belongs under MX2. $\int_{1}^{1+{x}^{2}}{\frac{1}{\sqrt{1+{e}^{t}}}\,dt}$ Also, you said differentiate? Did you mean you wanted to put d/dx in front of that integral? (or d/dt inside it?) yes that is the integral...and d/dx belongs in front of it...sorry... and ill go put it in MX2 now then :) Title: Re: Mathematics Question Thread Post by: Johny1234567 on April 03, 2016, 08:46:21 pm Hello y'all. How do I solve this question: Find the point T(a,lna) on y=lnx where the tangent passes through the origin, i.e. find the value of a. Title: Re: Mathematics Question Thread Post by: RuiAce on April 03, 2016, 08:56:47 pm Hello y'all. How do I solve this question: Find the point T(a,lna) on y=lnx where the tangent passes through the origin, i.e. find the value of a. $y=\ln{x}\Rightarrow\frac{dy}{dx}=\frac{1}{x}\\\text{When }x=a,\, \frac{dy}{dx}=\frac{1}{a}={m}_{tangent}$ \text{So the equation of the tangent is just }\\\begin{align*}y-\ln{a}&=\frac{1}{a}\left(x-a\right)\\\Leftrightarrow ay-a\ln{a}&=x-a\end{align*} \text{The tangent passes through (0,0), so}\\\begin{align*}0-a\ln{a}&=0-a\\\Leftrightarrow \ln{a}&=1\\\Leftrightarrow a&=e\end{align*} Title: Re: Mathematics Question Thread Post by: aqsarana_ on April 10, 2016, 05:11:45 pm Hi, So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help! Thank you :) Title: Re: Mathematics Question Thread Post by: jakesilove on April 10, 2016, 05:30:32 pm Hi, So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help! Thank you :) Hey! I really think it depends on what you want a textbook for. I had the Cambridge textbook, which (if I recall correctly) does a fairly good job of explaining concepts etc. It has a good range of past questions, which will challenge and extend any 2U student. However, most students don't use their textbook to explain concepts. Usually, students who actually USE their textbook are the super keen ones. In that case, they just want a billion past questions to smash through. I would always recommend doing HSC past questions rather than textbook ones. They give you a better idea of where you're at, what you need to prepare for, and exactly how you'll be assessed. However, if you've gone through all the past papers so many times that you feel like looking at them, or just want some variety, a textbook is a great option. If you're just looking for questions, I love the Coroneos textbooks. They are simple, to the point. There are a trillion billion questions to answer. The style takes a bit of getting used to (its like they used a type writer), but the range of questions is really great. It'll extend you as well as test your basic knowledge. But again, it depends on why you want the textbook. I don't know much about Fitzpatrick, but I would generally lean towards Cambridge of the two. Hope this helps! I'm sure others on here will have different opinions. Jake Title: Re: Mathematics Question Thread Post by: RuiAce on April 10, 2016, 08:14:37 pm Hi, So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help! Thank you :) Comparison between Fitzpatrick and Cambridge is somewhat hard. This is because the two textbooks are kinda on par with question quality. However, the Cambridge textbook's prime advantage is that it splits into foundation, development and extension, whereas Fitzpatrick doesn't mix up the topic order. Keep that in mind. Of course, when it comes exam time, as mentioned always use past papers. Title: Re: Mathematics Question Thread Post by: IkeaandOfficeworks on April 14, 2016, 07:34:45 pm Hi guys!, I seem to struggle with this integration question.Can you explain why the answer is 1/6(2x-1)3/2 + 1/2(2x-1)1/2 + C ?. Thank you! Title: Re: Mathematics Question Thread Post by: RuiAce on April 14, 2016, 08:28:33 pm Hi guys!, I seem to struggle with this integration question.Can you explain why the answer is 1/6(2x-1)3/2 + 1/2(2x-1)1/2 + C ?. Thank you! EDIT BY JAMON: Hey IkeaandOfficeworks! Glad you got an answer to your question, for any 2U students looking at this, don't stress, this is Extension 1 content! $u=2x-1\Leftrightarrow x=\frac{1}{2}\left(u+1\right)\Rightarrow \frac{1}{2}\,du=dx\\$ \begin{align*}\int{\frac{x}{\sqrt{2x+1}}\,dx}&=\int{\frac{\frac{1}{2} \left(u-1\right)}{\sqrt{u}}\cdot \frac{1}{2}\,du}\\&=\frac{1}{4}\int{\left({u}^{\frac{1}{2}}-{u}^{-\frac{1}{2}}\right)\,du}\\&=\frac{1}{4}\left(\frac{2{u}^{\frac{3}{2}}}{3}-2{u}^{\frac{1}{2}}\right)\\&=\frac{1}{6}{\left( 2x-1 \right)}^{\frac{3}{2}}-\frac{1}{2}{\left( 2x-1 \right)}^{\frac{1}{2}}\end{align*} Title: Re: Mathematics Question Thread Post by: jamesq on April 24, 2016, 02:14:23 am I need help with a rates and finance question. For the AP: 100 + 97 + 94 + ... Show that 68 is the minimum number of terms in which Sn is negative. Title: Re: Mathematics Question Thread Post by: RuiAce on April 24, 2016, 08:45:44 am I need help with a rates and finance question. For the AP: 100 + 97 + 94 + ... Show that 68 is the minimum number of terms in which Sn is negative. $\text{Generate the formula for the }n\text{th partial sum using the formula}\\{S}_{n}=\frac{n}{2}\left(2a+(n-1)d\right)$ $\text{The first term is 100 with common difference -3 so:}\\{S}_{n}=\frac{n}{2}\left(200-3(n-1)\right)$ \text{We need the first case such that }n\text{ will yield a negative value. Thus we solve:}\\ \begin{align*}0&=\frac{n}{2}\left(200-3(n-1)\right)\\0&=n(203-3n)\\203-3n&=0 \quad \text{since }n\neq 0\\n&=203/3=67.66666\dots\end{align*}\\ \text{Hence, we require }n\text{ to be the next integer above 67.666, that is, 68.} Title: Re: Mathematics Question Thread Post by: Phillorsm on May 08, 2016, 05:34:27 pm Hey Guys, Some help with this question please? https://goo.gl/3ZYQOC Title: Re: Mathematics Question Thread Post by: jakesilove on May 08, 2016, 06:07:16 pm Hey Guys, Some help with this question please? https://goo.gl/3ZYQOC Hey! Below are my answers :) The whole tan(2x) thing was a little bit nuanced, but really it ended up being just about playing around with what you're given, and trying to figure out what they're actually asking you to do. Worst case scenario with questions like this; you can always test it on a calculator, figure out the answer without the whole 'hence' part, and fudge the reasoning. (http://i.imgur.com/MiyII5t.png?1) (http://i.imgur.com/L1oN7Xc.png?1) Jake Title: Re: Mathematics Question Thread Post by: Phillorsm on May 08, 2016, 08:23:12 pm Hey Jake, Thanks for the help! But i'm a little confused, isn't there another intersection at -3pi/8? Which is before the endpoint at -pi/2... Title: Re: Mathematics Question Thread Post by: RuiAce on May 08, 2016, 08:43:29 pm $\text{The solution to }\tan{2x}\le 1 \text{ over the domain }-\frac{\pi}{2}< x<\frac{\pi}{2}\text{ is given by: }\\ -\frac{\pi}{2}< x \le -\frac{3\pi}{8}\\ -\frac{\pi}{4}< x \le \frac{\pi}{8}\\ \frac{\pi}{4} < x < \frac{\pi}{2}$ $\text{However, the link isn't working for me and I can't see the question. Therefore I will refrain from posting a method for now.}\\\text{These answers are courtesy of WolframAlpha anyway.}$ Title: Re: Mathematics Question Thread Post by: jakesilove on May 08, 2016, 09:09:24 pm Hey Jake, Thanks for the help! But i'm a little confused, isn't there another intersection at -3pi/8? Which is before the endpoint at -pi/2... Hmm this is true, not sure how I missed that (was cracking out an answer on the bus, so I'll blame it on that...). I'm unfortunately not home at the moment, but will just admit my mistake and let someone else correct me. Sorry about that! You may even be able to get to an answer based roughly on my method (although hopefully more correctly). Jake Title: Re: Mathematics Question Thread Post by: Phillorsm on May 08, 2016, 09:35:06 pm No worries Jake, yeah I used your method and was able to get a solution I believe to be right. You had the right idea, just missed one little thing. No problem, HAPPY BIRTHDAY BY THE WAY! Title: Re: Mathematics Question Thread Post by: jakesilove on May 09, 2016, 10:25:06 am No worries Jake, yeah I used your method and was able to get a solution I believe to be right. You had the right idea, just missed one little thing. No problem, HAPPY BIRTHDAY BY THE WAY! Thanks Phillorsm! Appreciate it :) Title: Re: Mathematics Question Thread Post by: Phillorsm on May 09, 2016, 02:55:56 pm No problem Jake, another question for you. I can do up to part iii, but part IV I am stuck. https://goo.gl/photos/wTkpMHqM6dVkFfSr7 Title: Re: Mathematics Question Thread Post by: jakesilove on May 09, 2016, 03:13:06 pm No problem Jake, another question for you. I can do up to part iii, but part IV I am stuck. https://goo.gl/photos/wTkpMHqM6dVkFfSr7 There ya go! Just think about what you need to find, use some random triangle to try and get the answer out! (http://i.imgur.com/HPOel05.png?1) Jake Title: Re: Mathematics Question Thread Post by: Phillorsm on May 10, 2016, 05:17:15 pm Hey Jake, wondering if we could revisit this question. For the final part, (the tan inequality) I seem to be getting different results from different places. The one I am currently thinking is correct is -pi/2<=x<=-3pi/8, -pi/4<x<=pi/8, pi/4<x<=pi/2 https://goo.gl/Hg97px Title: Re: Mathematics Question Thread Post by: RuiAce on May 10, 2016, 06:30:43 pm Hey Jake, wondering if we could revisit this question. For the final part, (the tan inequality) I seem to be getting different results from different places. The one I am currently thinking is correct is -pi/2<=x<=-3pi/8, -pi/4<x<=pi/8, pi/4<x<=pi/2 https://goo.gl/Hg97px Refer to my post on the previous page. I have already addressed this stating that yes that IS the solution, except for one thing. pi/4<x<pi/2 x can't equal to pi/2 Title: Re: Mathematics Question Thread Post by: Maz on May 10, 2016, 10:15:32 pm hey can someone please explain how to do this question for me? Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped? there is a second part of this question that i know how to do...but i don't know how to do this part.. any help will be much appreciated :) Title: Re: Mathematics Question Thread Post by: RuiAce on May 10, 2016, 10:27:45 pm hey can someone please explain how to do this question for me? Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped? there is a second part of this question that i know how to do...but i don't know how to do this part.. any help will be much appreciated :) $P(S)=0.8$ $\text{Since the 8th driver does not wear a seatbelt, we will have to multiply 0.2 to the rest of our working out.}$ $\text{This question can probably be more easily tackled by the Extension 1 mathematics}\\ \text{methods of counting principles in the HSC. Or even binomial probability. Consider 7 drivers.}$ $\text{Because we request that exactly one out of the seven drivers do not wear a seatbelt, we have}\\ {\left(0.8\right)}^6(0.2)\\ \text{But the problem is, either the 1st, 2nd, 3rd... OR... 7th driver can be the unbelted one.}$ $\text{Hence, we must address all SEVEN favourable outcomes.}\\ P(\text{1/7 is unbelted})=7{\left(0.8\right)}^6(0.2)$ $\text{Since we are indeed discussing conditional probabilities, our answer is}\\ P=(0.2)\left(7{\left(0.8\right)}^6(0.2\right)$ Title: Re: Mathematics Question Thread Post by: jakesilove on May 10, 2016, 10:28:02 pm hey can someone please explain how to do this question for me? Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped? there is a second part of this question that i know how to do...but i don't know how to do this part.. any help will be much appreciated :) Hey again! My solution is below. Essentially, you need to break up the question into its constituent parts (first, 1 in 7 drivers, then, 8th driver), find the probability of each part, and multiply them together :) (http://i.imgur.com/PlhUQpW.png?1) Jake Title: Re: Mathematics Question Thread Post by: jakesilove on May 10, 2016, 10:29:10 pm $P(S)=0.8$ $\text{Since the 8th driver does not wear a seatbelt, we will have to multiply 0.2 to the rest of our working out.}$ $\text{This question can probably be more easily tackled by the Extension 1 mathematics}\\ \text{methods of counting principles in the HSC. Or even binomial probability. Consider 7 drivers.}$ $\text{Because we request that exactly one out of the seven drivers do not wear a seatbelt, we have}\\ {\left(0.8\right)}^6(0.2)\\ \text{But the problem is, either the 1st, 2nd, 3rd... OR... 7th driver can be the unbelted one.}$ $\text{Hence, we must address all SEVEN favourable outcomes.}\\ P(\text{1/7 is unbelted})=7{\left(0.8\right)}^6(0.2)$ $\text{Since we are indeed discussing conditional probabilities, our answer is}\\ P=(0.2)\left(7{\left(0.8\right)}^6(0.2\right)$ Looks like Rui beat me too it :) Title: Re: Mathematics Question Thread Post by: RuiAce on May 10, 2016, 10:30:12 pm Looks like Rui beat me too it :) Ninjas. Ah sorry :) But yeah I basically also considered the binomial probability distribution. I just didn't exactly apply it since this question wasn't posted under the Extension 1 section of the forum. Title: Re: Mathematics Question Thread Post by: Maz on May 10, 2016, 10:47:27 pm Ninjas. Ah sorry :) But yeah I basically also considered the binomial probability distribution. I just didn't exactly apply it since this question wasn't posted under the Extension 1 section of the forum. Looks like Rui beat me too it :) thank you both of you :) i am doing binomial distribution but since I'm on the opposite side of the country as you guys i didn't know where to put which question :) sorry ??? but thank you again :) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on May 11, 2016, 12:01:06 am thank you both of you :) i am doing binomial distribution but since I'm on the opposite side of the country as you guys i didn't know where to put which question :) sorry ??? but thank you again :) All good! 2 Unit students in NSW can do this question with their knowledge anyway, just a little more difficult than it would be for yourself or a NSW Extension 1 student ;D Title: Re: Mathematics Question Thread Post by: jamesq on May 13, 2016, 11:19:39 pm A man takes out a250 000 mortgage. Interest is 7.2% per annum, compounded monthly. Repayments are $3 000 monthly. a) Find a formula for the balance on the loan after n months b) How many years does it take to repay the loan Thank you! Title: Re: Mathematics Question Thread Post by: RuiAce on May 14, 2016, 10:05:38 am A man takes out a$250 000 mortgage. Interest is 7.2% per annum, compounded monthly. Repayments are 3 000 monthly. a) Find a formula for the balance on the loan after n months b) How many years does it take to repay the loan Thank you! $\text{To convert the rate to per month: }$ $\frac{7.2\%}{12}=0.6\%$ \text{So we have}\\ \begin{align*}A_1&=250000{\left(1.006\right)}-3000\\A_2&=A_1{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^2-3000{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^2-3000\left(1+1.006\right)\\A_3&=A_2{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^3-3000{\left(1.006\right)}^2-3000{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^3-3000\left(1+1.006+{1.006}^2\right)\\&\vdots\\A_n&=250000{\left(1.006\right)}^n-3000\underbrace{\left(1+1.006+{1.006}^2+\dots+{1.006}^{n-1}\right)}_{\text{G.P. with }n\text{ terms, }a=1, r=1.006}\\&=250000{\left(1.006\right)}^n-3000\left(\frac{{1.006}^n-1}{1.006-1}\right)\end{align*} $\text{We seek to find the required number of months (and thus years) to pay off the loan.}\\ A_n=0$ \begin{align*}\Leftrightarrow 0&=250000{\left(1.006\right)}^n-3000\left(\frac{{1.006}^n-1}{0.006}\right)\\\Leftrightarrow 250000{\left(1.006\right)}^n&=500000\left({1.006}^n-1\right)\\ \Leftrightarrow -250000{\left(1.006\right)}^n&=-500000\\\Leftrightarrow{\left(1.006\right)}^n&=2\\\Leftrightarrow n \ln{1.006}&=\ln{2} \\ \Leftrightarrow n& = \frac{\ln{2}}{\ln{1.006}}\\ \Leftrightarrow n&=115.871\dots\end{align*} $\text{Hence roughly 10 years}$ Title: Re: Mathematics Question Thread Post by: jamonwindeyer on May 14, 2016, 05:37:45 pm $\text{To convert the rate to per month: }$ $\frac{7.2\%}{12}=0.6\%$ \text{So we have}\\ \begin{align*}A_1&=250000{\left(1.006\right)}-3000\\A_2&=A_1{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^2-3000{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^2-3000\left(1+1.006\right)\\A_3&=A_2{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^3-3000{\left(1.006\right)}^2-3000{\left(1.006\right)}-3000\\&=250000{\left(1.006\right)}^3-3000\left(1+1.006+{1.006}^2\right)\\&\vdots\\A_n&=250000{\left(1.006\right)}^n-3000\underbrace{\left(1+1.006+{1.006}^2+\dots+{1.006}^{n-1}\right)}_{\text{G.P. with }n\text{ terms, }a=1, r=1.006}\\&=250000{\left(1.006\right)}^n-3000\left(\frac{{1.006}^n-1}{1.006-1}\right)\end{align*} $\text{We seek to find the required number of months (and thus years) to pay off the loan.}\\ A_n=0$ \begin{align*}\Leftrightarrow 0&=250000{\left(1.006\right)}^n-3000\left(\frac{{1.006}^n-1}{0.006}\right)\\\Leftrightarrow 250000{\left(1.006\right)}^n&=500000\left({1.006}^n-1\right)\\ \Leftrightarrow -250000{\left(1.006\right)}^n&=-500000\\\Leftrightarrow{\left(1.006\right)}^n&=2\\\Leftrightarrow n \ln{1.006}&=\ln{2} \\ \Leftrightarrow n& = \frac{\ln{2}}{\ln{1.006}}\\ \Leftrightarrow n&=115.871\dots\end{align*} $\text{Hence roughly 10 years}$ Optionally, you can add a comment about the fact that a250 000 dollar mortgage is totally unrealistic for first home buyers in Sydney  ;)

Totally kidding, though it might give the marker a laugh.
Post by: ccarolineb on May 23, 2016, 08:32:57 pm
Struggling to understand how to do this!

- a population in a certain city is growing at a rate proportional to the population itself. after 3 years the population increases by 20%. How long will it take for the population to double?
Post by: RuiAce on May 23, 2016, 08:58:10 pm
Struggling to understand how to do this!

- a population in a certain city is growing at a rate proportional to the population itself. after 3 years the population increases by 20%. How long will it take for the population to double?
Inserted an edit to make Jamon's point a bit clearer. Let A=P0
$\frac{dP}{dt}=kP\text{ so }P=P_0e^{kt}$
$\text{When }t=3\text{, }P=1.2P_0$
\begin{align*}\therefore 1.2P_0 &= P_0e^{3t}\\ \Leftrightarrow \ln{1.2}&= 3t \\ \Leftrightarrow t&=\frac{\ln{1.2}}{3} \end{align*}
$\text{We need when }P=2P_0$
\begin{align*}2P_0&=P_0e^{t\frac{\ln{1.2}}{3}}\\ \Leftrightarrow t\frac{\ln{1.2}}{3}&=\ln{2} \end{align*}
Post by: jakesilove on May 23, 2016, 09:20:24 pm
Optionally, you can add a comment about the fact that a 250 000 dollar mortgage is totally unrealistic for first home buyers in Sydney ;) Totally kidding, though it might give the marker a laugh. This caused me much pain Title: Re: Mathematics Question Thread Post by: jamonwindeyer on May 23, 2016, 09:31:33 pm $\frac{dP}{dt}=kP\text{ so }P=Ae^{kt}$ $\text{When }t=3\text{, }P=1.2A$ \begin{align*}\therefore 1.2A &= Ae^{3t}\\ \Leftrightarrow \ln{1.2}&= 3t \\ \Leftrightarrow t&=\frac{\ln{1.2}}{3} \end{align*} $\text{We need when }P=2A$ \begin{align*}2A&=Ae^{t\frac{\ln{1.2}}{3}}\\ \Leftrightarrow t\frac{\ln{1.2}}{3}&=\ln{2} \end{align*} As an elaboration for you Caroline, in case you are still getting used to this style of problem, 'A' represents the initial value of the quantity in question (in this case, the population). You can prove this by substituting t=0 if you so choose, but Rui's solution is what you'd do in the HSC exam, a very standard process ;D Title: Re: Mathematics Question Thread Post by: angela99 on May 24, 2016, 06:43:22 pm Topic: Applications of Calculus to the Physical World Sub-topic: ￼Exponential growth and decay Formulas: ￼Q = Ae^kt, dQ/dt = kQ Can you help with this question please? not having obvious numbers to substitute into the formula is really throwing me off ??? The half-life of radium is 1600 years. (a) Find the percentage of radium that will be decayed after 500 years. (b) Find the number of years that it will take for 75% of the radium to decay. Thank you! :) :) Title: Re: Mathematics Question Thread Post by: jakesilove on May 24, 2016, 07:51:43 pm Topic: Applications of Calculus to the Physical World Sub-topic: ￼Exponential growth and decay Formulas: ￼Q = Ae^kt, dQ/dt = kQ Can you help with this question please? not having obvious numbers to substitute into the formula is really throwing me off ??? The half-life of radium is 1600 years. (a) Find the percentage of radium that will be decayed after 500 years. (b) Find the number of years that it will take for 75% of the radium to decay. Thank you! :) :) Hey! This is a standard method for exponential growth/decay questions that don't give you actual numbers, but rather percentages. You need to just LET Q equal something at time t=0 (eg. 100, like I used, or 1, which is what I usually used). Then, as time progresses, you just use an appropriate percentage of your original value. The explanation in the answer below should explain what I mean; once you've seen it once, you'll be able to do it again in a heartbeat! (http://i.imgur.com/7MoE2dZ.jpg) Jake Title: Re: Mathematics Question Thread Post by: angela99 on May 24, 2016, 08:10:28 pm Hey! This is a standard method for exponential growth/decay questions that don't give you actual numbers, but rather percentages. You need to just LET Q equal something at time t=0 (eg. 100, like I used, or 1, which is what I usually used). Then, as time progresses, you just use an appropriate percentage of your original value. The explanation in the answer below should explain what I mean; once you've seen it once, you'll be able to do it again in a heartbeat! (http://i.imgur.com/7MoE2dZ.jpg) Jake Thank you!! Title: Re: Mathematics Question Thread Post by: angela99 on May 27, 2016, 12:08:51 pm PLEASE HELP!! THANK YOU! Find, correct to 2 decimal places, the area enclosed by the curve y = log2 x, the x-axis and the lines x = 1 and x = 3 by using simpson’s rule with 3 function values. Title: Re: Mathematics Question Thread Post by: angela99 on May 27, 2016, 12:11:43 pm log2 is log base 2 btw Title: Re: Mathematics Question Thread Post by: jakesilove on May 27, 2016, 01:08:38 pm PLEASE HELP!! THANK YOU! Find, correct to 2 decimal places, the area enclosed by the curve y = log2 x, the x-axis and the lines x = 1 and x = 3 by using simpson’s rule with 3 function values. Hey! I've attached my answer below :) It's really important that you are comfortable using Simpson's rule, and I think it is also on your formula sheet in case you forget it. Just break up your area into the number of subdivisions you require, figure out the distance between each division, and then sub the values straight into the formula! For a comparison, the 'real' answer should be 1.87, so this is pretty damn close! (http://i.imgur.com/v5DAlKt.jpg) Jake Title: Re: Mathematics Question Thread Post by: Anika1098 on May 28, 2016, 10:28:16 pm Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand? 1 b ii 8 a iii 10 b 14 c ii 16 b iii 17 iii Iv 18 b i ii 20 22 a iii d ii e i ii iii I understand there are heaps so please just pick one or two that suit you best? Here are thre questions; sorry about my dodgy af collage. Title: Re: Mathematics Question Thread Post by: jakesilove on May 28, 2016, 11:12:11 pm Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand? 1 b ii 8 a iii 10 b 14 c ii 16 b iii 17 iii Iv 18 b i ii 20 22 a iii d ii e i ii iii I understand there are heaps so please just pick one or two that suit you best? Here are thre questions; sorry about my dodgy af collage. Hey! Unfortunately I can't decipher the questions you've sent over, they are way too blurry. If you could either send higher quality pictures, or just type up the important ones, that would be great! I'm happy to help with things that you're struggling with, but maybe do try to limit the questions to like one of each type. That way, hopefully I can help explain the general way to solve those kinds of questions, and then you can figure out the others yourself. Jake Title: Re: Mathematics Question Thread Post by: RuiAce on May 28, 2016, 11:31:05 pm Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand? 1 b ii 8 a iii 10 b 14 c ii 16 b iii 17 iii Iv 18 b i ii 20 22 a iii d ii e i ii iii I understand there are heaps so please just pick one or two that suit you best? Here are thre questions; sorry about my dodgy af collage. Hey! Unfortunately I can't decipher the questions you've sent over, they are way too blurry. If you could either send higher quality pictures, or just type up the important ones, that would be great! I'm happy to help with things that you're struggling with, but maybe do try to limit the questions to like one of each type. That way, hopefully I can help explain the general way to solve those kinds of questions, and then you can figure out the others yourself. Jake Or maybe just a larger collage. This is the solution to the only clear question. $\lim_{\theta \to 0}{\frac{\sin{2\theta}}{3\theta}}=\frac{2}{3}\lim_{\theta \to 0}{\frac{\sin{2\theta}}{2\theta}}=\frac{2}{3}\times 1 = \frac{2}{3}$ Title: Re: Mathematics Question Thread Post by: itswags98 on June 26, 2016, 09:12:59 pm Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand? Must be a pretty dodgy teacher if ur not allowed to ask them questions Title: Re: Mathematics Question Thread Post by: jakesilove on June 26, 2016, 09:26:00 pm Must be a pretty dodgy teacher if ur not allowed to ask them questions You got that right, surely that's just their entire job? Title: Re: Mathematics Question Thread Post by: RuiAce on June 26, 2016, 09:34:07 pm Must be a pretty dodgy teacher if ur not allowed to ask them questions You got that right, surely that's just their entire job? Yep. People like them should be fired. Title: Re: Mathematics Question Thread Post by: itswags98 on June 27, 2016, 08:45:09 pm Yep. People like them should be fired. I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D The difference between teachers can really make or break a subject IMO Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 27, 2016, 09:01:29 pm I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D The difference between teachers can really make or break a subject IMO Absolutely this. And the thing is, it's not just intelligence. I've had lecturers who were absolute wizards, knew their shit like absolutely nothing I've ever seen, but I still didn't like them as a lecturer. Then I have had tutors who know their course content really well and not too much beyond that, but holy crap they just knew how to explain it!! Being able to explain something well is just so important, something which I fear is overlooked in favour of pure academic skill. The best teachers in my experience are the ones who love their job, because it shows ;D Title: Re: Mathematics Question Thread Post by: RuiAce on June 27, 2016, 09:12:55 pm I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D The difference between teachers can really make or break a subject IMO My MX2 teacher did a Ph. D in some field of pure mathematics. His explanations were pretty much otherworldly. Maths was magic. Title: Re: Mathematics Question Thread Post by: conic curve on June 27, 2016, 09:22:20 pm How do I find the domain of y=(3-x)$Squareroot of x$? I want to do it without inspection Also how do I answer people's maths questions if I were to answer them? Title: Re: Mathematics Question Thread Post by: conic curve on June 27, 2016, 09:33:17 pm Let me retype my question above How do I find the domain of y=(3-x)squareroot of x Also how do I type up maths questions on this forum? Title: Re: Mathematics Question Thread Post by: jakesilove on June 27, 2016, 09:57:45 pm Let me retype my question above How do I find the domain of y=(3-x)squareroot of x Also how do I type up maths questions on this forum? Hey! So first, I'll answer your Maths question. To find the domain of $y=(3-x)\sqrt{x}$ We just need to consider which values of x are essentially 'not allowed'. Generally, to do this we just set whatever is in the square root to be greater than zero (noting that the square root of a negative number does not exist!). So, this one turns out the be easy; the (3-x) bit doesn't do anything, but the square root of x means that x must be greater than 0! In maths; $x\ge 0$ Not too difficult! Now, to answer your second question, you started on the right track (using the LaTeX button in the toolbar). However, various functions have various codes attributable to them. Check out Rui's incredible guide HERE to find out how to use it! I'm pretty shit at it, but Rui and Jamon are pros. Jake Title: Re: Mathematics Question Thread Post by: jakesilove on June 27, 2016, 09:59:36 pm :-\ How do I find the domain of y=(3-x)$Squareroot of x$? I want to do it without inspection Also how do I answer people's maths questions if I were to answer them? It's great to here that you want to answer maths questions! When you see something you want to take a crack at, just hit "reply" on their post, and type out a response. You can hand write an answer and attach it using the mona lisa button in the toolbar (you'll need to upload the image to Imgr, as you need a hyperlink, but that doesn't take too long) or use LaTeX to type using maths, as I've explained above! Looking forward to seeing more of you on the forum :) Jake Title: Re: Mathematics Question Thread Post by: conic curve on June 27, 2016, 10:07:35 pm :-\ It's great to here that you want to answer maths questions! When you see something you want to take a crack at, just hit "reply" on their post, and type out a response. You can hand write an answer and attach it using the mona lisa button in the toolbar (you'll need to upload the image to Imgr, as you need a hyperlink, but that doesn't take too long) or use LaTeX to type using maths, as I've explained above! Looking forward to seeing more of you on the forum :) Jake Thanks Jake Title: Re: Mathematics Question Thread Post by: conic curve on June 27, 2016, 10:14:40 pm Hey! So first, I'll answer your Maths question. To find the domain of $y=(3-x)\sqrt{x}$ We just need to consider which values of x are essentially 'not allowed'. Generally, to do this we just set whatever is in the square root to be greater than zero (noting that the square root of a negative number does not exist!). So, this one turns out the be easy; the (3-x) bit doesn't do anything, but the square root of x means that x must be greater than 0! In maths; $x\ge 0$ Not too difficult! Now, to answer your second question, you started on the right track (using the LaTeX button in the toolbar). However, various functions have various codes attributable to them. Check out Rui's incredible guide HERE to find out how to use it! I'm pretty shit at it, but Rui and Jamon are pros. Jake Once again thanks So basically (3-x) has no effect on the domain? If so why? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 27, 2016, 10:15:34 pm Once again thanks So basically (3-x) has no effect on the domain? If so why? Precisely! Because 3-x has no restrictions, we can put anything into it and it makes sense. It's only the square root term that causes issues (think Math error on your calculator) ;D Title: Re: Mathematics Question Thread Post by: conic curve on June 27, 2016, 10:17:31 pm Precisely! Because 3-x has no restrictions, we can put anything into it and it makes sense. It's only the square root term that causes issues (think Math error on your calculator) ;D Thanks Jamon Title: Re: Mathematics Question Thread Post by: itswags98 on June 27, 2016, 10:55:04 pm Absolutely this. And the thing is, it's not just intelligence. I've had lecturers who were absolute wizards, knew their shit like absolutely nothing I've ever seen, but I still didn't like them as a lecturer. Then I have had tutors who know their course content really well and not too much beyond that, but holy crap they just knew how to explain it!! Being able to explain something well is just so important, something which I fear is overlooked in favour of pure academic skill. The best teachers in my experience are the ones who love their job, because it shows ;D Theres that famous saying that goes along the lines of "If you cant explain it to a 6 year old in simple terms, you don't know it well enough" ahah Title: Re: Mathematics Question Thread Post by: RuiAce on June 28, 2016, 12:20:02 am Theres that famous saying that goes along the lines of "If you cant explain it to a 6 year old in simple terms, you don't know it well enough" ahah Try explaining linear algebra to a 6 year old Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 28, 2016, 09:10:51 am Try explaining linear algebra to a 6 year old Coordinates are steps to the left or right, or up and down. If you wanted the parametric version of a line you could explain it in terms of walking to a starting point, then walking so many steps in a given direction. Mappings would be taking longer steps, shorter steps, swapping the direction, etc. Vector spaces would be like the map of the park where you are walking. Etc. Of course none of this would give them any actual useful skills, but kind of fun to think about!! ;) I'm a big believer than you can teach anyone, anything (albeit at a very simple level) provided they have the right dedication. Of course, this relies on 6 year olds having the dedication to try and understand linear algebra... So maybe not ;) anyway, back to the math!! ;D Title: Re: Mathematics Question Thread Post by: RuiAce on June 28, 2016, 09:44:52 am Coordinates are steps to the left or right, or up and down. If you wanted the parametric version of a line you could explain it in terms of walking to a starting point, then walking so many steps in a given direction. Mappings would be taking longer steps, shorter steps, swapping the direction, etc. Vector spaces would be like the map of the park where you are walking. Etc. Of course none of this would give them any actual useful skills, but kind of fun to think about!! ;) I'm a big believer than you can teach anyone, anything (albeit at a very simple level) provided they have the right dedication. Of course, this relies on 6 year olds having the dedication to try and understand linear algebra... So maybe not ;) anyway, back to the math!! ;D You can probably explain a coordinate system but try to explain cross product hahaha Or more fun try explaining Gaussian elimination! Title: Re: Mathematics Question Thread Post by: itswags98 on June 29, 2016, 07:03:48 pm Im not sure if this is the right place to ask this but eh... Everywhere i look at Maths resources, i find study notes of maths... which is not something i want. Ive looked all over the place and i cant seem to find simple formula sheets of the stuff in the course. Unfortunately, this is something i really suck at doing. Ive written study notes for all my subjects but for maths I prefer a simple, clear list of formulas. Any idea where i can find something like this? Title: Re: Mathematics Question Thread Post by: RuiAce on June 29, 2016, 07:06:50 pm Im not sure if this is the right place to ask this but eh... Everywhere i look at Maths resources, i find study notes of maths... which is not something i want. Ive looked all over the place and i cant seem to find simple formula sheets of the stuff in the course. Unfortunately, this is something i really suck at doing. Ive written study notes for all my subjects but for maths I prefer a simple, clear list of formulas. Any idea where i can find something like this? You're literally given a formula sheet by BOSTES now for the exam. Title: Re: Mathematics Question Thread Post by: itswags98 on June 29, 2016, 07:16:37 pm You're literally given a formula sheet by BOSTES now for the exam. sure, but theres way too many things missing from it. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 29, 2016, 07:20:59 pm sure, but theres way too many things missing from it. Totally agree! If there is nothing out there, I'll make it my business to get one of these formula sheets written in the next week or two and uploaded to the site! Great idea ;D Title: Re: Mathematics Question Thread Post by: RuiAce on June 29, 2016, 07:47:07 pm sure, but theres way too many things missing from it. Totally agree! If there is nothing out there, I'll make it my business to get one of these formula sheets written in the next week or two and uploaded to the site! Great idea ;D What's missing that you guys want to see so badly? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 29, 2016, 08:29:00 pm What's missing that you guys want to see so badly? It's not something I'd want to see in an exam scenario (well this is given since I did without) but even with my tutoring students off the top of my head some missing ones: • Area of a Secant • Multiple Applications of Trapezoidal/Simpson's Rule • General Forms of the Trig Expansions I'm sure there are more ;D it's like Physics and Chem where you definitely have enough to get by, but it's nice to have something comprehensive around that covers absolutely everything!! And in a form that is easier to study from (I'm thinking this is a study resource more than anything) ;D Title: Re: Mathematics Question Thread Post by: RuiAce on June 29, 2016, 08:37:37 pm It's not something I'd want to see in an exam scenario (well this is given since I did without) but even with my tutoring students off the top of my head some missing ones: • Area of a Secant • Multiple Applications of Trapezoidal/Simpson's Rule • General Forms of the Trig Expansions I'm sure there are more ;D it's like Physics and Chem where you definitely have enough to get by, but it's nice to have something comprehensive around that covers absolutely everything!! And in a form that is easier to study from (I'm thinking this is a study resource more than anything) ;D Well I guess the HSC basically just gave you all the simplified formulae that you need cause Area of a segment = Area of a sector minus area of a triangle Trap/Simpson's rule can just be done recursively (tbh, the general Simpson's rule depends on if you start counting at x0 or x1) Trig is idk I just know it But yeah fair enough if you want sophistication I can see where you're coming at. But at least they give you the building blocks in the HSC now instead of just an integrals sheet! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 29, 2016, 10:14:10 pm Well I guess the HSC basically just gave you all the simplified formulae that you need cause Area of a segment = Area of a sector minus area of a triangle Trap/Simpson's rule can just be done recursively (tbh, the general Simpson's rule depends on if you start counting at x0 or x1) Trig is idk I just know it But yeah fair enough if you want sophistication I can see where you're coming at. But at least they give you the building blocks in the HSC now instead of just an integrals sheet! Oh absolutely, yes the current formula sheet is more than enough!! I'm interested to see whether the difficulty of the exam is any different in compensation for the sheet, or whether the sorts of questions will change at all :) PS - Different counting spots for Simpson's Rule was so annoying in my HSC year I literally refused to look at any formula sheets with it on it but my own posters :P Title: Re: Mathematics Question Thread Post by: conic curve on June 30, 2016, 07:20:00 pm For trigonometry limits, why do we have to multiply by the fraction? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on June 30, 2016, 07:39:03 pm For trigonometry limits, why do we have to multiply by the fraction? Hey conic, not quite sure what you mean! Could you give an example?? ;D Title: Re: Mathematics Question Thread Post by: anotherworld2b on July 03, 2016, 10:47:06 am Hi at school we've started trigonometry identities proofs I was wondering if i could get help in understanding what i can and can not do to prove trig identities and how i should interpret the proofs I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof Title: Re: Mathematics Question Thread Post by: RuiAce on July 03, 2016, 10:55:39 am Hi at school we've started trigonometry identities proofs I was wondering if i could get help in understanding what i can and can not do to prove trig identities and how i should interpret the proofs I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof Be clearer about the word "illegal". Only certain things are illegal and it always depends on what you're trying to prove and restrictions. $\text{The simple answer to the main question is that you can use any definition e.g. }\sec x = \frac{1}{\cos x}\\ \text{any ratio identity e.g.} \tan x =\frac{\sin x}{\cos x}\\ \text{and any Pythagorean identity e.g. } 1+\tan^2x=\sec^2x$ Title: Re: Mathematics Question Thread Post by: wyzard on July 03, 2016, 11:35:07 am Hi at school we've started trigonometry identities proofs I was wondering if i could get help in understanding what i can and can not do to prove trig identities and how i should interpret the proofs I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof What exactly do you mean by 'illegal'? If it means doing something that is mathematically wrong like dividing by zero on both sides on a equation, or assuming what you are trying to prove, then that's a big no no. However if you're using theorems, results and formulas that are outside of the syllabus and not taught covered in school, that's a big yes as it shows you have the initiative to learn more than you're expected to. Title: Re: Mathematics Question Thread Post by: RuiAce on July 03, 2016, 11:52:48 am What exactly do you mean by 'illegal'? If it means doing something that is mathematically wrong like dividing by zero on both sides on a equation, or assuming what you are trying to prove, then that's a big no no. However if you're using theorems, results and formulas that are outside of the syllabus and not taught covered in school, that's a big yes as it shows you have the initiative to learn more than you're expected to. Not necessarily in the HSC. Concepts such as L'Hopital's rule that used to get abused by students are now strictly forbidden and subject to being penalised. Title: Re: Mathematics Question Thread Post by: anotherworld2b on July 03, 2016, 12:55:14 pm I apologize for not being clear about what I meant by 'illegal' I was referring to my habit of constantly doing something that is mathematically wrong/not possible. How do you know what you can and can not do in regards to the trig identities? Are there particular rules that must be followed when proving trig identities? Title: Re: Mathematics Question Thread Post by: RuiAce on July 03, 2016, 12:58:22 pm I apologize for not being clear about what I meant by 'illegal' I was referring to my habit of constantly doing something that is mathematically wrong/not possible. How do you know what you can and can not do in regards to the trig identities? Are there particular rules that must be followed when proving trig identities? The rule that everything is not a mathematical mistake or misuse in assumption? Show me an example of where you messed up and did something "illegal" for a better idea. Title: Re: Mathematics Question Thread Post by: wyzard on July 03, 2016, 03:54:08 pm Not necessarily in the HSC. Concepts such as L'Hopital's rule that used to get abused by students are now strictly forbidden and subject to being penalised. Damn didn't realize the use L'Hopital rule got caught on, it's such a useful theorem to use though ;D Yeah I agree with you though, be mindful of what concepts you can't use too. I apologize for not being clear about what I meant by 'illegal' I was referring to my habit of constantly doing something that is mathematically wrong/not possible. How do you know what you can and can not do in regards to the trig identities? Are there particular rules that must be followed when proving trig identities? Unfortunately there's no easy way out of this, the process of learning mathematics is by making mistakes, and learning from them. So the more mistake you make, the better. It's best learned by experience, just jump in there, make many silly "illegal" moves, and study why you can't do that. Over time, you'll become more sensitive to the mistake and you'll avoid them. Title: Re: Mathematics Question Thread Post by: RuiAce on July 03, 2016, 06:03:30 pm Damn didn't realize the use L'Hopital rule got caught on, it's such a useful theorem to use though ;D Yeah I agree with you though, be mindful of what concepts you can't use too. Yep One of the more minor reasons as to why I wanted to do uni maths already was just to abuse the crap out of L'H ;D But the HSC is just the way it is. Always try to use what's in the syllabuses. I think one of the most lenient things they do is allow implicit differentiation in Ext 1...when it's really just properly taught in Ext 2. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 03, 2016, 07:59:05 pm Yep One of the more minor reasons as to why I wanted to do uni maths already was just to abuse the crap out of L'H ;D But the HSC is just the way it is. Always try to use what's in the syllabuses. I think one of the most lenient things they do is allow implicit differentiation in Ext 1...when it's really just properly taught in Ext 2. You'll get marks for the correct answer (if they are available based on the criteria) but not the working ;D you're totally right, stick to the method taught in that course, no point playing with fire :P Would like to say that L'Hopital was my favourite thing in first year math, now it's probably Cauchy-Goursat's Theorem for use in complex contour integration (beautiful math) ;) Title: Re: Mathematics Question Thread Post by: conic curve on July 04, 2016, 09:08:06 am Okay so I am having trouble with "quadratic identities" I don't seem to understand things like x^2+x+1=a(x-1)^2+b(x-1)+c How is a=1 c=3 b=3 in this case Same with this case 3x^2-5x-2=a(x-1)^2+b(x+1)+c How is a=3 c=-6 and b=1 If anyone here is good at explaining clearly then could you please help me Thanks ;D Title: Re: Mathematics Question Thread Post by: RuiAce on July 04, 2016, 09:26:27 am Okay so I am having trouble with "quadratic identities" I don't seem to understand things like x^2+x+1=a(x-1)^2+b(x-1)+c How is a=1 c=3 b=3 in this case Same with this case 3x^2-5x-2=a(x-1)^2+b(x+1)+c How is a=3 c=-6 and b=1 If anyone here is good at explaining clearly then could you please help me Thanks ;D $\text{Given you're still in prelim, I suggest the expand method, not the inspection method.}$ $\text{Consider your first question, we are saying that}\\ x^2+x+1 \equiv a(x-1)^2+b(x-1)+c$ $\text{That is, we're saying they are EXACTLY equal to each other. Just like how }\\ x^2-9 \equiv x^2-9\\ \text{So we will expand the RHS.}$ \begin{align*}x^2+x+1&\equiv a(x-1)^2+b(x-1)+c\\ &\equiv a(x^2-2x+1)+bx-b+c\\ &\equiv ax^2-2ax+a+bx-b+c\\ &\equiv ax^2+(-2a+b)x+(a-b+c)\end{align*} $\text{Two quadratics are IDENTICAL to each other}\\ \text{IF AND ONLY IF their COEFFICIENTS are equal. E.g.} \\ x^2+3x+2 \equiv \frac{1}{2}(2x^2+6x+4)\\ \text{Therefore, we start to EQUATE coefficients.}$ $\text{Coefficient of }x^2: 1=a \\ \text{Coefficient of }x: 1=-2a+b \implies 1=-2+b \iff b=3 \\ \text{Constant terms: }1=a-b+c \implies 1=1-3+c \iff c=3$ Title: Re: Mathematics Question Thread Post by: wyzard on July 04, 2016, 11:08:45 am Would like to say that L'Hopital was my favourite thing in first year math, now it's probably Cauchy-Goursat's Theorem for use in complex contour integration (beautiful math) ;) Oh that's such a sweet result :-* I'd say my the Residue Theorem's (again from Complex Analysis) my favorite, and how it can be used to find really nasty integrals and find inverse Laplace Transforms so efficiently. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 04, 2016, 08:41:07 pm Oh that's such a sweet result :-* I'd say my the Residue Theorem's (again from Complex Analysis) my favorite, and how it can be used to find really nasty integrals and find inverse Laplace Transforms so efficiently. Definitely up there, love the Residue Theorem, after I learned it I was like "Why did you even bother teaching us Cauchy's Integral Theorem then?" Title: Re: Mathematics Question Thread Post by: conic curve on July 11, 2016, 01:38:12 pm Need help with the following questions: 6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled? 7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km? 8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a" a. How far does he travel on the final part of his journey? b. Evaluate a 9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection? Title: Re: Mathematics Question Thread Post by: RuiAce on July 11, 2016, 05:12:41 pm Need help with the following questions: 6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled? 7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km? 8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a" a. How far does he travel on the final part of his journey? b. Evaluate a 9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection? The others are lecturing right now and I'm afraid I don't have paper on me at this moment. Have you tried drawing diagrams? If you could upload images of diagram I can tell you if it looks correct, and then just sketch out the method to attempting these questions. Title: Re: Mathematics Question Thread Post by: happy_turtle on July 12, 2016, 05:14:46 pm Need help with the following questions: 6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled? 7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km? 8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a" a. How far does he travel on the final part of his journey? b. Evaluate a 9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection? I have question 9... I'm not sure about the others though Title: Re: Mathematics Question Thread Post by: RuiAce on July 12, 2016, 07:03:58 pm I have question 9... I'm not sure about the others though This looks correct. I had glanced at Q9 and realised that it would throw some people off because it makes you go back to speed=distance/time, but judging by that you catered it correctly (and the cosine rule does not look incorrectly substituted) it appears correct Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 12, 2016, 09:36:01 pm I have question 9... I'm not sure about the others though Thanks for that answer happy_turtle, and welcome to the forums!! Absolute legend ;D conic I'll definitely come and give you an explanation of this style of question tomorrow, just need to finish this lecture series first ;) I'll run through one besides Question 9 ;D Title: Re: Mathematics Question Thread Post by: conic curve on July 12, 2016, 09:38:37 pm Thanks for that answer happy_turtle, and welcome to the forums!! Absolute legend ;D conic I'll definitely come and give you an explanation of this style of question tomorrow, just need to finish this lecture series first ;) I'll run through one besides Question 9 ;D Okay cheers I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning Title: Re: Mathematics Question Thread Post by: HighTide on July 13, 2016, 10:38:28 am Okay cheers I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning Just going to highlight that you can post multiple qs in the same comment and avoid a warning. You can also edit your comment by clicking modify. You don't need to post consecutively to ask multiple qs :) Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 10:41:32 am Just going to highlight that you can post multiple qs in the same comment and avoid a warning. You can also edit your comment by clicking modify. You don't need to post consecutively to ask multiple qs :) Can I just say, on a person note I really do not wish to be bombarded by a continuous streak of questions especially of the same topic. It begins to feel like a chore, instead of a generous act, in helping. Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 11:07:40 am Can I just say, on a person note I really do not wish to be bombarded by a continuous streak of questions especially of the same topic. It begins to feel like a chore, instead of a generous act, in helping. Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well? Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 12:59:55 pm Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well? It's becoming pushy when it's the same trigonometry based questions. If your diagram is wrong, then you don't make any progress altogether. Your answer: Because I leave the website open in its own separate tab. Sometimes I look through it, sometimes it sits there like my Facebook. As for the bit in the brackets, that is most likely assumed by some people because you seem to go a bit excessively on and on over it. Starts to feel like your only reason to be on here - boosting up the post count to have 50 or so essays marked. I'm sure that's not the case, but I reckon you wouldn't be too obsessed over your post count otherwise. Title: Re: Mathematics Question Thread Post by: heids on July 13, 2016, 01:24:09 pm I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well? I see you. This sounds quite like... 'elaborate trolling' to me... We don't want to be too strict but if I see you post about your post count or random useless irrelevancies, I'll just delete rather than discussing it, because it clutters the place irrelevantly so others can't give/receive useful help so easily. Fair? :) This is one last request to be respectful of others. The help RuiAce and others are giving is free, fast and simply amazing and it's incredibly rude for someone to start taking it as a given and keep demanding them to do more and more for that user alone; they aren't your private, paid tutors. So this is the last of us discussing your posting habits in public; please PM me if you think we're treating you unfairly :) Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 02:08:17 pm Whenever you differentatiate, you are finding the gradient of the tangent/finding the rate of change, right? For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this Thanks Title: Re: Mathematics Question Thread Post by: Floatzel98 on July 13, 2016, 02:34:12 pm For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case Nothing changes about differentiating cubic functions compared to any other power functions. I don't know exactly what you graphed, but the two graphs should be the same. Title: Re: Mathematics Question Thread Post by: EEEEEEP on July 13, 2016, 02:42:48 pm Whenever you differentatiate, you are finding the gradient of the tangent/finding the rate of change, right? For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this Thanks You differentiate and then sub in values to find the rate of change from A to B (or whatever).. EDIT By Jamon: Ensure all discussion is civil and respects the right of others to ask questions :) Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 02:45:11 pm Nothing changes about differentiating cubic functions compared to any other power functions. I don't know exactly what you graphed, but the two graphs should be the same. x^3 is a cubic and 3x^2 is a parabola. They are two separate graphs that do not touch each other except for when they touch each other at the origin Title: Re: Mathematics Question Thread Post by: WLalex on July 13, 2016, 02:51:12 pm Whenever you differentatiate, you are finding the gradient of the tangent/finding the rate of change, right? For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this Thanks I don't quite understand your question...they are obviously going to be different when graphed as the derivative and original function are now different, one a cubic and one a parabola? I'd like to be of more help :P Alex Title: Re: Mathematics Question Thread Post by: Floatzel98 on July 13, 2016, 02:54:46 pm x^3 is a cubic and 3x^2 is a parabola. They are two separate graphs that do not touch each other except for when they touch each other at the origin Yes x^3 and 3x^2 are indeed different graphs. When you differentiate x^3 you get 3x^2. If you graph x^3 against 3x^2 you will get different graphs. If you graph d/dx(x^3) against 3x^2 you will get the exact same graph. Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 02:57:14 pm The derivative measures the gradient of the tangent at that point. The derivative is never under any circumstances forced to intersect the original function. Do not, under any circumstances, assume that the graph of the derivative is supposed to be a consequence of the actual y-coordinates of the original function. The derivative is only a measure of the GRADIENT. Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 03:01:10 pm Just in case I am confusing, I hope this helps everyone out Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 03:05:42 pm Just in case I am confusing, I hope this helps everyone out EDIT by Jamon: Ensure all discussion is civil and respects others right to ask questions. Draw a tangent on y=x3. The GRADIENT of the TANGENT, is the y-coordinate on the derivative. Title: Re: Mathematics Question Thread Post by: EEEEEEP on July 13, 2016, 03:06:44 pm Just in case I am confusing, I hope this helps everyone out THe original equation is different from the first differentiation. The first differentiation which is 3x^2, shows the gradients (which also happens to be the rate of change). Both equations should look different. What else don't you get? Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 03:09:12 pm I told you, do NOT relate the y-coordinates of the derivative, to the y-coordinates of the original function. Draw a tangent on y=x3. The GRADIENT of the TANGENT, is the y-coordinate on the derivative. Using this: http://www.teacherschoice.com.au/Maths_Library/Calculus/tangents_and_normals.htm Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 03:10:52 pm Using this: http://www.teacherschoice.com.au/Maths_Library/Calculus/tangents_and_normals.htm According to the website you gave, the derivative is a measure of the GRADIENT of the TANGENT at the point (to the curve). Anything else? Obviously the gradient of the tangent at the point does not have to resemble that of the actual curve. The tangent to y=x2 at (1,1) is 2. What does that have to do with the original curve? Nothing. The derivative is not a tilted graph of the original thing. It ONLY measures the gradient of the tangent. If you want a function whose derivative is itself y=ex is your answer and you do that in Yr 12. Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 03:16:00 pm According to the website you gave, the derivative is a measure of the GRADIENT of the TANGENT at the point (to the curve). Anything else? Obviously the gradient of the tangent at the point does not have to resemble that of the actual curve. The tangent to y=x2 at (1,1) is 2. What does that have to do with the original curve? Nothing. Okay I think I am very confusing ANyways, going back to basics, if you were to differentaite y=x, you would get y=1 right? This is finding the gradient of the tangent (i.e. the rate of change). As you can see, the gradient of the tangent is "visible" on this graph For the example I posted originally, it was y=x^3 and if that was differentiated (i.e. the rate of change was found) it was y=3x^2 and that did not touch the graph at all except for the origin. Is this case shouldn't of 3x^2 of had been the gradient of the tanngent to x^3 If I am very confusing then just forget about this because I feel bad making all of you guys do this to me Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 03:18:09 pm Okay I think I am very confusing ANyways, going back to basics, if you were to differentaite y=x, you would get y=1 right? This is finding the gradient of the tangent (i.e. the rate of change). As you can see, the gradient of the tangent is "visible" on this graph For the example I posted originally, it was y=x^3 and if that was differentiated (i.e. the rate of change was found) it was y=3x^2 and that did not touch the graph at all except for the origin. Is this case shouldn't of 3x^2 of had been the gradient of the tanngent to x^3 If I am very confusing then just forget about this because I feel bad making all of you guys do this to me Ok. Let's clear one thing up. x is not necessarily a number. x is a variable. We have to let x = a certain number, to get the gradient of the tangent, at that number. The plot of the derivative shows the gradient of the tangent, EVERYWHERE on the original curve. It is not immediately obvious that 3x2 is the derivative of x3. The reason why this is the case is that $\text{The formula for the derivative is }\\ f^\prime(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ $\text{If we let }f(x)=x^3\text{, upon a tedious computation the final result is }f^\prime(x)=3x^2$ Title: Re: Mathematics Question Thread Post by: conic curve on July 13, 2016, 03:22:11 pm Ok. Let's clear one thing up. x is not necessarily a number. x is a variable. We have to let x = a certain number, to get the gradient of the tangent, at that number. The plot of the derivative shows the gradient of the tangent, EVERYWHERE on the original curve. It is not immediately obvious that 3x2 is the derivative of x3. The reason why this is the case is that $\text{The formula for the derivative is }\\ f^\prime(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ $\text{If we let }f(x)=x^3\text{, upon a tedious computation the final result is }f^\prime(x)=3x^2$ Oh....Right, I get it now thank you thank you thank you ;D Sorry for making you go through the pain in answering this :'( Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 03:24:22 pm Oh....Right, I get it now thank you thank you thank you ;D Sorry for making you go through the pain in answering this :'( Glad that we eventually got there Title: Re: Mathematics Question Thread Post by: shivali on July 13, 2016, 07:21:06 pm Question: "A radioactive substance decays by 10% after 80 years. By how much will it decay after 500 years?" My working: T0 = 100 T80 = 90 Tn=T0e^(kn) T80 = T0e^(80k) 90 = 100e^(80k) k = (ln(0.9))/80 T500 = 100e^500k T500 = 51.72655 Therefore the substance will decay by 51.7% in 500 years? Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 07:39:40 pm Question: "A radioactive substance decays by 10% after 80 years. By how much will it decay after 500 years?" My working: T0 = 100 T80 = 90 Tn=T0e^(kn) T80 = T0e^(80k) 90 = 100e^(80k) k = (ln(0.9))/80 T500 = 100e^500k T500 = 51.72655 Therefore the substance will decay by 51.7% in 500 years? $\text{Let }t\text{ be the time in years. Assume that the initial mass is }M_0$ $\text{When }t=80, M=0.9M_0\\ \text{as }0.1M_0\text{ has decayed.}$ Permit k to be negative in your working out. \text{We have}\\ \begin{align*}M&=M_0e^{kt}\\ \therefore 0.9M_0&=M_0e^{80k}\\ 0.9&=e^{80k}\\80k&=\ln 0.9 \\ k&=\frac{\ln 0.9}{80}\end{align*} $\text{When }t=500, M=M_0e^{500 \times \frac{\ln 0.9}{80}}=0.5176...M_0$ $\textbf{Therefore you are correct.}\text{ Just be sure to indicate that you have provided a rounded off answer.}$ Title: Re: Mathematics Question Thread Post by: shivali on July 13, 2016, 07:48:02 pm $\text{Let }t\text{ be the time in years. Assume that the initial mass is }M_0$ $\text{When }t=80, M=0.9M_0\\ \text{as }0.1M_0\text{ has decayed.}$ Permit k to be negative in your working out. \text{We have}\\ \begin{align*}M&=M_0e^{kt}\\ \therefore 0.9M_0&=M_0e^{80k}\\ 0.9&=e^{80k}\\80k&=\ln 0.9 \\ k&=\frac{\ln 0.9}{80}\end{align*} $\text{When }t=500, M=M_0e^{500 \times \frac{\ln 0.9}{80}}=0.5176...M_0$ $\textbf{Therefore you are correct.}\text{ Just be sure to indicate that you have provided a rounded off answer.}$ The correct answer was 48.2% ? In the textbook Title: Re: Mathematics Question Thread Post by: RuiAce on July 13, 2016, 08:15:41 pm Edit: No never mind. This is exactly what happens when you try to rush. 51.76...% of the original mass is whatever is left behind. 100% - 51.76%... of the original mass is what has actually decayed. Title: Re: Mathematics Question Thread Post by: jakesilove on July 14, 2016, 09:04:17 am Edit: No never mind. This is exactly what happens when you try to rush. 51.76...% of the original mass is whatever is left behind. 100% - 51.76%... of the original mass is what has actually decayed. Can we all just appreciate that Rui has answered about 100 questions in the past three days. An Atar Notes legend,who deserves all of our thanks. You're an absolute king, and hundreds of HSC students are better off with you around! Jake Title: Re: Mathematics Question Thread Post by: ehatton2016 on July 14, 2016, 12:12:21 pm Hey Jake! I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!! I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :( What do I do? I'm usually a band 6 student in maths, so this is really annoying me!! Thank you!! Title: Re: Mathematics Question Thread Post by: RuiAce on July 14, 2016, 12:14:53 pm Hey Jake! I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!! I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :( What do I do? I'm usually a band 6 student in maths, so this is really annoying me!! Thank you!! When you do past papers, do you struggle on questions? Because doing past papers "ineffectively" is the same thing as ineffective studying habits. Or are you saying that you struggle out of exam pressure. Title: Re: Mathematics Question Thread Post by: jakesilove on July 14, 2016, 12:17:35 pm Hey Jake! I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!! I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :( What do I do? I'm usually a band 6 student in maths, so this is really annoying me!! Thank you!! Hey! Firstly, welcome to the forums! Nice job studying for trials; sounds like you're really putting the hours in. Firstly, two and a bit weeks is ages. You could learn the entire Maths curriculum from scratch in that time. It looks like you've identified what areas you are weak in, because you have tried doing revision for topics that you lose marks in. What kind of sections are you struggling with? At the end of the day, if there is a section you're not very confident with, and you just do a billion questions of that type, you might not improve at all. That's because, in those sections, it might come down to UNDERSTANDING, before repetition. If you struggle to understand a concept, doing it wrong 100 times won't improve your mark. Past papers is really all you can do. Write out questions you get incorrect, and redo them once a day for a few days until you can do them with ease. This will solidify the methodology in your mind, and make sure that none of your time is wasted. Again, if you let us know where you are specifically struggling with, maybe we can help! Keep at it. You have so much time left, and doing past papers DOES help even if it feels like it doesn't. There's no like secret trick or anything, nothing I can tell you right now that will get you 100% in every paper you do. But past papers helps you improve, and identifying areas of weakness will help you improve. If you have any specific questions, post them here, and we can talk them through with you in a way that will hopefully help you understand the fundamental underlying mathematics as well as the answer itself. You'll be fine! Keep at it. Jake Title: Re: Mathematics Question Thread Post by: bing3 on July 14, 2016, 03:22:36 pm How do I do part (b) of this question? Title: Re: Mathematics Question Thread Post by: RuiAce on July 14, 2016, 03:26:41 pm How do I do part (b) of this question? $\text{You cannot get a completely accurate solution. The equation }\cos x - x=0\text{ is not solvable by algebra.}$ $\text{Draw your graph to a soundly good scale. Plot the point of intersection between the graphs }y=\cos 2x, y=\frac{2x}{3}$ $\text{Use your graph to find the }x\text{-coordinates of the point of intersection. These are the solutions.}$ Your answers should be somewhere around the value here. Title: Re: Mathematics Question Thread Post by: ehatton2016 on July 14, 2016, 07:05:20 pm Hey! Firstly, welcome to the forums! Nice job studying for trials; sounds like you're really putting the hours in. Firstly, two and a bit weeks is ages. You could learn the entire Maths curriculum from scratch in that time. It looks like you've identified what areas you are weak in, because you have tried doing revision for topics that you lose marks in. What kind of sections are you struggling with? At the end of the day, if there is a section you're not very confident with, and you just do a billion questions of that type, you might not improve at all. That's because, in those sections, it might come down to UNDERSTANDING, before repetition. If you struggle to understand a concept, doing it wrong 100 times won't improve your mark. Past papers is really all you can do. Write out questions you get incorrect, and redo them once a day for a few days until you can do them with ease. This will solidify the methodology in your mind, and make sure that none of your time is wasted. Again, if you let us know where you are specifically struggling with, maybe we can help! Keep at it. You have so much time left, and doing past papers DOES help even if it feels like it doesn't. There's no like secret trick or anything, nothing I can tell you right now that will get you 100% in every paper you do. But past papers helps you improve, and identifying areas of weakness will help you improve. If you have any specific questions, post them here, and we can talk them through with you in a way that will hopefully help you understand the fundamental underlying mathematics as well as the answer itself. You'll be fine! Keep at it. Jake Thank you! I will definitely keep doing questions, it's my nerdy form of procrastination! By the way, your chem lecture was great! Title: Re: Mathematics Question Thread Post by: jakesilove on July 14, 2016, 07:31:22 pm Thank you! I will definitely keep doing questions, it's my nerdy form of procrastination! By the way, your chem lecture was great! Really appreciate the positive feedback! Hope you learnt a thing or two, and definitely keep up the interaction on the forums! Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 12:57:52 pm Just wondering but how do you graph lines (without plotting the points) for the following (I don't remember learning this at school): x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator) Thanks Title: Re: Mathematics Question Thread Post by: EEEEEEP on July 15, 2016, 01:05:22 pm Just wondering but how do you graph lines (without plotting the points) for the following (I don't remember learning this at school): x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator) Thanks You can do it by letting x or y =0 and then find the intercepts. E.g. If we had an equation say... y = x + 1... Y intercept > >let x = 0 y =1 X intercept >> let y = 0 x = -1 Mark the points (-1, 0), (0, 1) and the join the two dots with a line. *Note, the lines can be joined in that fashion as the equation is linear.* Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 15, 2016, 01:15:56 pm Just wondering but how do you graph lines (without plotting the points) for the following (I don't remember learning this at school): x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator) Thanks You can also use the gradient-intercept form for a line if it happens to be in the correct form, or is easy to get there. If we have an equation that looks like this: $y=mx+b$ Then it is just a line with a y-intercept of b, and a gradient of m (remember gradient is just rise over run) ;D Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 02:05:02 pm You can also use the gradient-intercept form for a line if it happens to be in the correct form, or is easy to get there. If we have an equation that looks like this: $y=mx+b$ Then it is just a line with a y-intercept of b, and a gradient of m (remember gradient is just rise over run) ;D So for the equations, I plotted above you just assume they're all equal to 0 E.g. x-y+3, does this mean x-y+3=0? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 15, 2016, 02:30:00 pm So for the equations, I plotted above you just assume they're all equal to 0 E.g. x-y+3, does this mean x-y+3=0? We wouldn't assume it in all cases, but it is very common for equations to be expressed with everything on the left hand side equal to zero, called the general form. So that's how nerd1 interpreted it above ;D Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 02:33:06 pm We wouldn't assume it in all cases, but it is very common for equations to be expressed with everything on the left hand side equal to zero, called the general form. So that's how nerd1 interpreted it above ;D So if it's an equation like graph x+3 then would it be x+3=0 or y=x+3 because in the original question there's no y value? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 15, 2016, 02:34:28 pm So if it's an equation like graph x+3 then would it be x+3=0 or y=x+3 because in the original question there's no y value? The question itself is flawed, they'd say which, but I'd interpret it as y=x+3 in that case :) Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 03:12:37 pm The question itself is flawed, they'd say which, but I'd interpret it as y=x+3 in that case :) What if it was y+3? Title: Re: Mathematics Question Thread Post by: EEEEEEP on July 15, 2016, 03:23:20 pm Edit: nvm. Title: Re: Mathematics Question Thread Post by: RuiAce on July 15, 2016, 04:05:16 pm @conic Let's just clear one thing up. If in the exam they wanted you to graph, something is always EQUAL to something. Otherwise, as Jamon said, the question is flawed. Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 04:38:55 pm @conic Let's just clear one thing up. If in the exam they wanted you to graph, something is always EQUAL to something. Otherwise, as Jamon said, the question is flawed. Oh okay. I just tend to freak out at these questions because I've seen them in the past and don't know how to graph it Title: Re: Mathematics Question Thread Post by: RuiAce on July 15, 2016, 04:48:23 pm Without the equal sign we don't have an equation. We just have what's called an expression. An expression is arbitrary - it can be anything, so we have no rule for it. An equation forces one variable to be strictly defined, or form a relation between two or more variables. This is why it must be present to form a graph. (Or at least something similar to an equal sign. The < symbol will give us an inequality) Title: Re: Mathematics Question Thread Post by: conic curve on July 15, 2016, 09:24:56 pm Has anyone here learnt a topic called the unit circle where you know the exact values of the trig ratios and where you learnt the fundamental formula (I.e. +k360) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 15, 2016, 09:27:53 pm Has anyone here learnt a topic called the unit circle where you know the exact values of the trig ratios and where you learnt the fundamental formula (I.e. +k360) Yep I think I can help there ;D if/when you have questions let me know!! Title: Re: Mathematics Question Thread Post by: RuiAce on July 15, 2016, 10:07:18 pm Yep I think I can help there ;D if/when you have questions let me know!! Oh ok :P Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 15, 2016, 10:14:18 pm Oh ok :P I can now anyway! I went to a pretty decent 2U lecture on Trigonometry on Monday, lecturer was pretty good, think his name was James, or, Jameen, or something... ;) Title: Re: Mathematics Question Thread Post by: jakesilove on July 15, 2016, 10:27:44 pm I can now anyway! I went to a pretty decent 2U lecture on Trigonometry on Monday, lecturer was pretty good, think his name was James, or, Jameen, or something... ;) It was aright I guess Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 04:05:16 pm Anyone here know how to solve this trigonometry equation: Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 04:15:37 pm Anyone here know how to solve this trigonometry equation: $\text{Be clever about your three Pythagorean identities. (This question actually made me laugh.)}\\ \text{Note that csc just means cosec here.}$ \begin{align*}3\sin^2\theta+3\cos^2\theta+3\tan^2\theta+3\cot^2\theta+3\sec^2\theta+3\csc^2\theta&=29\\ 3(\sin^2\theta+\cos^2\theta)+3\tan^2\theta+3\cot^2\theta+3(1+\tan^2\theta)+3(1+\cot^2\theta)&=29\\ 6\tan^2\theta+6\cot^2\theta&=20\\ 3\tan^4\theta-10\tan^2\theta+3&=0\end{align*} \text{At this point, if it is too hard to see what's going on do yourself a favour and let }u=\tan^2\theta.\text{ Otherwise}\\ \begin{align*}3\tan^4\theta-10\tan^2\theta+3&=0\\ (3\tan^2\theta-1)(\tan\theta-3)&=0\\ \tan^2\theta=\frac{1}{3}\quad \text{or}\quad \tan^2\theta &= 3\\ \tan\theta =\pm\frac{1}{\sqrt{3}}\quad \text{or}\quad \tan\theta&=\pm\sqrt{3}\end{align*} $\text{So we have heaps of solutions.}\\ \theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6},\frac{7\pi}{6},\frac{4\pi}{3},\frac{5\pi}{3},\frac{11\pi}{6}$ Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 04:16:55 pm Anyone here know how to solve this trigonometry equation: Hey! That was very algebra intensive and required the use of trig identities; I'm sure there are heaps of ways to get to an answer, some of which will be quicker than mine. Still, when all I really did was sub in a bunch of identities, brute force doesn't take too long! (http://i.imgur.com/faFHmif.png?1) I also didn't get to the final answer (ie. what x/theta actually is); I'll leave that to you Jake Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 04:17:33 pm $\text{Be clever about your three Pythagorean identities. (This question actually made me laugh.)}\\ \text{Note that csc just means cosec here.}$ \begin{align*}3\sin^2\theta+3\cos^2\theta+3\tan^2\theta+3\cot^2\theta+3\sec^2\theta+3\csc^2\theta&=29\\ 3(\sin^2\theta+\cos^2\theta)+3\tan^2\theta+3\cot^2\theta+3(1+\tan^2\theta)+3(1+\cot^2\theta)&=29\\ 6\tan^2\theta+6\cot^2\theta&=20\\ 3\tan^4\theta-10\tan^2\theta+3&=0\end{align*} \text{At this point, if it is too hard to see what's going on do yourself a favour and let }u=\tan^2\theta.\text{ Otherwise}\\ \begin{align*}3\tan^4\theta-10\tan^2\theta+3&=0\\ (3\tan^2\theta-1)(\tan\theta-3)&=0\\ \tan^2\theta=\frac{1}{3}\quad \text{or}\quad \tan^2\theta &= 3\\ \tan\theta =\pm\frac{1}{\sqrt{3}}\quad \text{or}\quad \tan\theta&=\pm\sqrt{3}\end{align*} $\text{So we have heaps of solutions.}\\ \theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6},\frac{7\pi}{6},\frac{4\pi}{3},\frac{5\pi}{3},\frac{11\pi}{6}$ Dammit Rui Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 04:20:46 pm Damm, it required that much working out. Woah, you guys are awesome. Cheers ;D COuldn't of you have divided by 3 first? Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 04:22:13 pm Damm, it required that much working out. Woah, you guys are awesome. Cheers ;D COuldn't of you have divided by 3 first? That would've made the factorisation process suck Also if you think that's a lot of working out, check out the Ext 2 qn thread :P Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 04:25:02 pm Jake what is that typing program you use to type up the answer to the question? Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 04:27:29 pm Jake what is that typing program you use to type up the answer to the question? It's a nifty little program called Word; I suck at LaTeX (at least when it comes to slightly more complicated stuff, or typing quickly), so I prefer to use Word, screenshot, upload to Imgur and attach it here. Totally impractical, but I'm an idiot when it comes to coding-related languages. Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 04:29:54 pm It's a nifty little program called Word; I suck at LaTeX (at least when it comes to slightly more complicated stuff, or typing quickly), so I prefer to use Word, screenshot, upload to Imgur and attach it here. Totally impractical, but I'm an idiot when it comes to coding-related languages. You mean microsoft word? Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 04:31:58 pm You mean microsoft word? Uh huh. Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 04:43:25 pm How do I upload photos to this forum when the file is too big? I tried to upload an image from my phone but it won't let me :( Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 04:45:24 pm How do I upload photos to this forum when the file is too big? I tried to upload an image from my phone but it won't let me :( You might have to do what I do, and upload the file to Imgur. I'm not super techy though, so I'll leave it to someone else to answer properly Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 05:09:01 pm You might have to do what I do, and upload the file to Imgur. I'm not super techy though, so I'll leave it to someone else to answer properly Do you need a imgur account? Also if anyone can help me with these Q's it would be great: 4. g(x)=2x-1.find x if a. g(x+1)=3 b. [g(x)]^2=3 Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 05:11:58 pm Do you need a imgur account? Also if anyone can help me with these Q's it would be great: 4. g(x)=2x-1.find x if a. g(x+1)=3 b. [g(x)]^2=3 \begin{align*}g(x+1)&=3\\ 2(x+1)-1&=3\\ x&=1 \end{align*} \begin{align*}[g(x)]^2&=3\\ (2x-1)^2&=3\\ 2x-1&=\pm \sqrt{3}\\ x&= \frac{1}{2}\pm \frac{\sqrt{3}}{2}\end{align*} The second one could've been done by quadratic formula Also, no. Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 05:13:34 pm \begin{align*}g(x+1)&=3\\ 2(x-1)&=3\\ x=\frac{5}{2} \end{align*} \begin{align*}[g(x)]^2&=3\\ (2x-1)^2&=3\\ 2x-1&=\pm \sqrt{3}\\ x&= \frac{1}{2}\pm \frac{\sqrt{3}}{2}\end{align*} The second one could've been done by quadratic formula Also, no. Thanks Also sorry for the overburdening of questions: 5. a. g(x)= x +(1/x) Prove that g(x)g(x+(1+x))=g(x^2)+3 b. f(x)= (a^x+a^(-x))/2 Show that f(2x)=2[f(x)]^2-1 c.g(x)=log⁡((1+x)/(1-x) .Show that g(2x/(1+x^2 ))=2g(x) d. f(x)=log⁡((x+1)/x) Show that: I. F(1)+f(2)+f(3)=log4 II. F(1)+f(2)+…+f(n)=log(n+1) Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 05:14:26 pm Thanks Also sorry for the overburdening of questions: 5. a. g(x)= x +(1/x) Prove that g(x)g(x+(1+x))=g(x^2)+3 b. f(x)= (a^x+a^(-x))/2 Show that f(2x)=2[f(x)]^2-1 c.g(x)=log⁡((1+x)/(1-x) .Show that g(2x/(1+x^2 ))=2g(x) d. f(x)=log⁡((x+1)/x) Show that: I. F(1)+f(2)+f(3)=log4 II. F(1)+f(2)+…+f(n)=log(n+1) Before tackling these questions, can I ask about if you've revised how functions actually work? Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 05:16:17 pm \begin{align*}g(x+1)&=3\\ 2(x+1)-1&=3\\ x&=1 \end{align*} \begin{align*}[g(x)]^2&=3\\ (2x-1)^2&=3\\ 2x-1&=\pm \sqrt{3}\\ x&= \frac{1}{2}\pm \frac{\sqrt{3}}{2}\end{align*} The second one could've been done by quadratic formula Also, no. Given that Rui has now shown you very clearly how these kinds of questions work, I would recommend giving the new ones a go yourself. Try to work through them slowly, and if you really need more help then upload the work you've done. Then, we can help you figure out the next step, so you can keep working through them yourself. Giving you a billion answers won't help you learn, if you don't try to do some of the work yourself. Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 05:21:16 pm Before tackling these questions, can I ask about if you've revised how functions actually work? I did it last term for school but these were unanswered questions from a while ago which I found out today. The reason why it was unanswered was because I wasn't sure of how to do it back then Title: Re: Mathematics Question Thread Post by: RuiAce on July 16, 2016, 05:23:42 pm I did it last term for school but these were unanswered questions from a while ago which I found out today. The reason why it was unanswered was because I wasn't sure of how to do it back then So have you tried looking over your resources and figuring how to do them now instead of blindly giving us your unfinished homework? Title: Re: Mathematics Question Thread Post by: conic curve on July 16, 2016, 05:25:36 pm So have you tried looking over your resources and figuring how to do them now instead of blindly giving us your unfinished homework? I did back then. Just curious but do these questions seem easy to you all? Title: Re: Mathematics Question Thread Post by: jakesilove on July 16, 2016, 05:28:03 pm I did back then. Just curious but do these questions seem easy to you all? Obviously to Rui and myself, who did Mathematics a while ago, have continued to do Maths at uni, and studied pretty damn hard for the HSC, year 11 questions will be pretty straight forward. That doesn't mean it should be, or would be, easy for you. Ideally, it should become easy over time. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 16, 2016, 09:04:40 pm Moderator Action: Removed posts to keep things on track. A friendly reminder to all users that posts which are clearly "spam" in nature, and do not contribute to discussion (EG - literally repeating what others have said, or simply replying "ha") will be deleted ;D Title: Mathematics Question Thread Post by: EEEEEEP on July 16, 2016, 10:52:51 pm I did back then. Just curious but do these questions seem easy to you all? That's because we have all studied it before. I recommend you grab your textbook and read over things before you try any questions. Feeding answers does not help. Title: Re: Mathematics Question Thread Post by: dreamdog10 on July 17, 2016, 11:48:20 am Hi! So i was wondering what the difference between a solution and a root is. For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that. Title: Re: Mathematics Question Thread Post by: RuiAce on July 17, 2016, 11:57:29 am Hi! So i was wondering what the difference between a solution and a root is. For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that. $\text{A solution is just a number that satisfies the equation}$ $\text{Roots are important for the study of polynomials as behaviour featuring double roots and other multiple roots needs investigation.}\\ \text{The }x\text{-intercept of a polynomial }f(x)\text{ are indeed the roots of }f(x)=0\\ \text{However that root may be iterated.}$ $\text{Notice how }y=x\text{ just passes through the origin, but }y=x^2\text{ exhibits a turning point there.}\\ y=x^3\text{ even exhibits a horizontal point of inflexion.}$ $\text{Study into polynomials is not required at the 2U level (beyond that of the quadratic).}$ _______________________________ $\text{If you want an example relating to the behaviour of quadratics only, note how }y=(x+1)(x-3)\text{ just passes through }(-1,0), (3,0)\\ \text{But instead something like }y=(x-1)^2\text{ just bounces off }(1,0)\text{ smoothly.}$ Title: Re: Mathematics Question Thread Post by: EEEEEEP on July 17, 2016, 12:20:34 pm Hi! So i was wondering what the difference between a solution and a root is. For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that. Alternative solution/explanation to rui's, but I think it is more friendly to beginners. Look at it this way to find the roots (you find out the x intercepts) or what x values make it equal to 0 For an equation with an discriminant of 0, such as (x +2)^2 ... DELTA = 16 - 4 x 1 x 2 = 0 Lets look at the roots (x+2)(x+2) = 0 That means the first bracket equals 0 or the second bracket equals 0, so either x=-2 or x=-2. Again, you really only have one root here: 1. But in this situation you can call it a "repeating root" or "equal roots". "Real roots" simply mean that you HAVE roots. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 17, 2016, 02:13:06 pm Hi! So i was wondering what the difference between a solution and a root is. For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that. As stated above, a solution is just a value (or values) that satisfies a given condition (EG - an equation). A root signifies a value that puts an expression equal to zero, usually, an x-value where a given function is equal to zero. When you look at a function on the number plane, the roots represent the x-intercepts!! I totally understand the confusion about the discriminant. Let me explain it this way. The discriminant is a particular tool we can use to determine the number of roots of a quadratic function. This stems directly from the quadratic formula (the thing inside the square root sign): $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ If the discriminant is positive, then the quadratic formula has two solutions, and thus the quadratic has two roots (two x-intercepts). If the discriminant is equal to zero, then the plus and minus in the formula means nothing (since we are adding and subtracting zero), it has a single solution. This means that the quadratic has a single root, the parabola would just be resting on the x-axis, touching with its vertex at a single point. Finally, if the discriminant is negative, we can't have the square root of a negative, and thus the formula is invalid (at least, for our level of study). Thus, at this level of Mathematics, we say that there are no solutions to the equation, and the parabola will never touch the x-axis (no roots). This corresponds to positive/negative definite expressions. Just from your question, when it says two real equal roots, that really means just one root. Like, there are two roots, but they are the same root, so we say there is a single root. Saying that there are two is kind of confusing if they are the same! ;D I hope these answers help!! Mine is probably a solid conceptual basis, nerd gave a great practical example, and Rui gave you the more complex terminology and ideas ;D Title: Re: Mathematics Question Thread Post by: dreamdog10 on July 17, 2016, 02:44:19 pm Thanks so much! its good to know the background info and that made heaps of sense :) Title: Re: Mathematics Question Thread Post by: anotherworld2b on July 22, 2016, 01:01:52 am Hi, at school we have started probability. I was wondering would anyone have tips on how to remember whether event A and B are independent or mutually exclusive. I am confused about the probability rules in general. There are so many so how would you know which rule to use? TTT ^ TTT I have tried to answer this question but i didnt understand how to do it Title: Re: Mathematics Question Thread Post by: RuiAce on July 22, 2016, 01:10:12 am Hi, at school we have started probability. I was wondering would anyone have tips on how to remember whether event A and B are independent or mutually exclusive. I am confused about the probability rules in general. There are so many so how would you know which rule to use? TTT ^ TTT I have tried to answer this question but i didnt understand how to do it Generally, you can use your intuition to determine if multiple events are mutually exclusive or not. If two events are mutually exclusive, they cannot occur at the same time (simultaneously). If you want, you can provide some examples and I will attempt to explain which are and which are not. $\text{I'm not sure if you have been taught this, however by the definition of conditional probability}\\ Pr(X|Y)=\frac{Pr(X \cap Y)}{Pr(Y)}$ $\text{So if we have }Pr(X|Y)=Pr(X)\\ \text{Upon substitution and rearranging:} \\Pr(X)=\frac{Pr(X \cap Y)}{Pr(Y)} \iff Pr(X)Pr(Y)=Pr(X \cap Y)$ $\text{Also we have }Pr(Y)=\frac{Pr(X \cap Y)}{Pr(Y)}=\frac{Pr(Y \cap X)}{Pr(Y)} =Pr(Y|X)$ $\text{The proof of the formula I used is simple. It comes from the multiplication principle:}\\ Pr(\text{A and B happens}) = Pr(\text{B has happened}) \times Pr(\text{A happens after B})\\ \therefore Pr(A \cap B) = Pr(B) Pr(A|B)$ _______________________ $\text{The reason why }Pr(X)=Pr(X|Y)\text{ demonstrates statistic independence is that }\\ \text{If the probability of }X\text{ happening is the same regardless of whether or not }Y\text{ happened,}\\ \text{then logically they must be two completely seperate events.}$ _______________________ Before I try the Venn diagram proof, can I have a look at what you have learnt thus far? Because the peculiar nature of the question makes me uncertain as to where I should start and how limited my mathematical communication should be. Note that in 2U Venn Diagram analysis are basically NEVER asked. Title: Re: Mathematics Question Thread Post by: anotherworld2b on July 23, 2016, 01:51:27 am We've gone through the rules of mutually exclusive and jndependent rules. Its the first time we've used these rules so i consider myself as a novice. It takes me a while to understand a topic so i will mosy likely require further explanation and guidance of that is okay. I will try to upload some questions that we have done tomorrow :) Title: Re: Mathematics Question Thread Post by: RuiAce on July 23, 2016, 08:52:03 am Like I said, as far as the 2U course goes use of set theory and Venn Diagrams is practically never seen. Nor is it forbidden though, as far as I'm aware of. $\text{The definitions are}$ $\text{If two events are mutually exclusive, they simply cannot occur at the same time.}\\ Pr(A \cap B)=0\text{ as a result of this.}\\ \\ \text{By the definition of conditioning, this naturally also implies }Pr(A|B)=0$ Examples of mutually exclusive events are the probabilities it is raining and sunny at the same time. $\text{If two events are statistically independent, then the probability one occurs does not affect the probability another occurs.}\\ \text{The main basis of conditional probability (something happens first) does not matter}\\ \text{Hence we have }Pr(A|B)=Pr(A)$ Examples of statistically independent events include turning on the computer but also playing a PSP game simultaneously. ________________________________________ $\textbf{Edit: VERY IMPORTANT QUESTION. Did you know this formula in advance?}\\ Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}$ Title: Re: Mathematics Question Thread Post by: jackcornwell on July 23, 2016, 12:54:03 pm solve 2e^2x -7e^x +3 i let u = e^x then found u but cant remember where to go from here Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on July 23, 2016, 12:56:21 pm solve 2e^2x -7e^x +3 i let u = e^x then found u but cant remember where to go from here Thanks I can assume you meant e^(2x) but more importantly is the expression equal to 0? Without it being equal to something it's just an expression, not an equation. ______________________________________________________________________________ If so \text{The equation becomes a quadratic in }u\\ \begin{align*}2u^2-7u+3&=0\\ (u-3)(2u-1)&=0\\u-3=0 \iff u=3\quad \text{ o}&\text{r }\quad2u-1=0 \iff u=\frac{1}{2}\end{align*} \begin{align*}u&=3\\ e^x&=3\\ x&=\ln 3\end{align*} \begin{align*}u&=\frac{1}{2}\\ e^x&=\frac{1}{2}\\ x&=\ln \frac{1}{2}\\ &=-\ln 2\end{align*} Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 23, 2016, 01:36:35 pm solve 2e^2x -7e^x +3 i let u = e^x then found u but cant remember where to go from here Thanks Welcome to the forums Jack! Glad Rui could help you out! Let me know if you need any help finding things, be sure to stop by our New User's Lounge if you haven't already ;D Title: Re: Mathematics Question Thread Post by: jamie anderson on July 23, 2016, 10:57:10 pm Hey, this may be a simple question but i cant seem to grasp the idea of domain and range, for example a common question would be the multiple choice ones where it says find the value for x that satisfy this equation ( sorry dont have one on me ) i can solve it but i wont know what the domain is e.g 4<= x <= 2, 4< x while x> 2etc Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on July 23, 2016, 11:10:49 pm Hey, this may be a simple question but i cant seem to grasp the idea of domain and range, for example a common question would be the multiple choice ones where it says find the value for x that satisfy this equation ( sorry dont have one on me ) i can solve it but i wont know what the domain is e.g 4<= x <= 2, 4< x while x> 2etc Thanks Find the values for x that satisfy this equation sounds like just solving something such as 5x+3>5, i.e. an equation (or inequality) Domain and range is much more different. Domain is the set of x-values that are allowed for a function you're given. E.g. y=x: you can allow for any x-value whatsoever, so you have all real x E.g. y=x^2: you can allow for any x-value as well, so you also have all real x E.g. y=sqrt(x): you can only square root a positive number, so you have x≥0 E.g. y=1/x: you can't divide by zero, so you have all real x except x≠0 If you just started doing domain and range, draw graphs. Going by intuition at the very start is a dangerous idea. The range on the other hand is the set of y-values allowed FOR the x-values you're given. E.g. y=x: you can have any y-value whatsoever, so you have all real y E.g. y=x2: when you square something it can no longer be negative. So you have y≥0 E.g. y=sqrt(x): taking a square root cannot spit out a negative value. So you have y≥0 E.g. y=1/x: 0=1/x has no solution, so you hvae all real y except y≠0 Once again, draw graphs The concept of restricting the domain is what might get some people. Restricting the domain is when you limit the domain of something even further than what it already is. E.g. y=x for x≥1: You chopped off every bit of the graph where x is less than 1. That no longer exists. If you sketch the graph of y=x, you'll also find a new range of y≥1 E.g. y=x2 for -1≤x≤1: You chopped off every bit of the graph where x is greater than 1, and less than -1. You only have a tiny u-shape left. If you sketch this graph, you'll also find a new range of 0≤y≤1 If you need more specific help, please find some questions. Title: Re: Mathematics Question Thread Post by: dtinaa on July 24, 2016, 02:50:53 pm Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions?? Title: Re: Mathematics Question Thread Post by: Jakeybaby on July 24, 2016, 03:20:30 pm Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions?? I'll just go through what I know from SACE, it may be a little different for you, but I'm sure that the fundamentals are still there. You'll always be given the function for s, displacement, at the stage of the course. This tells you the displacement (how far away) the particle is from the origin. When: s(t) = 0, the particle is at the origin s(t) < 0, the particle is to the left of the origin s(t) > 0, the particle is to the right of the origin Hopefully you know that the derivative of the displacement function results in the velocity function. This is due to the velocity being how quickly the particle is moving / how quickly it's displacement is changing. Hence: s'(t) = v(t) When: v(t) = 0, the particle is at rest/not moving v(t) < 0, the particle is moving to the left v(t) > 0, the particle is moving to the right And the rate of which the velocity changes, is the acceleration, as if I am accelerating in the same direction as I am travelling, my velocity will also increase at said rate. Therefore acceleration is how quickly the velocity of the particle is increasing/decreasing. v'(t) = a(t) a(t) = 0, velocity may be at a maximum or minimum Now, the most fundamental concept in one of these questions is determining : (https://i.gyazo.com/34a6e612aeb9646f23ce8b2e215f037a.png) Another key concept is the be able to determine using sign tests, when the speed is increasing or decreasing. If the signs of v(t) and a(t) are the same, then the speed of particle P is increasing If the signs of v(t) and a(t) are different, then the speed of P is decreasing. This is due to the fact that if I am travelling to the left, but I am being pushed to the right, my speed will decrease as I won't be able to walk to the left for much longer. Title: Re: Mathematics Question Thread Post by: RuiAce on July 24, 2016, 03:43:21 pm $\text{Whilst having }v(t)\text{ instead of just }v\text{ being a feasible notation, for displacement}\\ \text{it is generally tagged as }x(t)$ That aside, here's the add-on. Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions?? $\text{It is easy to determine whether we have a maximum velocity or minimum velocity by testing the nature of the stationary points.}\\ \text{I think this is what the allusion was with 'sign tests'}$ $\text{Also keep in mind that we are talking about the displacement, velocity and acceleration here.}\\ \text{The 'speed' the particle travels at is the magnitude of the velocity.}\\ \text{Keep in mind that speed is therefore independent of direction.}$ $\text{If a particle turns around, its displacement will go the other way but it's }\textit{distance}\text{ keeps increasing.}\\ \text{If you have to calculate the distance, you need to be careful of where the particle turns around.}\\ \text{If the particle is travelling to the left, you may need absolute values.}\\ \text{(An alternate way to find distance travelled is to just find the area under the velocity curve}\\ \text{taking absolute values for negative areas.)}$ $\textit{Often, there will be initial conditions when you are required to integrate.}\\ \textit{Take the time to actually understand what is going on. It helps you find the constant of integration.}$ $\textbf{If you're still stuck, please send example questions.}$ Title: Re: Mathematics Question Thread Post by: Jakeybaby on July 24, 2016, 04:13:56 pm $\text{Whilst having }v(t)\text{ instead of just }v\text{ being a feasible notation, for displacement}\\ \text{it is generally tagged as }x(t)$ Oh alright, just different notation between the states :) Title: Re: Mathematics Question Thread Post by: RuiAce on July 24, 2016, 06:27:07 pm Oh alright, just different notation between the states :) Yep. All good here; I'm not entirely why sure why x, r and s get juggled to represent displacement. (Incidentally the latter two are found in the HSC physics courses.) But yeah, at the 2U level the information you provided formed the bulk of what is necessary. The one other thing to keep in mind (regarding notation) though is how v as a function of t really gets defined. In Extension 1, because they also do velocity as a function of displacement (so v(x)), we might have problems if we just say v(t)=2t so v(x)=2x which, depending on the situation can easily be mistaken. It's probably why Leibniz's notation (or just Newton's notation - the dot) is more favourable. (Of course, in 2U this is hardly problematic in my opinion) Title: Re: Mathematics Question Thread Post by: isaacdelatorre on July 24, 2016, 07:18:55 pm Hey, This question might be super simple, but I've been having some trouble with this question, mainly part ii. Could someone please explain how to do it? Title: Re: Mathematics Question Thread Post by: RuiAce on July 24, 2016, 07:49:18 pm Hey, This question might be super simple, but I've been having some trouble with this question, mainly part ii. Could someone please explain how to do it? $\text{The point }A\text{ lies on the tangent for }x=0\\ y=(-t-b)(t-b)=b^2-t^2\\ \text{The point }B\text{ lies on the tangent for }y=0\\ 0=(2x-t-b)(t-b)\iff 2x-t-b=0 \iff x=\frac{t+b}{2}\\ \text{Keeping in mind that }t\neq b\text{ so we can safely cancel out the }t-b$ \text{Now, the area of }\triangle ABO\text{ using }A=\frac{1}{2}bh\text{ is:}\\ \begin{align*}A&=\frac{1}{2}\times OA \times OB\\ &= \frac{1}{2}\left((b^2-t^2)-0\right)\left(\frac{t+b}{2}-0\right)\\ &= \frac{(b^2-t^2)(b+t)}{4}\\ &=\frac{1}{4}\left(b^3+b^2t-bt^2-t^3\right)\\ \therefore \frac{dA}{dt}&=\frac{1}{4}\left(b^2-2bt-3t^2\right)\\ &= \frac{1}{4}(b-3t)(b+t)\end{align*} $\text{To find where }A\text{ is maximised, set }\frac{dA}{dt}=0\\ \frac{1}{4}(b-3t)(b+t)=0\\ \therefore t=\frac{b}{3}\text{ or }t=-b\\ \text{But since both }t\text{ and }b\text{ are positive and non-zero, reject the latter.}\\ \therefore\text{ the area is possibly maximised } t=\frac{b}{3}$ $\frac{d^2A}{dt^2}=\frac{1}{4}(-2b-6t)< 0 \text{ for all positive }t\\ \text{Hence given the concavity of }A\text{ as a function of }t\text{, the area is indeed maximised when }t=\frac{b}{3}$ $\text{Substitute this in back for }T\text{ to get }\left(\frac{b}{3}, \frac{4b^2}{9}\right)$ For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation y-(t-b)^2 = 2(t-b)(x-t) So y = (t-b)(t-b)+(t-b)(2x-2t) So y=(t-b)(t-b+2x-2t) =(t-b)(2x-t-b) Title: Re: Mathematics Question Thread Post by: isaacdelatorre on July 25, 2016, 06:42:13 pm $\text{The point }A\text{ lies on the tangent for }x=0\\ y=(-t-b)(t-b)=b^2-t^2\\ \text{The point }B\text{ lies on the tangent for }y=0\\ 0=(2x-t-b)(t-b)\iff 2x-t-b=0 \iff x=\frac{t+b}{2}\\ \text{Keeping in mind that }t\neq b\text{ so we can safely cancel out the }t-b$ \text{Now, the area of }\triangle ABO\text{ using }A=\frac{1}{2}bh\text{ is:}\\ \begin{align*}A&=\frac{1}{2}\times OA \times OB\\ &= \frac{1}{2}\left((b^2-t^2)-0\right)\left(\frac{t+b}{2}-0\right)\\ &= \frac{(b^2-t^2)(b+t)}{4}\\ &=\frac{1}{4}\left(b^3+b^2t-bt^2-t^3\right)\\ \therefore \frac{dA}{dt}&=\frac{1}{4}\left(b^2-2bt-3t^2\right)\\ &= \frac{1}{4}(b-3t)(b+t)\end{align*} $\text{To find where }A\text{ is maximised, set }\frac{dA}{dt}=0\\ \frac{1}{4}(b-3t)(b+t)=0\\ \therefore t=\frac{b}{3}\text{ or }t=-b\\ \text{But since both }t\text{ and }b\text{ are positive and non-zero, reject the latter.}\\ \therefore\text{ the area is possibly maximised } t=\frac{b}{3}$ $\frac{d^2A}{dt^2}=\frac{1}{4}(-2b-6t)< 0 \text{ for all positive }t\\ \text{Hence given the concavity of }A\text{ as a function of }t\text{, the area is indeed maximised when }t=\frac{b}{3}$ $\text{Substitute this in back for }T\text{ to get }\left(\frac{b}{3}, \frac{4b^2}{9}\right)$ For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation y-(t-b)^2 = 2(t-b)(x-t) So y = (t-b)(t-b)+(t-b)(2x-2t) So y=(t-b)(t-b+2x-2t) =(t-b)(2x-t-b) Wow, thanks RuiAce. I don't know why I was struggling with it. Thanks again :D Title: Re: Mathematics Question Thread Post by: Deng on July 28, 2016, 11:23:43 pm Hey was looking for help with a couple questions ive been stuck on Part ii for the area under the curve questions Thanks! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 29, 2016, 01:32:15 am Hey was looking for help with a couple questions ive been stuck on Part ii for the area under the curve questions Thanks! Hey Deng!! Interestingly, I'd actually approach that first question not as an integration area question, but with some geometry! Let's have a think. The area we want is actually just the area of the triangle enclosed by the axes and the tangent, then subtract the area of the quarter circle! Let me know if that's unclear what I mean there! ;D So, we need the intercepts of the tangent so we can find the area of that triangle. Use whatever method you like, I'll use substitution: $x=0 \implies -4y=15 \quad \therefore y=\frac{-15}{4} \\ y=0 \implies 3x=15 \quad \therefore x=5$ So the triangle has a width of 5 and a height of 15/4, let's use that to find the area we need: $A = A_\text{triangle}-A_{quarter-circle} \\ =\frac{1}{2}bh-\frac{1}{4}\pi r^2 \\=\left(\frac{1}{2}\times\frac{15}{4}\times5\right)-\frac{9\pi}{4}$ I'll let you take it from there ;D For Question 2, we can massively simplify that expression by recognising that sine squared plus cosine squared of the same value is just 1. So, using that Pythagorean Identity, the expression becomes: $\frac{30}{3\sin^2{\theta}+2\cos^2{\theta}}=\frac{30}{2+\sin^2{\theta}}$ Now we just need to think about this intuitively (easier than rigorous mathematics). We want the smallest value of this expression over the given domain, which means we want the denominator as large as possible. Well, the maximum value of sin^2 is 1, so that means the minimum value of the expression is: $\frac{30}{2+1} = 10$ Part (ii) of that last question is a little tricky! Basically, we can get it by understanding that the integral of the first derivative will get us back to the function itself! I'll leave you to have the full attempt. But let me get you started. So, from your sketch in Part (i), you know that the graph of f'(x) has intercepts at x=-2, x=-1, and x=3. First, let's find the area enclosed between x=3 and x=-1. Use our regular method, and remember, we are integrating a derivative! The result will be our original function: $A_1=\int^3_{-1}f'(x)dx = \left[f(x)\right]^3_{-1}$ We now just need to evaluate this at 3 and -1, which we can do, because we are given these values in the graph: $\therefore A_1 = f(3)-f(-1) = 5+3 = 8u^2$ The same principle would apply to the other part of the area between x=-1 and x=-2, but with an absolute value sign as is normal for these sorts of questions ;D then just add the two areas together to get your answer! That last one was a bit of a doozy, does that make sense Deng? :D Title: Re: Mathematics Question Thread Post by: Deng on July 29, 2016, 01:14:39 pm Yeah it does Thanks! Title: Re: Mathematics Question Thread Post by: conic curve on July 29, 2016, 02:59:52 pm What does it mean by "magnitude" in maths? I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing Thanks :D Title: Re: Mathematics Question Thread Post by: jakesilove on July 29, 2016, 03:04:37 pm What does it mean by "magnitude" in maths? I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing Thanks :D Magnitude is literally the 'size' of something. The magnitude of 2 is 2. It just strips a vector value of it's direction; so, say you had a value which was 2km East. The MAGNITUDE of that value is 2km. The DIRECTION is East. Title: Re: Mathematics Question Thread Post by: anotherworld2b on July 30, 2016, 02:12:10 am Hi I have a few questions to ask if it is okay :) i wasnt sure how to do q1 and q3. I tried to prove q2 but im not sure whether it is right. For q11 I was able to get the answer by using a tree diagram but im am unsure how to use a venn diagram (how to put the info in the venn diagram) and how to use The rules approach? Title: Re: Mathematics Question Thread Post by: RuiAce on July 30, 2016, 08:17:21 am Once again, I need to emphasise the fact that study of set theory is NOT required in the HSC mathematics courses. These diagrams are pretty small and I can't see everything typed on Q11. Don't shrink diagrams to upload them; just upload them to imgur or something and post the links to the images next time please. Also, try not to ask too many questions on the exact same topic; it's burdening $\text{Q1 is mostly just formula work.}\\ \text{By the addition rule of probability we start on}\\ Pr(A \cup B)=Pr(A)+Pr(B)-Pr(A \cap B)\\ \text{Using the definition of complementary probabilities we have }Pr(\overline{A \cap B}) = 1-Pr(A \cap B)\\ \text{The definition of conditional probability states that }Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)}$ $\text{For part d), you start with }Pr(\overline{A}|B)=\frac{Pr(\overline{A}\cap B)}{Pr(B)}\\ \text{However by the law of total probability }Pr(B)=Pr(A \cap B)+Pr(\overline{A} \cap B)\\ \text{so you need to find the probability of not A and B first.}\\ \text{The last one can be done similarly.}$ $\text{For Q2, recall that statistical independence occurs when }Pr(A|B)=Pr(A)\\ Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)}=\frac{Pr(A)+Pr(B)-Pr(A \cup B)}{0.2} = \frac{0.45+0.2+0.56}{0.2}=0.45=Pr(A)$ $\text{The fact that they are all mutually exclusive in Q3 means that probability of any two equals }0\\ \text{So if you add the probabilities they immediately equal 1 as we have no overlap.}\\ 2p+3p+5p=1 \iff p=\frac{1}{10}$ Title: Re: Mathematics Question Thread Post by: dreamdog10 on July 30, 2016, 02:09:24 pm Hi! just wondering how you would go about graphing dv/dt=1−(√2)sin(πt/60) between 0<=t<=120 Title: Re: Mathematics Question Thread Post by: RuiAce on July 30, 2016, 02:21:24 pm Hi! just wondering how you would go about graphing dv/dt=1−(√2)sin(πt/60) between 0<=t<=120 $\text{Without information to find the constant of integration, the graph of acceleration v.s. time is assumed.}\\ \text{The amplitude of the graph is clearly }\sqrt{2}\\ \text{The period of the graph will be }T=\frac{2\pi}{\pi/60}=120\\ \text{So we know the graph completes one full cycle for every }120\text{ seconds}$ $\text{There is no shift in phase, so the graph definitely starts at }1\\ \text{Clearly, the extreme cases will therefore be }\ddot{x}=1+\sqrt{2},\, \ddot{x}=1-\sqrt{2}$ $\text{Hence, the graph will look like this.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps841cmddc.png) $\text{Note that because we have negative sine, the graph was flipped about the line }\ddot{x}=1$ $\text{The dotted lines }\ddot{x}=1+\sqrt{2}\text{ and }\ddot{x}=1-\sqrt{2}\text{ are present for reference only.}$ Title: Re: Mathematics Question Thread Post by: dreamdog10 on July 30, 2016, 02:55:34 pm wow okay fantastic thank you so much :) Title: Re: Mathematics Question Thread Post by: vamshimadas on July 30, 2016, 04:58:46 pm Hi, I'm just having trouble working through methodology of this question. By substitution i can assume the answer, but I want to know how to actually work it out. Find the radius of the circle that has its centre at the origin and a tangent with equation given by 4x - 3y - 5 =0 Title: Re: Mathematics Question Thread Post by: jakesilove on July 30, 2016, 05:03:16 pm Hi, I'm just having trouble working through methodology of this question. By substitution i can assume the answer, but I want to know how to actually work it out. Find the radius of the circle that has its centre at the origin and a tangent with equation given by 4x - 3y - 5 =0 I think this question requires the use of the distance between a line and a point formula! Since the radius will be the line perpendicular to the tangent, to the origin, you can literally plug in the values and a radius should pop out. (http://images.tutorcircle.com/cms/images/108/formula-of-distance-from-a-point-to-line.png) Where A=4, B=-3, C=-5, x=0 and y=0 I think the radius turns out to be 1! Jake Title: Re: Mathematics Question Thread Post by: RuiAce on July 30, 2016, 05:15:52 pm The above method is endorsed in 2U however is not completely clear as to why it works, so a brief, non-rigorous (and probably not as easy to follow) explanation is provided here for the sake of reference. $\text{For convenience, let the circle's centre be at }(0,0)\\ \text{The two semicircles are therefore }y=\pm \sqrt{r^2-x^2}\\ \implies \frac{dy}{dx}=\mp \frac{x}{\sqrt{r^2-x^2}}\\ \text{Take any tangent }ax+by+c=0\\ \text{The gradient of the tangent is }-\frac{a}{b}$ $\text{Because the derivative is just the gradient of the tangent, we have }\frac{a}{b}=\pm \frac{x}{\sqrt{r^2-x^2}}\\ \text{Of course, in an actual question we'd know what the relevant point on the circle is, so }x\text{ and }y\text{ are strictly defined.}\\ \text{We'd also know the values of }a\text{ and }b\\ \text{By substituting the relevant values in, we can ascertain the value for }r\\ \text{In fact, if we substitute the correct value for }r\text{ in, that's how we can ensure the equality holds true.}$ On the other hand, it would be more easier to identify for a 3U student due to the circle geometry theorem "the radius is perpendicular to the tangent drawn to point of contact". The formalised proof is found here. An alternate informal proposal (that is, however, probably much easier to understand,) is that since the tangent only meets the circle once, the distance from the tangent to the centre is forcibly the radius. Basically, the tangent meets the circumference. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on July 30, 2016, 08:45:57 pm I think this question requires the use of the distance between a line and a point formula! Since the radius will be the line perpendicular to the tangent, to the origin, you can literally plug in the values and a radius should pop out. (http://images.tutorcircle.com/cms/images/108/formula-of-distance-from-a-point-to-line.png) Where A=4, B=-3, C=-5, x=0 and y=0 I think the radius turns out to be 1! Jake This is definitely the correct method, since we know the centre of the circle we just need the radius, which can be obtained through this formula, and it does indeed end up as 1! Rui's method above is a cool proof, but far beyond what is required here. At this level, intuition is absolutely fine. The perpendicular distance to a line is the shortest distance to said line. Since it is the shortest distance, a circle with a radius equal to that distance will clearly only touch the line once. So basically, in a 2 Unit Exam, just the formula would suffice, but if you are doing 3 unit or above Rui's proof is good to understand ;D Title: Re: Mathematics Question Thread Post by: conic curve on July 31, 2016, 08:19:42 am I don't know whether or not this is a 2U or 3U mahs question, so I just posted it here if secθ-tanθ= 3/5 show sinθ=8/17 Title: Re: Mathematics Question Thread Post by: RuiAce on July 31, 2016, 09:04:45 am \begin{align*}\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}&=\frac{3}{5}\\ 5(1-\sin \theta)&=3\cos \theta\\ 25(1-2\sin\theta-\sin^2\theta)&=9\cos^2\theta\\ &=9(1-\sin^2\theta)\\ 34\sin^2\theta - 50\sin\theta + 16 &= 0\\ 2(17\sin\theta - 8)(\sin \theta-1)&=0\\ \therefore \sin \theta=\frac{8}{17}\text{ o}&\text{r }\sin \theta=1\end{align*} $\text{However }\cos \theta \neq 0\text{ from the start of the question}\\ \therefore \sin \theta \neq 1\\ \therefore \sin \theta = \frac{8}{17}$ Title: Re: Mathematics Question Thread Post by: conic curve on July 31, 2016, 09:07:35 am \begin{align*}\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}&=\frac{3}{5}\\ 5(1-\sin \theta)&=3\cos \theta\\ 25(1-2\sin\theta-\sin^2\theta)&=9\cos^2\theta\\ &=9(1-\sin^2\theta)\\ 34\sin^2\theta - 50\sin\theta + 16 &= 0\\ 2(17\sin\theta - 8)(\sin \theta-1)&=0\\ \therefore \sin \theta=\frac{8}{17}\text{ o}&\text{r }\sin \theta=1\end{align*} $\text{However }\cos \theta \neq 0\text{ from the start of the question}\\ \therefore \sin \theta \neq 1\\ \therefore \sin \theta = \frac{8}{17}$ Why is cos theta not equal to zero at the beginning of the question? Title: Re: Mathematics Question Thread Post by: RuiAce on July 31, 2016, 09:11:17 am Why is cos theta not equal to zero at the beginning of the question? $\text{If }\cos \theta=0\text{ then }\sec\theta=\frac{1}{\cos\theta}\text{ and }\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ are undefined.}$ Title: Re: Mathematics Question Thread Post by: conic curve on August 01, 2016, 08:31:44 am $\text{If }\cos \theta=0\text{ then }\sec\theta=\frac{1}{\cos\theta}\text{ and }\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ are undefined.}$ Thanks I get it now In this equation: (x-3)(x+5)=<0 I got x=<5 and x>=-3 as the answer but the answer says -3=<x=<5. What am I doing wrong? For this question 4x^2-12x+10>0 how is the answer all real x Thanks guys ;D Title: Re: Mathematics Question Thread Post by: RuiAce on August 01, 2016, 08:39:53 am Thanks I get it now In this equation: (x-3)(x+5)=<0 I got x=<5 and x>=-3 as the answer but the answer says -3=<x=<5. What am I doing wrong? For this question 4x^2-12x+10>0 how is the answer all real x Thanks guys ;D For the first one your solution is actually the same as the answer. But the answer makes it tidier, which is good practice that you should be getting use to. Note how x≤5 can stay as x≤5. But x≥-3 can become -3≤x. Just combine them together to get -3≤x≤5. This answer is tidier because you have an enclosed domain, not one that's going on infinitely and forever. It shows that x is between two things where possible. $\text{For the second one, notice how you can't factorise }4x^2-12x+10\text{ any better than }2(2x^2-6x+5)\\ \text{If you attempt to factorise, you will realise you can't.}\\ \text{If you attempt to bring out the quadratic formula, the thing under the square root will be negative.}$ $\text{The idea is that the quadratic discriminant }\Delta = (-12)^2 - 4\times 4\times 10 = -16 < 0 \\ \text{We know that when the discriminant is negative, }4x^2-12x+10=0\text{ has no real roots.}$ $\text{When we look at the coefficient on }x^2\text{ (which is 4), we realise that it is positive}\\ \text{so the equation defines a positive definite quadratic.}\\ \text{By definition, a positive definite function is ALWAYS greater than }0.$ Title: Re: Mathematics Question Thread Post by: conic curve on August 01, 2016, 01:10:21 pm For the first one your solution is actually the same as the answer. But the answer makes it tidier, which is good practice that you should be getting use to. Note how x≤5 can stay as x≤5. But x≥-3 can become -3≤x. Just combine them together to get -3≤x≤5. This answer is tidier because you have an enclosed domain, not one that's going on infinitely and forever. It shows that x is between two things where possible. $\text{For the second one, notice how you can't factorise }4x^2-12x+10\text{ any better than }2(2x^2-6x+5)\\ \text{If you attempt to factorise, you will realise you can't.}\\ \text{If you attempt to bring out the quadratic formula, the thing under the square root will be negative.}$ $\text{The idea is that the quadratic discriminant }\Delta = (-12)^2 - 4\times 4\times 10 = -16 < 0 \\ \text{We know that when the discriminant is negative, }4x^2-12x+10=0\text{ has no real roots.}$ $\text{When we look at the coefficient on }x^2\text{ (which is 4), we realise that it is positive}\\ \text{so the equation defines a positive definite quadratic.}\\ \text{By definition, a positive definite function is ALWAYS greater than }0.$ Thanks How do you do the question I attached below: I know it involves substituting in the function but I find it confusing and very hard to understand Title: Re: Mathematics Question Thread Post by: RuiAce on August 01, 2016, 01:21:15 pm Thanks How do you do the question I attached below: I know it involves substituting in the function but I find it confusing and very hard to understand $\text{That's exactly it. Substitute it in.}\\f(x)=x^2+2x\\ \text{so }f(x+h)=(x+h)^2+2(x+h)$ \begin{align*}f^\prime(x) &= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to 0}\frac{\left[(x+h)^2+2(x+h)\right]-(x^2+2x)}{h}\\ &= \lim_{h\to 0}\frac{x^2+2xh+h^2+2x+2h-x^2-2x}{h}\\ &= \lim_{h\to 0}\frac{2xh+2h+h^2}{h}\\ &= \lim_{h\to 0} 2x+2+h\\ &= 2x+2+0\\ &=2x+2\end{align*} Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 01, 2016, 02:06:43 pm Thanks How do you do the question I attached below: I know it involves substituting in the function but I find it confusing and very hard to understand And remember that you can always check yourself/give yourself some idea of where to go by just differentiating the function normally ;D Title: Re: Mathematics Question Thread Post by: conic curve on August 01, 2016, 03:53:53 pm Could you please elaborate further for what I have attached below because I don't seem to understand that Thanks ;D Title: Re: Mathematics Question Thread Post by: RuiAce on August 01, 2016, 04:02:39 pm Could you please elaborate further for what I have attached below because I don't seem to understand that Thanks ;D Applied the definition of a function? Where there's x I replace it with x+h? If f(x)=x^2 Then f(a)=a^2 f(2)=2^2 = 4 f(x+h)=(x+h)^2 = x^2+2xh+h^2 And then I just expanded the whole thing out. Title: Re: Mathematics Question Thread Post by: isaacdelatorre on August 01, 2016, 07:21:20 pm Hey, I'm just having a bit of trouble with this question. Could someone please give me an explanation of how to do this. Thank you!!! Let g(t) be another function defined for all t ≥ 0. The gradient function is given by g′(t) = 33/10 te−kt . It is given that g′(t) attains the greatest value at t = 7.5 and g(0) = 0. g′(t) where k is a positive constant. The diagram below shows a sketch of g′(t) against t. (i) Show that k = 2/15 (ii) Use the trapezoidal rule with four sub-intervals to estimate the shaded area in the diagram above. (iii) Explain why your answer to (ii) is an estimate for g(12). Title: Re: Mathematics Question Thread Post by: RuiAce on August 01, 2016, 09:37:00 pm Hey, I'm just having a bit of trouble with this question. Could someone please give me an explanation of how to do this. Thank you!!! Let g(t) be another function defined for all t ≥ 0. The gradient function is given by g′(t) = 33/10 te−kt . It is given that g′(t) attains the greatest value at t = 7.5 and g(0) = 0. g′(t) where k is a positive constant. The diagram below shows a sketch of g′(t) against t. (i) Show that k = 2/15 (ii) Use the trapezoidal rule with four sub-intervals to estimate the shaded area in the diagram above. (iii) Explain why your answer to (ii) is an estimate for g(12). $\text{From }g^\prime(t)=\frac{33}{10}te^{-kt}\text{, we use the product rule to deduce that }\\ g^{\prime\prime}(t)=\frac{33}{10}\left(e^{-kt}-kte^{-kt}\right)=\frac{33}{10}e^{-kt}(1-kt)$ $\text{We are saying that }g^\prime(t)\text{ is maximised when }t=7.5\\ \text{This means that it's derivative }g^{\prime\prime}(t)=0\text{ when }t=7.5$ \begin{align*}g^{\prime\prime}(7.5)=\frac{33}{10}(1-7.5k)e^{-7.5k}&=0\\ 1-7.5k&=0\\ k&=\frac{2}{15}\end{align*}\\ \text{Cancelling out the exponential was permitted as }e^{-7.5k}\text{ never equals to }0 $\text{So we now have }g^\prime(t)=\frac{33}{10}te^{-\frac{2t}{15}}$ \text{By the trapezoidal rule (taking four sub-intervals):}\\ h=\frac{12-0}{4}=3 \\ \begin{align*}\therefore \int_0^{12} g^\prime(t)\, dt &\approx \frac{3}{2}\left[g^\prime(0) + g^\prime(12) + 2\left(g^\prime(3)+g^\prime(6)+g^\prime(9)\right) \right]\\ &= \frac{3}{2}\left[0+\frac{198}{5}e^{-\frac{8}{5}}+2\left(\frac{99}{10}e^{-\frac{2}{5}}+\frac{99}{5}e^{-\frac{4}{5}}+\frac{297}{10}e^{-\frac{6}{5}}\right) \right] \end{align*}\\ \text{which you can tidy up or put into your calculator.} \text{Now:}\\ \begin{align*}\int_0^{12}g^\prime(t)\, dt &= \left[ g(t)\right]_0^{12}\\ &= g(12)-g(0)\\ &=g(12)-0\tag{from question} \\ &= g(12) \end{align*}\\ \text{So clearly, the trapezoidal rule approximation is trying to approximate }g(12) Title: Re: Mathematics Question Thread Post by: liiz on August 02, 2016, 05:22:08 pm Hey there, wondering whether someone could please help me out with part b) of this question that I've attached. Not quite sure how to approach it! Thanks so much :)) Title: Re: Mathematics Question Thread Post by: RuiAce on August 02, 2016, 06:06:05 pm Hey there, wondering whether someone could please help me out with part b) of this question that I've attached. Not quite sure how to approach it! Thanks so much :)) $\text{This is your ordinary trigonometric curve that you are trying to sketch.}\\ \text{The amplitude is obviously }50\\ \text{Because the graph is translated up by }400\text{, we know that its equilibrium is at }400\text{ and the extremes at }350, 450$ $\text{The period of the function is }\\ T=\frac{2\pi}{\pi/6}=12\\ \text{So the curve repeats once every }12\text{ seconds thus we are just drawing one oscillation.}$ $\text{So you just have your ordinary cosine curve as shown.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsqkvdjlmm.png) Title: Re: Mathematics Question Thread Post by: wesadora on August 03, 2016, 11:27:42 pm 2U Question. Halp. Title: Re: Mathematics Question Thread Post by: RuiAce on August 04, 2016, 07:11:20 am 2U Question. Halp. This does not look like a 2U question given that I do not believe spirals are so easy to analyse. However this is the best I could come up with. An assumption is made that the inclination of the string is constant. The basis of this is that the midpoint is collinear to the endpoints of the string, and the line between them is perpendicular to the base. $\text{Half of half of the height is }\frac{1}{4}h\\ \text{Half of the circumference of the base is }\pi r$ $\text{Draw a right-angled triangle reflecting the solid up to one quarter of its height.}\\ \text{The base has length }\pi r\text{ and the height }\frac{1}{4}h\\ \text{The hypotenuse is }\frac{1}{4}\text{ the length of the string.}$ \text{Using Pythagoras' Theorem}\\ \begin{align*}\frac{l^2}{16}&=\frac{h^2}{16}+\pi^2 r^2\\ l&=\sqrt{h^2+16\pi^2 r^2}\end{align*} Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 04, 2016, 09:03:45 am 2U Question. Halp. Just to extend on Rui's answer (I get the same outcome), you could consider it in a similar way. This is definitely 2U content, but a really weird one, they like to do strange stuff with Pythag for some reason :P it could show up in an Extension exam I suppose, but I see it more as a Band 6 2U question :) The string is wrapped around a cylinder right, so what we can almost do (bear with me) is assign a cylindrical coordinate system. This sounds complicated, but it's actually easy if we think about it practically. Picture grabbing the end of the string at the top and unravelling it from the cylinder by pulling directly outward, not changing it's height, and leaving the other end fixed on the ground. This forms a right angled triangle where the string is the hypotenuse (and hence comes back in Rui's method, but I'm considering the whole cylinder at once). The height of this triangle is still the same as the height of the cylinder (remember I said don't change the height), so that is h. The base is whatever distance the string travelled around the circumference. What I've almost done is initially considered the x (horizontal) axis as wrapped around the cylinder, and then when I unravel it, I've moved back to our regular Cartesian system. But the distance is the same, that string has wrapped itself around that cylinder twice, so the 'horizontal' distance travelled is double the circumference. I've taken the horizontal distance as measured around the cylinder (double the circumference), and stretched it out into a straight line. Then, as Rui said, we use Pythagoras: $l=\sqrt{h^2+(2\times\text{Circumference})^2}=\sqrt{h^2+(4\pi r)^2} = \sqrt{h^2+16\pi^2r^2}$ To simplify all that, I copied Rui's method but considered the whole cylinder at once, just to make it more practical. You COULD consider it in terms of changing coordinate systems, but that's unnecessary if you can picture unravelling the string. I threw it in to make myself sound smart ;) Title: Re: Mathematics Question Thread Post by: conic curve on August 04, 2016, 06:51:37 pm How do I differentiate this? Title: Re: Mathematics Question Thread Post by: RuiAce on August 04, 2016, 06:55:25 pm How do I differentiate this? $\text{That's deceptive lol. Once you see it once you should be able to figure it out everything similar.}\\ \text{The basic idea is to split the fraction up.}$ \begin{align*}\frac{d}{dx}\left(\frac{10x^2+4x+10}{2\sqrt{x}}\right)&= \frac{d}{dx}\left(5x^{\frac{3}{2}}+2x^{\frac{1}{2}}+5x^{-\frac{1}{2}}\right)\\ &= \frac{15}{2}x^{\frac{1}{2}}+x^\frac{1}{2}-\frac{5}{2}x^{-\frac{3}{2}}\end{align*} Title: Re: Mathematics Question Thread Post by: jakesilove on August 04, 2016, 06:56:15 pm Alternatively, you can just use a straight application of the Quotient rule! Title: Re: Mathematics Question Thread Post by: leila_ameli on August 05, 2016, 08:38:56 pm Heyyy :) :) I need some help with how i should approach the following; The parabola P has the equation y^2=8(x+2). i) write down the coordinates of the vertex of P. ii) find the coordinates of the focus of P. iii) write down the equation of the directrix of P. iv) sketch the parabola P showing the features found above, as well as any x or y intercepts. cheers Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 05, 2016, 08:50:03 pm Heyyy :) :) I need some help with how i should approach the following; The parabola P has the equation y^2=8(x+2). i) write down the coordinates of the vertex of P. ii) find the coordinates of the focus of P. iii) write down the equation of the directrix of P. iv) sketch the parabola P showing the features found above, as well as any x or y intercepts. cheers Hey there!! No worries I'll give you a hand! ;D Before we do, we notice that the equation is in the typical form of a 'sideways' parabola: $(y-k)^2=4a(x-h)$ If we consider it this way, we can immediately read off the coordinates of the vertex (h,k) as (-2, 0): $\text{Vertex}=(-2,0)$ We can also determine the focal length: $4a=8\\a=2$ Now, remember that the focus is 'a' units away from the vertex. This is a right facing parabola (try some random points to check), so it is 2 units to the right of the vertex: $\text{Focus}=(0,0)\quad\text{(Origin)}$ The directrix is 2 units the other way (remember, in this case, the directrix is a vertical line since the parabola is sideways): $\text{Directrix}: x=-4$ All of this should help you sketch ;D does that help? Title: Re: Mathematics Question Thread Post by: leila_ameli on August 05, 2016, 10:30:59 pm Very much so, yes!!!! Thank you so much :D Hey there!! No worries I'll give you a hand! ;D Before we do, we notice that the equation is in the typical form of a 'sideways' parabola: $(y-k)^2=4a(x-h)$ If we consider it this way, we can immediately read off the coordinates of the vertex (h,k) as (-2, 0): $\text{Vertex}=(-2,0)$ We can also determine the focal length: $4a=8\\a=2$ Now, remember that the focus is 'a' units away from the vertex. This is a right facing parabola (try some random points to check), so it is 2 units to the right of the vertex: $\text{Focus}=(0,0)\quad\text{(Origin)}$ The directrix is 2 units the other way (remember, in this case, the directrix is a vertical line since the parabola is sideways): $\text{Directrix}: x=-4$ All of this should help you sketch ;D does that help? Title: Re: Mathematics Question Thread Post by: conic curve on August 06, 2016, 02:30:13 am If they a function is monotonic increasing does that mean f(x)>0 or f(x)>=0? If so why, I can't seem to distinguish between the two Title: Re: Mathematics Question Thread Post by: RuiAce on August 06, 2016, 08:48:19 am If they a function is monotonic increasing does that mean f(x)>0 or f(x)>=0? If so why, I can't seem to distinguish between the two f'(x)≥0 Don't forget the dash A stationary point doesn't necessarily impede it being "monotonic" increasing. f(x)>0 still satisfies monotonic increasing but more specifically it means strictly increasing. Title: Re: Mathematics Question Thread Post by: conic curve on August 06, 2016, 05:03:59 pm f'(x)≥0 Don't forget the dash A stationary point doesn't necessarily impede it being "monotonic" increasing. f(x)>0 still satisfies monotonic increasing but more specifically it means strictly increasing. I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples) Thanks Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 06, 2016, 05:11:51 pm I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples) Thanks $\text{Sure thing! }f'(x)>0\text{ means that the gradient of your curve is ALWAYS positive.}\\\text{That is, the curve is ALWAYS increasing, it has no stationary points.}\\\text{For }f'(x)\ge0\text{, there could be a stationary point, but besides these,}\\\text{ the curve is always increasing. It never turns around (turning point)}\\\text{ and decreases, it is always going up. It just might pause along the way.}$ Title: Re: Mathematics Question Thread Post by: conic curve on August 06, 2016, 05:13:15 pm $\text{Sure thing! }f'(x)\ge0\text{ means that the gradient of your curve is ALWAYS positive. That is, the curve is ALWAYS increasing, it has no stationary points. For }f'(x)>0\text{, there could be a stationary point, but besides these, the curve is always increasing. It never turns around (turning point) and decreases, it is always going up. It just might pause along the way.}$ Finally this cleared things up for me ;D Thanks When would you actually use it (with an example question) Title: Re: Mathematics Question Thread Post by: RuiAce on August 06, 2016, 05:14:28 pm I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples) Thanks $\textbf{By DEFINITION}\text{, the derivative }f^\prime(x)\text{ measures the gradient of the tangent at a point.}\\ \text{The slope of the tangent determines whether the curve is increasing or decreasing.}$ $\text{Without using the rigorous definition, if a function is }\textit{monotone}\text{ then it behaves in such a way}\\ \text{that as }x\text{ increases, }f(x)\text{ only travels in one direction.}\\ f(x)\text{ can either stay still /get more positive, OR stay still or get more negative.}$ $\text{Because }f^\prime(x)\text{ measures the gradient of the tangent, if }f^\prime(x)=0\text{ we have a scenario where the tangent is horizontal.}\\ \text{This is what we call a 'stationary point' - where }f(x)\text{ does not get larger or smaller.}$ $\text{Due to the definition of monotonicity, if a function has a horizontal tangent at a point, or a whole set of points, it is still monotone.}\\ \text{An example of a function that has a horizontal tangent EVERYWHERE is }f(x)=2\\ \text{This is an example of a function that is both monotonic increasing, and monotonic decreasing.}$ $\text{A piecewise function that is monotonic increasing could be}\\ f(x)=\begin{cases}-(x+1)^2 & \text{for }x<1\\ 0&\text{for }-1 \le x < 1\\ (x-1)^2&\text{for }x\ge 1\end{cases}\\ \text{This function is increasing on both domains }x<1\text{ and }x>1\\ \text{however is stationary for all }-1 \le x \le 1$ Further reading: Wikipedia $\textit{In the event a function is STRICTLY increasing, we take a special case of monotonic functions.}\\ \text{A function that is strictly increasing can NOT satisfy }f^\prime(x)=0\text{, that is, we force }f^\prime(x)\text{ to be strictly greater than }0.\\ \text{In other words, a stationary point FAILS to satisfy the criteria of 'strict'.} \\ \text{A strict inequality means that we take out the equal sign, i.e. we replace }\le, \ge\text{ with }<, >$ _________________________________ Edited on: $\text{In addition, another function that is monotonic increasing is }f(x)=x^3\\ \text{Another function that is monotonic increasing is }f(x)=x$ $\text{The latter is ALSO an example of a function that is STRICTLY increasing, because }f^\prime(x)=1>0\text{ ALWAYS}\\ \text{The former FAILS to be strictly increasing, because }f^\prime(0)=0$ Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 06, 2016, 05:16:35 pm Finally this cleared things up for me ;D Thanks When would you actually use it (with an example question) Check that post again, I made a slight error, fixed now :) (and Rui goes into more detail if it helps). No problem, to be honest it is a tiny thing, it is unlikely to be pressed hard. If you are asked whether a function is monotonic increasing (they probably wouldn't use this language in 2U tbh), then just remember the dash, that's all ;D Title: Re: Mathematics Question Thread Post by: olivercutbill on August 07, 2016, 12:50:41 pm Can someone explain 'b)' and 'c)'? Thank you! Title: Re: Mathematics Question Thread Post by: zsteve on August 07, 2016, 01:38:07 pm Can someone explain 'b)' and 'c)'? Thank you! For (b), you'll have to differentiate for the velocity as a function of time. Then solve for zeros. For (c), you'll have to use a definite integral. Assumingt$can only be positive,$xis defined over the interval $$[0, 7]$$, so 'altogether' means from t=0 to t=7. These are standard procedures in kinematics :) Title: Re: Mathematics Question Thread Post by: RuiAce on August 07, 2016, 01:41:27 pm Can someone explain 'b)' and 'c)'? Thank you! \text{We may utilise a trick to differentiate, however I will not use it here.}\\ \text{By the product rule (and chain rule):}\\ \begin{align*}x&= t\sqrt{49-t^2}\\ v&= \sqrt{49-t^2} - t\left(\frac{-t}{\sqrt{49-t^2}}\right)\\ &= \sqrt{49-t^2}-\frac{t^2}{\sqrt{49-t^2}}\end{align*} \text{We know that the particle is at rest when }v=0\\ \sqrt{49-t^2}-\frac{t^2}{\sqrt{49-t^2}}=0 \\ \text{So just multiplying left and right by }\sqrt{49-t^2} \\ \begin{align*}\sqrt{49-t^2}-\frac{t^2}{\sqrt{49-t^2}}&=0 \\ (49-t^2)-t^2&=0\\ 2t^2&=49\\ t&=\frac{7}{\sqrt{2}}\end{align*}\\ \text{Keeping in mind that time is positive.} $\text{Now, to find the distance}\textbf{ we need to watch out for wherever the particle turns.}\\ \text{Observe how when }t<\frac{7}{\sqrt{2}}\, v > 0\\ \text{But instead when }t > \frac{7}{7\sqrt{2}}, \, v < 0\\ \text{Thus, indeed when }t=\frac{7}{\sqrt{2}}\text{ the particle, being at rest, }\textbf{turns around.}\\ \text{Therefore, we consider the distances seperately for }0 < t < \frac{7}{\sqrt{2}}\text{ and }\frac{7}{\sqrt{2}} < t < 7$ $\text{When }t=0, x=0\\ \text{When }t=\frac{7}{\sqrt{2}}, x= \frac{49}{2} \textbf{ check this.}\\ \text{When }t=7, x=0$ $\text{So in the first bit the particle travels a distance of }\frac{7}{\sqrt{2}}\\ \text{And in the second bit the particle travels a distance of }\frac{7}{\sqrt{2}}\text{ as well.}$ $\text{Hence, the total distance travelled is }\frac{2\times 7}{\sqrt{2}}=7\sqrt{2}$ For (b), you'll have to differentiate for the velocity as a function of time. Then solve for zeros. For (c), you'll have to use a definite integral. Assumingt$can only be positive,$xis defined over the interval $$[0, 7]$$, so 'altogether' means from t=0 to t=7. These are standard procedures in kinematics :) The statement is true but I disagree with its suitability. Integrating that velocity function is way outside 2U level. Actually it's fine, forgot that we had the antiderivative given, sorry :) But we don't need to. We can just rely on x(t). Title: Re: Mathematics Question Thread Post by: jakesilove on August 07, 2016, 01:45:27 pm Can someone explain 'b)' and 'c)'? Thank you! Dammit guys, I was just posting a solution. But yeah got to the same thing; I better pick up my game Title: Re: Mathematics Question Thread Post by: olivercutbill on August 07, 2016, 07:32:26 pm I won't quote everything but cheers guys! Problem was I forgot to use the product rule in part b). Classic error. Title: Re: Mathematics Question Thread Post by: conic curve on August 08, 2016, 10:51:04 am Need help with this Q Title: Re: Mathematics Question Thread Post by: jakesilove on August 08, 2016, 11:02:02 am Need help with this Q Let's start by simplifying some stuff $((p-q)(p+q))^2+pq=p^3+q^3$ By difference of two squares. But, p+q=1, so $(p-q)^2+pq=p^3+q^3$ $p^2-2pq+q^2+pq=p^3+q^3$ $p^2-pq+q^2=p^3+q^3$ $(p-q)^2=p^3+q^3$ Now, let's use sum of two cubes $(p-q)^2=(p+q)(p^2-pq+q^2)$ As p+q=1 $(p-q)^2=(p^2-pq+q^2)$ $(p-q)^2=(p-q)^2$ As required Title: Re: Mathematics Question Thread Post by: conic curve on August 08, 2016, 01:04:27 pm Let's start by simplifying some stuff $((p-q)(p+q))^2+pq=p^3+q^3$ By difference of two squares. But, p+q=1, so $(p-q)^2+pq=p^3+q^3$ $p^2-2pq+q^2+pq=p^3+q^3$ $p^2-pq+q^2=p^3+q^3$ $(p-q)^2=p^3+q^3$ Now, let's use sum of two cubes $(p-q)^2=(p+q)(p^2-pq+q^2)$ As p+q=1 $(p-q)^2=(p^2-pq+q^2)$ $(p-q)^2=(p-q)^2$ As required Thanks Jake How would I do question iii and iv Title: Re: Mathematics Question Thread Post by: RuiAce on August 08, 2016, 01:08:11 pm Thanks Jake How would I do question iii and iv $\text{For iii), note that the gradient of the perpendicular to the line }AB\text{ can be found using}\\ m_1m_2=-1\\ \text{Hence, the required gradient is }m_2=-\frac{1}{m_{AB}}$ $\text{Then since you already know the midpoint, just use your point-grad formula }y-y_1=m_2(x-x_1)$ $\textbf{The trick for iv) is that if P is equidistant from A and B}\\ \textbf{P MUST lie on the perpendicular bisector of AB.}\\ \text{Hence, for the equation of the line you found in iii), solve simultaneous equations with }y=2x-9$ $\text{If you can't visualise it, draw a diagram.}$ Title: Re: Mathematics Question Thread Post by: conic curve on August 09, 2016, 10:58:46 am How do I differentiate each function with respect to x? Title: Re: Mathematics Question Thread Post by: jakesilove on August 09, 2016, 11:24:45 am How do I differentiate each function with respect to x? Far out buddy, those are some killer questions. Still, both are simple applications of the Chain rule: I'll only do the second, because it is substantially harder, and understanding it should let you differentiate the first one. We want to differentiate $y=((3x^3+1)^2+2)^4$ Now, we know that if $y=(f(x))^n$ $\frac{dy}{dx}=nf(x)^{n-1}*f'(x)$ So, for the question above, we have to let $f(x)=(3x^3+1)^2+2$ At first, I was thinking we needed to do two iterations of the Chain rule. But, since it's only raised to the power of 2, let's just expand it. This will make the solution much easier to come to; if you had a power higher than 2, you would have to chain rule again. $f(x)=9x^6+6x^3+3$ $f'(x)=54x^5+18x^2$ So, for the entire thing, $\frac{dy}{dx}=4((3x^3+1)^2+2)^3*(54x^5+18x^2)$ Title: Re: Mathematics Question Thread Post by: Jakeybaby on August 09, 2016, 07:19:39 pm Sorry, but I haven't become familiar with LaTeX yet, but I will do soon. Could someone assist me with the integral of (e2x-1 +3)2 dx? Thankyou Title: Re: Mathematics Question Thread Post by: RuiAce on August 09, 2016, 07:43:05 pm Sorry, but I haven't become familiar with LaTeX yet, but I will do soon. Could someone assist me with the integral of (e2x-1 +3)2 dx? Thankyou $\text{Expand it out and integrate term by term.}\\ \text{Note that }(e^{2x-1})^2=e^{4x-2}$ \begin{align*}\int (e^{2x-1}+3)^2dx&=\int (e^{4x-2}+6e^{2x-1}+9)dx\\ &= \frac{1}{4}e^{4x-2}+3e^{2x-1}+9x+C\end{align*} Title: Re: Mathematics Question Thread Post by: Jakeybaby on August 09, 2016, 07:58:33 pm $\text{Expand it out and integrate term by term.}\\ \text{Note that }(e^{2x-1})^2=e^{4x-2}$ \begin{align*}\int (e^{2x-1}+3)^2dx&=\int (e^{4x-2}+6e^{2x-1}+9)dx\\ &= \frac{1}{4}e^{4x-2}+3e^{2x-1}+9x+C\end{align*} Alright, did it a different way, would you always recommend expanding? Title: Re: Mathematics Question Thread Post by: RuiAce on August 09, 2016, 08:04:52 pm Alright, did it a different way, would you always recommend expanding? $\text{Unless it's linear and you can just do }\int (ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}\\ \text{(or at Ext 1 and you can do a substitution), I don't know any scenario where factorised is easier than expanded.}$ Title: Re: Mathematics Question Thread Post by: conic curve on August 10, 2016, 01:18:10 pm How do I find the deriative of the following? Title: Re: Mathematics Question Thread Post by: jakesilove on August 10, 2016, 01:46:33 pm How do I find the deriative of the following? Hey Conic, A lot of your questions appear to be fairly normal applications of the Chain Rule, Product rule or Quotient rule. The rules are described below. (http://mathcs.slu.edu/history-of-math/images/history/thumb/7/7f/Pqr-rules.jpg/350px-Pqr-rules.jpg) I don't think that it is very helpful if we just keep answering your questions, because that doesn't seem to be helping you very much. Rather, I think I would prefer if you post up your attempt at a solution, so that we can correct you and help you as you go. My main advice is to (for product/quotient) identify u, v, u' and v', write it all down then simply apply the formula. Give these questions a go; will be happy to help once I can see that you've made an attempt! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 10, 2016, 01:49:32 pm How do I find the deriative of the following? Hey conic! To show you what Jake means, if you do the derivatives separately like so: $u=(x^2+1)^3\quad v=x^2\\u'=6x(x^2+1)^2\quad v'=2x$ Note that u' in the question above was obtained with the chain rule! Pull the power out the front, subtract one from the power, the multiply by the derivative of the inside. You then literally sub this into the chain rule formula: $y'=\frac{u'v-v'u}{v^2}$ If you try breaking your big problems down into smaller ones, they will become much easier ;D as Jake said feel free to come back to us with your working if it isn't quite coming together and we can help you figure out what went wrong! Title: Re: Mathematics Question Thread Post by: RuiAce on August 10, 2016, 02:20:37 pm Hey Conic, A lot of your questions appear to be fairly normal applications of the Chain Rule, Product rule or Quotient rule. The rules are described below. (http://mathcs.slu.edu/history-of-math/images/history/thumb/7/7f/Pqr-rules.jpg/350px-Pqr-rules.jpg) I don't think that it is very helpful if we just keep answering your questions, because that doesn't seem to be helping you very much. Rather, I think I would prefer if you post up your attempt at a solution, so that we can correct you and help you as you go. My main advice is to (for product/quotient) identify u, v, u' and v', write it all down then simply apply the formula. Give these questions a go; will be happy to help once I can see that you've made an attempt! This. And I also feel that in the previous accounts you've posted up a solution, most of the mistakes were in algebraic manipulation, not the actual computation. You should put more emphasis on the basics because your grasp over the advanced concepts is superior to those as evidenced. Title: Re: Mathematics Question Thread Post by: conic curve on August 10, 2016, 04:15:26 pm (http://i.imgur.com/CTH9ycL.jpg) Title: Re: Mathematics Question Thread Post by: RuiAce on August 10, 2016, 04:37:21 pm (http://i.imgur.com/CTH9ycL.jpg) $\text{If you're using the }\textbf{product rule}\text{, you take }u=(x^2+1)^3\text{ and }v=\frac{1}{x^2}$ $\text{If you're using the }\textbf{quotient rule}\text{, you take }u=(x^2+1)^3\text{ and }v=x^2$ $\text{Know how the rule works. In the quotient rule, }u\text{ is the numerator whilst }v\text{ is the denominator.}\\ \text{As opposed to the product rule where }u\text{ and }v\text{ are the factors.}$ $\text{For the other one, you must use the }\textbf{chain rule} \text{ to find your }u^\prime, v^\prime\\ u=(x^2+1)^2 \implies u^\prime = 4x(x^2+1)\\ v= (4x+1)^3\implies v^\prime = 12(4x+1)^2$ Title: Re: Mathematics Question Thread Post by: conic curve on August 10, 2016, 05:35:29 pm $\text{If you're using the }\textbf{product rule}\text{, you take }u=(x^2+1)^3\text{ and }v=\frac{1}{x^2}$ $\text{If you're using the }\textbf{quotient rule}\text{, you take }u=(x^2+1)^3\text{ and }v=x^2$ $\text{Know how the rule works. In the quotient rule, }u\text{ is the numerator whilst }v\text{ is the denominator.}\\ \text{As opposed to the product rule where }u\text{ and }v\text{ are the factors.}$ $\text{For the other one, you must use the }\textbf{chain rule} \text{ to find your }u^\prime, v^\prime\\ u=(x^2+1)^2 \implies u^\prime = 4x(x^2+1)\\ v= (4x+1)^3\implies v^\prime = 12(4x+1)^2$ Oh yeah I was meant to use the product rule. Apologies. Anyways here is my working to the question Spoiler (http://i.imgur.com/EqDdbz2.jpg) (http://i.imgur.com/zGV36Yr.jpg) Someone please correct me if I am wrong Moderator edit: Put images in spoiler Title: Re: Mathematics Question Thread Post by: RuiAce on August 10, 2016, 06:46:00 pm Oh yeah I was meant to use the product rule. Apologies. Anyways here is my working to the question Spoiler (http://i.imgur.com/EqDdbz2.jpg) (http://i.imgur.com/zGV36Yr.jpg) Someone please correct me if I am wrong Moderator edit: Put images in spoiler Here's my feedback: In lines 2 and 3 you quoted the product rule correctly In line 4, suddenly all the x's turned into u's. Please fix this. (There are several instances where this occurs) In line 6, your derivatives are correct. In lines 7 and 8, your substitution is correct In line 9, your factorisation is correct. However you can go (at least one step) further. \begin{align*}&\qquad \frac{1}{x^2}(x^2+1)^2\left[(x^2+1)\cdot -\frac{2}{x}+6x\right]\\ &= \frac{1}{x^3}(x^2+1)\left[-2(x^2+1)+6x^2\right]\\ &= \frac{1}{x^3}(x^2+1)(4x^2-2)\\ &= \frac{2}{x^3}(x^2+1)(x^2-1)\end{align*} __________________________ Once again, too many of those x's look like u's. Your answer is certainly correct according to WolframAlpha Except they factorised the 4 out. Title: Re: Mathematics Question Thread Post by: olivercutbill on August 13, 2016, 04:39:01 pm Hey guys- Can anyone help me understand this question? Title: Re: Mathematics Question Thread Post by: RuiAce on August 13, 2016, 05:30:46 pm Hey guys- Can anyone help me understand this question? $\text{The differential equation is the equation demonstrating the behaviour of logistical growth,}\\ \text{which is a growth that decreases in rate as the population goes to a maximum carrying capacity.}$ $\text{We are asked to show that the ugly thing satisfies the equation. So we just do the same process.}\\ \textit{The only difference is that this question is appreciably harder than that of natural growth due to the messy algebra.}$ $\text{Factorise the solution so that what we're trying to prove is slightly clearer.}\\ \frac{dN}{dt}=N(k-bN)\\ \text{Keep in mind we want to FORCE out the }N$ \begin{align*}N&=\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\\ &= kN_0\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-1}\\ \frac{dN}{dt}&=-kN_0\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-2} \times -k\left(k-bN_0\right)e^{-kt}\\ &= \frac{k^2N_0(k-bN_0)e^{-kt}}{\left({bN_0+(k-bN_0)e^{-kt}}\right)^{-2}}\end{align*} \text{If we're careful enough, we can perfectly factorise out the }N\\ \begin{align*}\implies \frac{dN}{dt}&=\left(\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &= N\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\end{align*} $\text{So now, we just deal with what's in the brackets.}\\ \text{For the sake of my sanity, I worked backwards on paper. But once you figure out how to work backwards, it's easy to work forwards.}\\ \text{The add-subtract trick comes in handy here in working forwards.}$ \begin{align*}\frac{dN}{dt}&= N\left(\frac{k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &=N\left(\frac{kbN_0+k(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}-\frac{kbN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\\ &= N\left(k\left(\frac{bN_0+(k-bN_0)e^{-kt}}{bN_0+(k-bN_0)e^{-kt}}\right)-b\left(\frac{kN_0}{bN_0+(k-bN_0)e^{-kt}}\right)\right)\\ &= N(k-bN)\\ &= kN - bN^2\end{align*} Title: Re: Mathematics Question Thread Post by: olivercutbill on August 15, 2016, 07:17:27 pm (from above) Thanks for that guys! Title: Re: Mathematics Question Thread Post by: conic curve on August 17, 2016, 10:52:50 am Help with iv and v Title: Re: Mathematics Question Thread Post by: RuiAce on August 17, 2016, 11:13:02 am Help with iv and v $\text{Using trigonometry in }\triangle PQS\text{ we have that }\\ \sin \theta = \frac{PS}{\ell} \implies PS=\ell \sin \theta$ $\text{Then the area of the triangle can be computed using just }A=\frac{1}{2}bh\\ A=\frac{1}{2} \times (\ell \sin \theta) \times (2\ell \cos \theta) = \ell^2 \cos \theta \sin \theta$ \text{We keep in mind that }0 < \ell < \frac{\pi}{2}\text{ when we do our calculus, as otherwise we have no triangle.}\\ \begin{align*}A&=\ell^2 \cos \theta \sin \theta\\\frac{dA}{d\ell}&= \ell^2 \left(\cos^2 \theta - \sin^2 \theta\right)\end{align*} $\text{To maximise the area, set }\frac{dA}{d\ell}=0\\ \ell^2 (\cos^2\theta - \sin^2\theta) = 0 \implies \tan^2 \theta = 1\\ \text{Keeping in mind }0<\theta < \frac{\pi}{2}\text{, solving this gives }\theta = \frac{\pi}{4}$ $\text{You can test that it is a maximum (and not a minimum) yourself}$ Title: Re: Mathematics Question Thread Post by: conic curve on August 17, 2016, 01:04:05 pm Need help for sketching the following: y= -x^3/6 + x^2/3 - x/6 I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in? Thanks guys :D Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 17, 2016, 01:42:19 pm Need help for sketching the following: y= -x^3/6 + x^2/3 - x/6 I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in? Thanks guys :D Hey conic! Derivative looks good, remember that solving for y'=0 isn't solving for x-intercepts, it is solving for turning points: $\frac{dy}{dx}=\frac{-3x^2}{6}+\frac{4x}{6}-\frac{x}{6} \\ \frac{dy}{dx}=0 \implies -3x^2+4x-1=0 \\ (1-3x)(x-1)=0\\\therefore x=\frac{1}{3}\text{ or }1$ We now test their nature. You can use one of the y' tables as you call it, or more conveniently, we can use the second derivative test: $\frac{d^2y}{dx^2}=-x+\frac{2}{3} \\\text{At }x=1\text{, }\frac{d^2y}{dx^2}<0\text{, therefore a maximum exists at x=1}\\\text{At }x=\frac{1}{3}\text{, }\frac{d^2y}{dx^2}>0\text{, therefore a minimum exists at }x=\frac{1}{3}$ Is this a test you are familiar with? What you suggested is comparing the sign of the derivative either side of the turning point, but there is an easier way. Basically, it is using a concavity test to determine the nature of a stationary point. If it is concave down, it is a maximum, if it is concave down, it is a minimum ;D Finding the x-intercepts is done by equating the initial expression to 0. Remember, the x intercept is where y=0: $\frac{-x^3}{6}+\frac{x^2}{3}-\frac{x}{6}=0 \\\therefore -x^3+2x^2-x=0\\-x(x^2-2x+1)=0 \\\therefore x=0,1$ Putting all of that (x-intercepts and turning points) together, you should be able to sketch something like this ;D Hope that makes sense ;D Title: Re: Mathematics Question Thread Post by: anotherworld2b on August 19, 2016, 12:58:22 am Hi :) Could I please have help with solving these three questions? I apologise if i have put too many questions at once Title: Re: Mathematics Question Thread Post by: conic curve on August 19, 2016, 06:34:13 am (http://i.imgur.com/oDcI5Ll.jpg) Title: Re: Mathematics Question Thread Post by: RuiAce on August 19, 2016, 08:08:23 am (http://i.imgur.com/oDcI5Ll.jpg) Nice, but I have a few comments. $\text{All of your }x\text{'s look like }u\text{'s...}\\ \text{For the first one, I'm sure you saw and just didn't bother writing it but }\frac{2}{8}=\frac{1}{4}\\ \text{The most important one is part 3. This is only half correct. The answer is }x=\pm 3\\ \text{Note that for an even power, the plus or minus sign is important.}$ Also for part 3, I'm not sure if you picked up on it but you don't have to FOIL out the whole thing. It's just difference of two squares. Title: Re: Mathematics Question Thread Post by: anotherworld2b on August 19, 2016, 11:22:38 pm thank you for the help conic curve :D I was wondering how exactly did the surds disappear to equal 2/8? I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2? (http://i.imgur.com/oDcI5Ll.jpg) Mod edit: Swapped over the quote. Title: Re: Mathematics Question Thread Post by: RuiAce on August 19, 2016, 11:50:18 pm thank you for the help :D I was wondering how exactly did the surds disappear to equal 2/8? I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2? Mod edit: Took out the quote. I'm not the one who deserves the thanks here. \text{This is just because }2\sqrt{2}=\sqrt8\\ \text{So by rewriting:}\\ \begin{align*}2\sqrt{2x}&=\sqrt{2}\\ \sqrt{8x}&=\sqrt{2}\\ 8x&=2\\ x&=\frac{2}{8}=\frac{1}{4}\end{align*} Title: Re: Mathematics Question Thread Post by: anotherworld2b on August 20, 2016, 12:31:07 am I was also wondering how would you do these 2 questions? I am not sure how to approach them Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 20, 2016, 12:59:06 am I was also wondering how would you do these 2 questions? I am not sure how to approach them You might make both a bit easier by making a substation for yourself! In the first: $m=5^x\implies (25m-1)(m-1)=0\\\therefore m=\frac{1}{25},1\\\therefore5^x=\frac{1}{25},1\\x=-2,1$ Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me ;D let me know if you need more of a hand with it ;D Title: Re: Mathematics Question Thread Post by: anotherworld2b on August 20, 2016, 01:48:05 am Thank you for your help :D i was able to get the answer for the second one by following your example. I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin ??? You might make both a bit easier by making a substation for yourself! In the first: $m=5^x\implies (25m-1)(m-1)=0\\\therefore m=\frac{1}{25},1\\\therefore5^x=\frac{1}{25},1\\x=-2,1$ Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me ;D let me know if you need more of a hand with it ;D Title: Re: Mathematics Question Thread Post by: conic curve on August 20, 2016, 07:55:48 am thank you for the help conic curve :D I was wondering how exactly did the surds disappear to equal 2/8? I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?Mod edit: Swapped over the quote. nws Usually if there is a constant (number) to the power of an even index (i.e. power) it is usually even e.g. -1 to the power of 4 is equal to 1 right because -1 x -1 x -1 x-1 =1 (remember two negatives make a positive) if it was -1 to the power of 5 the answer would be -1 (obviously) 2Squareroot of 2x=8 Title: Re: Mathematics Question Thread Post by: RuiAce on August 20, 2016, 08:49:49 am Thank you for your help :D i was able to get the answer for the second one by following your example. I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin ??? $\text{The key characteristic of your unaltered exponential is that }\lim_{x\to -\infty}a^x=0\\ \text{This is all we need to find the vertical translation.}$ $\text{So since the first one features }\lim_{x\to -\infty}a^x=2\text{ we have }y=a^{x-b}+2$ \text{Then all we do is use the grid (which is quite rare for an HSC maths question)}\\ \text{The graph passes through }(0,5)\text{ and }(-1,3)\\ \text{So we sub these points in}\\ \begin{align*}3&=a^{-1+b}+2\\ 5&=a^b+2\end{align*}\\ \text{Solving this gives }a=3, b=1\text{ which you can sub back in} Title: Re: Mathematics Question Thread Post by: anotherworld2b on August 20, 2016, 10:15:46 am How do you know that lim a^x = 2? I have never used or hear about lim before $\text{The key characteristic of your unaltered exponential is that }\lim_{x\to -\infty}a^x=0\\ \text{This is all we need to find the vertical translation.}$ $\text{So since the first one features }\lim_{x\to -\infty}a^x=2\text{ we have }y=a^{x-b}+2$ \text{Then all we do is use the grid (which is quite rare for an HSC maths question)}\\ \text{The graph passes through }(0,5)\text{ and }(-1,3)\\ \text{So we sub these points in}\\ \begin{align*}3&=a^{-1+b}+2\\ 5&=a^b+2\end{align*}\\ \text{Solving this gives }a=3, b=1\text{ which you can sub back in} Title: Mathematics Question Thread Post by: RuiAce on August 20, 2016, 11:10:05 am How do you know that lim a^x = 2? I have never used or hear about lim before This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial. The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0. Because fact is if you raise a number to a really negative number it just goes down to 0 However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 20, 2016, 12:37:01 pm This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial. The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0. Because fact is if you raise a number to a really negative number it just goes down to 0 However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course. Another world is a WACE student ;D you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value :) Title: Re: Mathematics Question Thread Post by: RuiAce on August 20, 2016, 12:49:38 pm Another world is a WACE student ;D you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value :) Hmm. Yeah I thought that VCE also do have limits. But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 20, 2016, 03:02:46 pm Hmm. Yeah I thought that VCE also do have limits. But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course. Eh, maybe not covered yet, s'all gee in the bee :o Title: Re: Mathematics Question Thread Post by: olivercutbill on August 24, 2016, 05:09:24 pm Got a foolish question. Indefinite integral of (3)^(2x-1) Thanks guys Title: Re: Mathematics Question Thread Post by: RuiAce on August 24, 2016, 05:26:41 pm Got a foolish question. Indefinite integral of (3)^(2x-1) Thanks guys \text{We use the fact that }e^{\ln x}=x\text{ for all }x>0\\ \begin{align*}\int 3^{2x-1}dx&=\int e^{\ln \left(3^{2x-1}\right)}dx\\ &= \int e^{(2x-1)\ln 3}dx\\ &= \frac{1}{2\ln 3}e^{(2x-1)\ln 3}+C\\ &= \frac{1}{2\ln 3}3^{2x-1}+C\end{align*} $\text{Observe that }\ln 3\text{ is constant and our integral was of the form }\int e^{ax+b}dx = \frac{1}{a}e^{ax+b}+C\\ \text{and that we reverse engineer steps at the end.}\\ \text{Recall that }\log a^m = m \log a$ Title: Re: Mathematics Question Thread Post by: olivercutbill on August 24, 2016, 07:32:14 pm \text{We use the fact that }e^{\ln x}=x\text{ for all }x>0\\ \begin{align*}\int 3^{2x-1}dx&=\int e^{\ln \left(3^{2x-1}\right)}dx\\ &= \int e^{(2x-1)\ln 3}dx\\ &= \frac{1}{2\ln 3}e^{(2x-1)\ln 3}+C\\ &= \frac{1}{2\ln 3}3^{2x-1}+C\end{align*} $\text{Observe that }\ln 3\text{ is constant and our integral was of the form }\int e^{ax+b}dx = \frac{1}{a}e^{ax+b}+C\\ \text{and that we reverse engineer steps at the end.}\\ \text{Recall that }\log a^m = m \log a$ ah - more complicated than I thought but I got it. Awesome, you're a legend Rui! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 24, 2016, 11:23:22 pm ah - more complicated than I thought but I got it. Awesome, you're a legend Rui! Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick) ;D Title: Re: Mathematics Question Thread Post by: olivercutbill on August 27, 2016, 12:00:15 pm Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick) ;D Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of: -equations reducible to quadratics -factoring -simultaneous equations -the limiting sum -graphing the problem etc etc That being said, I have another question for probability: a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn? b) Dealt five cards, what is the probability of all cards being of the same suit? I've been trying a four of a kind +1 kind of approach but it hasn't worked. Cheers guys Title: Re: Mathematics Question Thread Post by: RuiAce on August 27, 2016, 12:36:17 pm Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of: -equations reducible to quadratics -factoring -simultaneous equations -the limiting sum -graphing the problem etc etc That being said, I have another question for probability: a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn? b) Dealt five cards, what is the probability of all cards being of the same suit? I've been trying a four of a kind +1 kind of approach but it hasn't worked. Cheers guys $\text{Assuming no replacement.}$ $\text{If all four aces must be drawn, then in a deck of 52 we have }\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{1}{49}\\ \text{And then the last card can be any of the remaining. }\left(\frac{48}{48}\right)$ $\text{Within one suit, if all five cards are of the same suit then we have }\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{49}\times\frac{9}{48}\\ \text{But we have to multiply this by 4 because there are four suits in a deck of cards.}$ Title: Re: Mathematics Question Thread Post by: olivercutbill on August 27, 2016, 02:48:53 pm for a) the answer the textbook gives is 1/54145. I had the same logic as you and I'm not sure how to do it. On my calculator, it gives a much smaller answer. Weird Title: Re: Mathematics Question Thread Post by: RuiAce on August 27, 2016, 02:54:11 pm $\text{The card that isn't an ace can be drawn before, after, or in between }\\ \text{successive draws of the aces. There are thus 5}\\ \text{positions to draw the aces.}$ $\text{So the correct answer multiplies my incorrect expression by }5.$ The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.) Title: Re: Mathematics Question Thread Post by: olivercutbill on August 27, 2016, 05:19:26 pm $\text{The card that isn't an ace can be drawn before, after, or in between }\\ \text{successive draws of the aces. There are thus 5}\\ \text{positions to draw the aces.}$ $\text{So the correct answer multiplies my incorrect expression by }5.$ The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.) Ok! I thought so but didn't full grasp the logic. Title: Re: Mathematics Question Thread Post by: Deng on August 30, 2016, 02:24:17 pm Hey guys i was wondering how to effectively study maths in the sense that i do past hsc papers and review my mistakes but i feel like it isnt necessarily improving my maths skills. Like even doing maths everyday i manage to forget small details and concepts of questions, im not sure how to explain it but an example would be like one of my maths exams i did well in, when i sat the same exam again 2 months later i manage to forget how to do some of the questions despite doing at least 30mins to 2 hours of maths everyday. Also, when a question asks for maximum velocity e.g attached how would i do part (ii) Thanks Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 30, 2016, 02:38:33 pm Hey guys i was wondering how to effectively study maths in the sense that i do past hsc papers and review my mistakes but i feel like it isnt necessarily improving my maths skills. Like even doing maths everyday i manage to forget small details and concepts of questions, im not sure how to explain it but an example would be like one of my maths exams i did well in, when i sat the same exam again 2 months later i manage to forget how to do some of the questions despite doing at least 30mins to 2 hours of maths everyday. Also, when a question asks for maximum velocity e.g attached how would i do part (ii) Thanks Hey Deng! To address your question first, the answer is to differentiate! We seek a maxima for the velocity function, so: $v=1-2\cos{t} \\ \frac{dv}{dt}=2\sin{t} \\ \frac{dv}{dt}=0 \implies t=0,\pi...$ Any subsequent values will be irrelevant because they will give the same answer. Test those values in the original function: $t=0\implies v=1-2=-1\\t=\pi\implies v=1+2=3$ Clearly, the second is larger, and you can use typical tests to deduce that there is a maxima at t equal to pi. Note that you could, in this case, do this conceptually. A cosine function will always range between -1 and 1, and so, a -2cost function will always range between -2 and 2. Pick the case which maximises velocity, and you get the same answer as above: $\text{Max Velocity }=3\text{ms}^{-1}$ As for your study dilemma, let me ask you something. When you do those papers and forget concepts, when was the last time you had studied that specific topic, rather than Mathematics in general? Often, if you haven't done something in a while, you'll mess things up. I do it all the time if/when I revisit old math concepts. Hell, not just math, it applies to pretty much everything (writing a bit of code in a coding language you've not used in ages, or driving a manual after a year of driving an auto, for example). Your brain will forget things, are you consistently revising the old concepts? :) Title: Re: Mathematics Question Thread Post by: jakesilove on August 30, 2016, 02:39:25 pm Hey guys i was wondering how to effectively study maths in the sense that i do past hsc papers and review my mistakes but i feel like it isnt necessarily improving my maths skills. Like even doing maths everyday i manage to forget small details and concepts of questions, im not sure how to explain it but an example would be like one of my maths exams i did well in, when i sat the same exam again 2 months later i manage to forget how to do some of the questions despite doing at least 30mins to 2 hours of maths everyday. Also, when a question asks for maximum velocity e.g attached how would i do part (ii) Thanks Hey! I'm on the run so can only quickly answer your Maths question, will get to your study question a bit later (unless someone else jumps in!). I'll just quickly say that your experience with maths is not abnormal, and there are definitely techniques to help you out! For your question, you just need to recall that $-1 < cos t < 1$ In this case, velocity will be at a MAXIMUM when cos(t) is at its most NEGATIVE (as it is a negative term). Therefore, when cos(t) is -1, velocity will be 3. This is the maximum value! Title: Re: Mathematics Question Thread Post by: Deng on August 30, 2016, 02:50:50 pm @Jamon, ah i assumed you had to differentiate since it said maximum however i had never seen a question like this so i was confused on what to do I havent touched the topics individually since ive been taught them so i guess it does make sense, should probably get down with the basics before doing the HSC papers to be more effective? @Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way Title: Re: Mathematics Question Thread Post by: olivercutbill on August 30, 2016, 03:48:29 pm Specifically iii) please! This one is a series/probability one, not sure how to get to the answer. Thanks! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 30, 2016, 03:57:08 pm @Jamon, ah i assumed you had to differentiate since it said maximum however i had never seen a question like this so i was confused on what to do I havent touched the topics individually since ive been taught them so i guess it does make sense, should probably get down with the basics before doing the HSC papers to be more effective? @Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way That absolutely makes sense then, so yep, perhaps tackle some old chapter reviews before hitting HSC past papers? ;D I always did Paper first, review second if required, but it might work better the other way around for you!! ;D Title: Re: Mathematics Question Thread Post by: jamonwindeyer on August 30, 2016, 04:08:24 pm Specifically iii) please! This one is a series/probability one, not sure how to get to the answer. Thanks! Hey! This is a fairly common ace up the sleeve for BOSTES, the same style of question was at the end of my MX1 exam: First question is fairly easy, remember that there are 36 possible outcomes for throwing two dice, and only one yields a win! $P(\text{Pat Wins First Turn})=\frac{1}{36}$ For the second bit, remember that for Pat to have a second turn, Chandra must not roll a double six on her turn, otherwise she would win, AND Pat must not roll a double six on his first turn either. This creates the following: \begin{align*} P(\text{First or Second}) &=P(\text{First})+P(\text{Second})\\&=\frac{1}{36}+\left(\frac{35}{36}\right)^2\times\frac{1}{36}\\&=\frac{71}{1296} \end{align*} Now we get technical. For Pat to eventually win the game, we need to consider an infinite number of turns. After all, the game could go on forever, couldn't it? Let's look at the series for just the 1st-3rd turns: $P(\text{eventually})=\frac{1}{36}+\frac{1}{36}\left(\frac{35}{36}\right)^2+\frac{1}{36}\left(\frac{35}{36}\right)^4 ... \\P=\frac{1}{36}\left[1+\left(\frac{35}{36}\right)^2+\left(\frac{35}{36}\right)^4...\right]$ Notice what's inside the brackets? An infinite series! We can find it's sum using the formula: $S_\infty=\frac{a}{1-r}$ Note that we are allowed to use this formula because the common ratio is less than 1 ;D Using that formula should get the answer you need! ;D $\text{My Answer (Subject to Error, I'm in a Lecture, lol)}\implies P = \frac{36}{71}$ Title: Re: Mathematics Question Thread Post by: RuiAce on August 30, 2016, 10:57:27 pm @Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way $-1\le \cos t \le 1\text{ is just the range of cos. You have to adapt to this.}$ $\text{The manipulation is pretty minute. It is not hard.}\\ -1\le \cos t \le 1\\ 1 \ge -\cos t \ge -1$ $\text{Fix it up.}\\ -1 \le -\cos t \le 1\\ -2 \le -2\cos t \le 2\\ -1 \le 1-2\cos t \le 3$ Jamon's method is your standard method that you turn to when you're stuck. But something like this is something you should be able to cope with. Hey! This is a fairly common ace up the sleeve for BOSTES, the same style of question was at the end of my MX1 exam: First question is fairly easy, remember that there are 36 possible outcomes for throwing two dice, and only one yields a win! $P(\text{Pat Wins First Turn})=\frac{1}{36}$ For the second bit, remember that for Pat to have a second turn, Chandra must not roll a double six on her turn, otherwise she would win, AND Pat must not roll a double six on his first turn either. This creates the following: \begin{align*} P(\text{First or Second}) &=P(\text{First})+P(\text{Second})\\&=\frac{1}{36}+\left(\frac{35}{36}\right)^2\times\frac{1}{36}\\&=\frac{71}{1296} \end{align*} Now we get technical. For Pat to eventually win the game, we need to consider an infinite number of turns. After all, the game could go on forever, couldn't it? Let's look at the series for just the 1st-3rd turns: $P(\text{eventually})=\frac{1}{36}+\frac{1}{36}\left(\frac{35}{36}\right)^2+\frac{1}{36}\left(\frac{35}{36}\right)^4 ... \\P=\frac{1}{36}\left[1+\left(\frac{35}{36}\right)^2+\left(\frac{35}{36}\right)^4...\right]$ Notice what's inside the brackets? An infinite series! We can find it's sum using the formula: $S_\infty=\frac{a}{1-r}$ Note that we are allowed to use this formula because the common ratio is less than 1 ;D Using that formula should get the answer you need! ;D $\text{My Answer (Subject to Error, I'm in a Lecture, lol)}\implies P = \frac{36}{71}$ This looks good (y) Title: Re: Mathematics Question Thread Post by: anotherworld2b on September 02, 2016, 07:10:19 pm Hi i was wonering how do I solve these questions? Title: Re: Mathematics Question Thread Post by: jakesilove on September 02, 2016, 07:18:11 pm Hi i was wonering how do I solve these questions? I'll just help you out with the first, one, because once you've done one you can definitely figure out the rest yourself! You need to start with the relevant formula for the type of sequence (ie. geometric or arithmetic). For arithmetic sequences; $T_n=a+d(n-1)$ Where n is the term number, a is the first term, and d is the difference between subsequent terms. So, we know two pieces of information. $T_{50}=a+d(49)=1853$ $T_{70}=a+d(69)=1793$ We can quickly subtract one equation from the other (to eliminate the a term) $49d-69d=1853-1793$ $d=-3$ So, subbing back into an equation above $1853=a-3(49)$ $a=2000$ Giving us the final formula $T_n=2000+(-3)(n-1)$ We've already found the first term (a=2000). The 51st term will be $T_{51}=2000+(-3)(50)=1850$ The same approach (simultaneous equations to fill in the relevant equation) should be taken for your other questions. If you can't get them, post up your attempt at a solution and I'll help you out! Jake Title: Re: Mathematics Question Thread Post by: anotherworld2b on September 02, 2016, 11:27:24 pm thank you for your help with Q29 but for Q30 I'm not sure how to do it. I got a really large number that is completely off the correct answer I'll just help you out with the first, one, because once you've done one you can definitely figure out the rest yourself! You need to start with the relevant formula for the type of sequence (ie. geometric or arithmetic). For arithmetic sequences; $T_n=a+d(n-1)$ Where n is the term number, a is the first term, and d is the difference between subsequent terms. So, we know two pieces of information. $T_{50}=a+d(49)=1853$ $T_{70}=a+d(69)=1793$ We can quickly subtract one equation from the other (to eliminate the a term) $49d-69d=1853-1793$ $d=-3$ So, subbing back into an equation above $1853=a-3(49)$ $a=2000$ Giving us the final formula $T_n=2000+(-3)(n-1)$ We've already found the first term (a=2000). The 51st term will be $T_{51}=2000+(-3)(50)=1850$ The same approach (simultaneous equations to fill in the relevant equation) should be taken for your other questions. If you can't get them, post up your attempt at a solution and I'll help you out! Jake Title: Re: Mathematics Question Thread Post by: jamonwindeyer on September 02, 2016, 11:39:06 pm thank you for your help with Q29 but for Q30 I'm not sure how to do it. I got a really large number that is completely off the correct answer Show us your working mate! ;D wait, are you definitely using geometric sequence terms? $T_n=ar^{n-1}\\\therefore ar^9=98415\\\therefore ar^{12}=2657205$ Try solving that simultaneously to find the parameters, then proceed! If you still have trouble then definitely pop up your working and we'll see if we can spot the problem ;D Title: Re: Mathematics Question Thread Post by: anotherworld2b on September 02, 2016, 11:50:24 pm I found why I was getting it wrong yay ;D Title: Re: Mathematics Question Thread Post by: jamonwindeyer on September 03, 2016, 12:19:11 am I found why I was getting it wrong yay ;D Well done! ;D glad to hear it :) Title: Re: Mathematics Question Thread Post by: anotherworld2b on September 03, 2016, 12:58:46 am Thank you very much jamonwindeyer and jakesilove for your help ;D Title: Re: Mathematics Question Thread Post by: onepunchboy on September 03, 2016, 08:15:36 pm (http://uploads.tapatalk-cdn.com/20160903/9c8e20f1a0050a56205d57886442b93f.jpg) Hey guys im not very good at doing probability qs can someone help me out with this one? Thanks ! Title: Re: Mathematics Question Thread Post by: jakesilove on September 03, 2016, 10:21:35 pm (http://uploads.tapatalk-cdn.com/20160903/9c8e20f1a0050a56205d57886442b93f.jpg) Hey guys im not very good at doing probability qs can someone help me out with this one? Thanks ! Hey! So the smartest way to consider this question is to think about what happens if NONE of the runners finish the race in under 10 seconds. Then, we can take 1-P(none) to find the probability that AT LEAST one finishes within 10 seconds! Let me know if you need me to explain why in any more detail. Now, the probability of NONE of the runners finishing within 10 seconds will be each of their respective probability (of not completing) multiplied together. If we want events A AND B AND C to happen simultaneously, then their probability will be P(A)*P(B)*P(C). The probability of not finishing will be 1-P(finishing) $(1-\frac{1}{4})(1-\frac{1}{6})(1-\frac{2}{5})=\frac{3}{8}$ Therefore, the probability of AT LEAST ONE of the runners finishing within 10 seconds will be $1-\frac{3}{8}=\frac{5}{8}$ Title: Re: Mathematics Question Thread Post by: Ali_Abbas on September 03, 2016, 10:29:53 pm [IMG]http://uploads.tapatalk-cdn.com/20160903/9c8e20f1a0050a56205d57886442b93f.jpg[/IMG Hey guys im not very good at doing probability qs can someone help me out with this one? Thanks ! I just noticed that this question has already been answered by Jake but since mine has already been typed up I will still post it as an additional reference to potentially consolidate your understanding. The probabilities of each competitor finishing the race in under 10 seconds are given as: $p = \frac{1}{4}, \space q = \frac{1}{6}, \space and \space r = \frac{2}{5}$ respectively. So these are the probabilities of "success". To solve this question, we make note of the sample space i.e. the list of all possible outcomes (keeping in mind the probabilities of all instances sum to 1). These are: No successes and three failures, one success and two failures, two successes and one failure, three successes and no failures. Notice however, that the latter three cases can be captured collectively by the single statement that at least one success occurs. We thus obtain the relationship: $P(at \space least \space one \space success) + P(no \space successes) = 1$ $\therefore P(at \space least \space one \space success) = 1 - P(no \space successes) = 1 - P(all \space failures)$ To calculate the probability that each competitor fails, we simply take the complements of p, q, and r and multiply them together. This is using two applications of the rule that P(A and B) = P(A)P(B). Thus, $P(at \space least \space one \space success) = 1 - (1 - \frac{1}{4})(1 - \frac{1}{6})(1 - \frac{2}{5}) = ... = \frac{5}{8}$ and the answer is D. Title: Re: Mathematics Question Thread Post by: jakesilove on September 04, 2016, 12:01:27 pm For anyone looking to absolutely smash their HSC exam, understanding what the question actually expects of you is vital. Check our Rui's beast guide of Maths verbs HERE, and get an edge in your final exam! As always, thanks must go out to the legend himself, RuiAce; Improving Atars Since 2015. Title: Re: Mathematics Question Thread Post by: onepunchboy on September 08, 2016, 06:50:13 pm (http://uploads.tapatalk-cdn.com/20160908/fbd35324c152443c57020f766686223e.jpg) Hey guys im not really sure how to do question i) , i got out cos(pi on 6) = -1/2 but not really sure where to go from there.. Thanks ! Title: Re: Mathematics Question Thread Post by: RuiAce on September 08, 2016, 07:01:22 pm (http://uploads.tapatalk-cdn.com/20160908/fbd35324c152443c57020f766686223e.jpg) Hey guys im not really sure how to do question i) , i got out cos(pi on 6) = -1/2 but not really sure where to go from there.. Thanks ! $\text{Observe that }0\le t \le 12\\ \implies 0 \le \frac{\pi}{6}t \le 2\pi$ \begin{align*}375&=400+50\cos \left(\frac{\pi}{6}t\right)\\ -\frac{1}{2}&=\cos \left(\frac{\pi}{6}t\right)\end{align*} $\text{So for some reason, in your working out you dropped a very important }'t'$ \text{Using our rule of ASTC, to solve this consider the second and third quadrants:}\\ \begin{align*}\cos \left(\frac{\pi}{6}t\right)&=-\frac{1}{2}\\ \frac{\pi}{6}t&=\frac{2\pi}{3}, \, \frac{4\pi}{3}\\ t&= 4,\, 8\end{align*} Title: Re: Mathematics Question Thread Post by: onepunchboy on September 11, 2016, 11:09:17 am (http://uploads.tapatalk-cdn.com/20160911/db59b46aafce63ccd1e9b1a018625847.jpg) Hello again :), i was wondering how to get the equation for example 10.. dont really know what the question is asking. Thanks !! Im not sure if this is past hsc Title: Re: Mathematics Question Thread Post by: RuiAce on September 11, 2016, 11:16:00 am (http://uploads.tapatalk-cdn.com/20160911/db59b46aafce63ccd1e9b1a018625847.jpg) Hello again :), i was wondering how to get the equation for example 10.. dont really know what the question is asking. Thanks !! Im not sure if this is past hsc $\text{Observe that the distance between the hill and the take-off path is }PQ\\ \text{The distance of }PQ\text{ is vertical and is thus given as:}\\ PQ=\frac{1}{5}x-\frac{1}{20}(200-x)(x-240)$ $\text{Note that we wish to }\textit{minimise}\text{ the distance }PQ\text{ (read the question again!)}\\ \text{Expanding and rearranging we have }\\ PQ=\frac{x^2}{20}-\frac{109x}{5}+2400$ $\text{So we have }\frac{d(PQ)}{dx}=\frac{x}{10}-\frac{109}{5}\\ \frac{d(PQ)}{dx}=0\text{ when }x=218\\ \text{Check that this is indeed a minimum and we have our answer.}$ Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 11:42:43 am How do you do this question (http://i.imgur.com/DiZub5k.jpg) Title: Re: Mathematics Question Thread Post by: jakesilove on September 13, 2016, 12:03:11 pm How do you do this question (http://i.imgur.com/DiZub5k.jpg) You just need to think about the physical interpretation of the derivative. If the derivative is positive (ie. above the x axis), the gradient will be positive. If the derivative is negative (ie. below the x-axis), the gradient will be negative. If the derivative is zero (ie. on the x axis), the gradient will be zero, and so there will be a turning point. Have a go at sketching the function, and I'll let you know if you're correct! Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 12:26:25 pm This is what I got I think it may be wrong (http://i.imgur.com/gOFmDSR.jpg) Title: Re: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 12:36:39 pm This is what I got I think it may be wrong (http://i.imgur.com/gOFmDSR.jpg) You didn't understand what Jake said. Think about what the derivative is. If the derivative f'(x) is negative (below the x-axis), then the original function f(x) is DECREASING. If the derivative is positive (above the x-axis), then the original function f(x) is INCREASING. If the derivative is 0 (x-intercept), then the original function has a STATIONARY POINT. (Also it doesn't matter where your graphs start. They can start however low and however high you like. All we care about is that your curve's stationary points are placed at a correct x-coordinate and it increases/decreases where it should be.) As an exercise, you may wish to sketch y=4x2(x-3) and an antiderivative y=x3(x-4). Use GeoGebra or just Desmos to ensure your graph is correct. Compare where the DERIVATIVE is ABOVE the x-axis, to where the ORIGINAL CURVE is INCREASING. Title: Re: Mathematics Question Thread Post by: jakesilove on September 13, 2016, 02:19:41 pm This is what I got I think it may be wrong (http://i.imgur.com/gOFmDSR.jpg) As an example, I've plotted the first two derivatives of the basic cubic graph. Can you see what's going on? (http://i.imgur.com/tK6XbBG.png) Jake Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 05:08:17 pm As an example, I've plotted the first two derivatives of the basic cubic graph. Can you see what's going on? (http://i.imgur.com/tK6XbBG.png) Jake Nah not really. Which one would be f'(x)? Title: Re: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 05:26:27 pm Nah not really. Which one would be f'(x)? $\text{First, IGNORE the line, treat the cubic as }f(x)\text{ and the quadratic as }f^\prime(x)$ $\text{Then, IGNORE the cubic, treat the PARABOLA as }f(x)\text{ and the line as }f^\prime(x)$ $\textbf{Compare WHERE the stationary points are on }f(x)\text{, to the intercepts of }f^\prime(x)$ Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 05:53:50 pm $\text{First, IGNORE the line, treat the cubic as }f(x)\text{ and the quadratic as }f^\prime(x)$ $\text{Then, IGNORE the cubic, treat the PARABOLA as }f(x)\text{ and the line as }f^\prime(x)$ $\textbf{Compare WHERE the stationary points are on }f(x)\text{, to the intercepts of }f^\prime(x)$ They interesect at the origin Title: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 06:12:11 pm They interesect at the origin Not the point. Here I might have a favouritism towards the graphs I suggested (Jake picked the most coincidental) but read the above posts 5 times. Where f(x) is INCREASING, f'(x) is ABOVE THE X-AXIS etc. Know what you're comparing. To know what you're comparing, read all of the above posts 5 times. Read and compare simultaneously Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 06:24:02 pm Not the point. Here I might have a favouritism towards the graphs I suggested (Jake picked the most coincidental) but read the above posts 5 times. Where f(x) is INCREASING, f'(x) is ABOVE THE X-AXIS etc. Know what you're comparing. To know what you're comparing, read all of the above posts 5 times. Read and compare simultaneously So imagine f'(x) was the cubic in the image I have attached above Is it the fact that f(x), f'(x) and f''(x) are all monotonic increasing Also for this question below, for 16 and 17 I know how to do it in my head but am not sure how to geometrically prove it Also for 18, the answer is 6,6 but I got 3 (http://i.imgur.com/W3VpSJs.jpg) Title: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 06:29:11 pm So imagine f'(x) was the cubic in the image I have attached above Is it the fact that f(x), f'(x) and f''(x) are all monotonic increasing Also for this question below, for 16 and 17 I know how to do it in my head but am not sure how to geometrically prove it Also for 18, the answer is 6,6 but I got 3 (http://i.imgur.com/W3VpSJs.jpg) For the sake of this question, STOP considering f"(x). Like I said, WHEN f(x) is INCREASING, the DERIVATIVE f'(x) is POSITIVE. ALL I want you to note is when f(x) is increasing/decreasing/stationary that f'(x) is above/below/intersecting the x-axis respectively. I think you do not understand what our explanations meant. For Q18, the correct answer should technically be 6,3. 6,6 is not right in my opinion as there's double counting. Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 06:39:02 pm To prevent confusion I decided to redo the question given the information you and Jake gave me If it is above the x axis, it is monotonic increasing and if it is below the x axis, it is monotonic decreasing (http://i.imgur.com/bAbUgni.jpg) Title: Re: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 06:57:49 pm To prevent confusion I decided to redo the question given the information you and Jake gave me If it is above the x axis, it is monotonic increasing and if it is below the x axis, it is monotonic decreasing (http://i.imgur.com/bAbUgni.jpg) How you labelled your +,s and -'s are correct but the graph you ended up drawing makes no sense. (http://uploads.tapatalk-cdn.com/20160913/5102c53b9e2230b3dd561d295cba8f43.jpg) Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 09:32:53 pm WHat is the answer to this Title: Re: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 09:38:43 pm WHat is the answer to this Well what was your attempt? Title: Re: Mathematics Question Thread Post by: conic curve on September 13, 2016, 10:04:59 pm Well what was your attempt? I found the domain first and got x is not equal to 4 and subbed back in to the original equation in order to find the range but then I get 1/0 as the range (which is infinity which is all y) Am I right? Title: Re: Mathematics Question Thread Post by: RuiAce on September 13, 2016, 10:06:36 pm I found the domain first and got x is not equal to 4 and subbed back in to the original equation in order to find the range but then I get 1/0 as the range (which is infinity which is all y) Am I right? That's only a part of the answer. Anything under a square root also has to be positive. $\text{So the domain isn't just }8-2x\neq 0\\ \text{It is }8-2x>0$ Title: Re: Mathematics Question Thread Post by: kaufoou on September 17, 2016, 08:52:00 am Hey! So I was wondering if it is worth buying a Mathomat or Math-Aid to take into the exam? (btw they're permitted by BOSTES) Title: Re: Mathematics Question Thread Post by: RuiAce on September 17, 2016, 08:59:30 am Hey! So I was wondering if it is worth buying a Mathomat or Math-Aid to take into the exam? (btw they're permitted by BOSTES) They are most certainly permitted and I bought one in for each of my maths exams. But their value isn't that great. They just help you draw a really neat sine curve and circles and what not. I'm fairly positive BOSTES is lenient on certain diagrams being decent enough. If you spend too much time fussing over graphing like me, however, then you may as well because they're not "expensive" either. The math-aid is more powerful than the mathomat, but the latter comes with a protractor Title: Re: Mathematics Question Thread Post by: olivercutbill on September 17, 2016, 04:00:42 pm How am I supposed to do vi) Answer for v) is (40)^1/2 Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on September 17, 2016, 04:31:33 pm How am I supposed to do vi) Answer for v) is (40)^1/2 Thanks $\text{Keep in mind that if your triangle has an obtuse angle in it}\\ \text{The area of the triangle is always }A=\frac{1}{2}bh$ $\text{Here, }b=\text{ length of }BC=\sqrt{40}\\ h=\text{ perp. distance from }A\text{ to }BC$ Title: Re: Mathematics Question Thread Post by: lha on September 19, 2016, 06:50:33 pm How many past papers shpuld we complete before hsc? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on September 19, 2016, 06:56:19 pm How many past papers shpuld we complete before hsc? I think that depends on the person! I personally did over 20, but that was probably excessive, do however many you think you need to!! Your aim should be to be able to know how to do almost any question on sight ;D Title: Re: Mathematics Question Thread Post by: jakesilove on September 19, 2016, 07:05:15 pm I think that depends on the person! I personally did over 20, but that was probably excessive, do however many you think you need to!! Your aim should be to be able to know how to do almost any question on sight ;D Like Jamon says, it's as many as you think works. That being said, if you did two past papers a day (for different subjects), 5 days a week, for two weeks, that's already 20 past papers. You've got way more than two weeks for most of your subjects, so plenty of time to do heaps of papers! Title: Re: Mathematics Question Thread Post by: conic curve on September 19, 2016, 07:38:10 pm Just curious but how do I change my username? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on September 19, 2016, 08:08:14 pm Just curious but how do I change my username? You message a national moderator (like me), but it does need to be a good reason for us to agree to change it for you :) Title: Re: Mathematics Question Thread Post by: itswags98 on September 21, 2016, 09:41:41 pm How do i attempt this question? Havent done anything like it before. Dumb brain immediately thought of pythag to work out the slope lol Title: Re: Mathematics Question Thread Post by: RuiAce on September 21, 2016, 09:58:18 pm How do i attempt this question? Havent done anything like it before. Dumb brain immediately thought of pythag to work out the slope lol $\text{Assuming that }t\le 8\\ \text{Which should be fine because the graph only goes to about }t=5$ $\text{Consider the given motion graph.}\\ \text{The velocity is depicted as a function of time.}\\ \text{Recall that the velocity represents how fast the particle moves, and in what direction.}$ $\text{Observe that for all }0\le t \le 4\text{ the particle has a positive velocity.}\\ \text{However, once }t\ge 4\text{, the velocity becomes negative.}$ $\text{Hence the particle moves in the positive direction (assume to the right) for the first 4 seconds}\\ \text{and then in the negative direction (assume left) from here on.}$ __________________________ $\text{We wish to maximise the displacement, }x\\ \text{Recall that }v=\frac{dx}{dt}\\ \text{i.e. when }v=0\text{, the particle is at rest.}$ $\text{The significance of the particle being at rest is that}\\ \textit{we have a stationary point at }t=4$ $\text{Indeed, once the velocity goes negative, the displacement starts decreasing.}\\ \text{Hence at }x=4\text{, the displacement is maximised.}$ __________________________ $\text{Of course, we can't actually find it yet because we don't know what }x\text{ is.}$ $\text{We must find the equation of }v\text{ to proceed:}\\ v=8-2t$ $\text{So upon integrating, }x=8t-t^2+C\\ \text{And since when }t=0, \, x=2\text{ we have }\\ C=2$ $\text{Hence we have }x=8t-t^2+2\\ \text{Sub }t=4 \text{ and we're done}$ Answers to the 2015 HSC can be found here. Title: Re: Mathematics Question Thread Post by: jakesilove on September 21, 2016, 10:00:50 pm $\text{Assuming that }t\le 8\\ \text{Which should be fine because the graph only goes to about }t=5$ $\text{Consider the given motion graph.}\\ \text{The velocity is depicted as a function of time.}\\ \text{Recall that the velocity represents how fast the particle moves, and in what direction.}$ $\text{Observe that for all }0\le 4 \le 4\text{ the particle has a positive velocity.}\\ \text{However, once }t\ge 4\text{, the velocity becomes negative.}$ $\text{Hence the particle moves in the positive direction (assume to the right) for the first 4 seconds}\\ \text{and then in the negative direction (assume left) from here on.}$ __________________________ $\text{We wish to maximise the displacement, }x\\ \text{Recall that }v=\frac{dx}{dt}\\ \text{i.e. when }v=0\text{, the particle is at rest.}$ $\text{The significance of the particle being at rest is that}\\ \textit{we have a stationary point at }t=4$ $\text{Indeed, once the velocity goes negative, the displacement starts decreasing.}\\ \text{Hence at }x=4\text{, the displacement is maximised.}$ __________________________ $\text{Of course, we can't actually find it yet because we don't know what }x\text{ is.}$ $\text{We must find the equation of }v\text{ to proceed:}\\ v=8-2t$ $\text{So upon integrating, }x=8t-t^2+C\\ \text{And since when }t=0, \, x=2\text{ we have }\\ C=2$ $\text{Hence we have }x=8t-t^2+2\\ \text{Sub }t=4 \text{ and we're done}$ Answers to the 2015 HSC can be found here. Alternatively (or, rather, using the exact same method but without the actual integration), you can remember that the area under the curve is the change in displacement. Assuming that, prior to t=4, the particle continues to move away from the origin, you can easily find the area under the curve (16) and add that to the original displacement (2). I would have done Rui's way, and they're exactly the same. Title: Re: Mathematics Question Thread Post by: RuiAce on September 21, 2016, 10:01:57 pm Alternatively (or, rather, using the exact same method but without the actual integration), you can remember that the area under the curve is the change in displacement. Assuming that, prior to t=4, the particle continues to move away from the origin, you can easily find the area under the curve (16) and add that to the original displacement (2). I would have done Rui's way, and they're exactly the same. I was going to do it that way, but then I remembered the original displacement was 2 and that's just a bugger. Don't feel safe just adding 2 on for most questions. It's safer here cause most of the graph is positive. In general though, the absolute extrema might change due to the new ordinates of the stationary points Title: Re: Mathematics Question Thread Post by: itswags98 on September 21, 2016, 10:18:23 pm $\text{Assuming that }t\le 8\\ \text{Which should be fine because the graph only goes to about }t=5$ $\text{Consider the given motion graph.}\\ \text{The velocity is depicted as a function of time.}\\ \text{Recall that the velocity represents how fast the particle moves, and in what direction.}$ $\text{Observe that for all }0\le t \le 4\text{ the particle has a positive velocity.}\\ \text{However, once }t\ge 4\text{, the velocity becomes negative.}$ $\text{Hence the particle moves in the positive direction (assume to the right) for the first 4 seconds}\\ \text{and then in the negative direction (assume left) from here on.}$ __________________________ $\text{We wish to maximise the displacement, }x\\ \text{Recall that }v=\frac{dx}{dt}\\ \text{i.e. when }v=0\text{, the particle is at rest.}$ $\text{The significance of the particle being at rest is that}\\ \textit{we have a stationary point at }t=4$ $\text{Indeed, once the velocity goes negative, the displacement starts decreasing.}\\ \text{Hence at }x=4\text{, the displacement is maximised.}$ __________________________ $\text{Of course, we can't actually find it yet because we don't know what }x\text{ is.}$ $\text{We must find the equation of }v\text{ to proceed:}\\ v=8-2t$ $\text{So upon integrating, }x=8t-t^2+C\\ \text{And since when }t=0, \, x=2\text{ we have }\\ C=2$ $\text{Hence we have }x=8t-t^2+2\\ \text{Sub }t=4 \text{ and we're done}$ Answers to the 2015 HSC can be found here. I appreciate it. Makes sense and the other way suggested makes sense too. Thanks :3. Hope you wont get sick of me asking questions until HSC time ahah Title: Re: Mathematics Question Thread Post by: lozil on September 25, 2016, 03:26:56 pm Hi, I was just wondering how you find the limit approaching infinity (I get how to do it for the limit approaching an actual number). E.g. Idk just a simple equation like lim x->infinity (x+1)/x Title: Re: Mathematics Question Thread Post by: RuiAce on September 25, 2016, 04:04:44 pm Hi, I was just wondering how you find the limit approaching infinity (I get how to do it for the limit approaching an actual number). E.g. Idk just a simple equation like lim x->infinity (x+1)/x $\text{I'm fairly positive that these limits are NOT in the 2U course.}\\ \text{To approach these questions, however, we must always consider the result}\\ \lim_{x\to \infty}\frac{1}{x}=0$ $\text{To force the }\frac{1}{x}\text{ in, we must divide top and bottom appropriately.}\\ \text{We choose the highest leading power. Here, it's 1.}$ \begin{align*}\lim_{x\to \infty}\frac{x+1}{x} &=\lim_{x\to \infty}\frac{1+\frac{1}{x}}{1}\\ &=\frac{1+0}{1}\\ &=1\end{align*} Title: Re: Mathematics Question Thread Post by: lozil on September 25, 2016, 04:27:29 pm $\text{I'm fairly positive that these limits are NOT in the 2U course.}\\ \text{To approach these questions, however, we must always consider the result}\\ \lim_{x\to \infty}\frac{1}{x}=0$ $\text{To force the }\frac{1}{x}\text{ in, we must divide top and bottom appropriately.}\\ \text{We choose the highest leading power. Here, it's 1.}$ \begin{align*}\lim_{x\to \infty}\frac{x+1}{x} &=\lim_{x\to \infty}\frac{1+\frac{1}{x}}{1}\\ &=\frac{1+0}{1}\\ &=1\end{align*} Ok, thanks! Title: Re: Mathematics Question Thread Post by: olivercutbill on September 27, 2016, 12:32:36 pm How is this done? Title: Re: Mathematics Question Thread Post by: RuiAce on September 27, 2016, 12:40:44 pm How is this done? $\text{The discriminant is }\Delta=(k-2)^2-4(2)(8)$ $\text{If the parabola does }\textit{not}\text{ intersect the line}\\ \text{Then there should be }\textit{NO points of intersection.}\\ \text{Meaning, if we tried to find any, we'd get }\textit{none.}$ \text{Suppose we wanted to find points of intersection.}\\ \text{Then we'd just sub it in.}\\ \begin{align*}2x^2+kx+9&=2x+1\\ 2x^2+(k-2)x+8&=0\end{align*} $\text{Well what do you know? This is the same equation as given in part (i)!}\\ \text{So if there are no solutions,}\textbf{ the discriminant must be negative.}\\ \text{So you end up solving a quadratic inequality }(k-2)^2 - 64 < 0$ Title: Re: Mathematics Question Thread Post by: lha on September 27, 2016, 02:21:49 pm Like Jamon says, it's as many as you think works. That being said, if you did two past papers a day (for different subjects), 5 days a week, for two weeks, that's already 20 past papers. You've got way more than two weeks for most of your subjects, so plenty of time to do heaps of papers! Okay thank you! Can anyone help me with question 9c)iii) in the 2010 CSSA HSC Trial paper? Its not letting me attach a picture but the paper should be online. Mod Edit: I've already told you how to edit posts. If you have further issues such as images, take the time to read this article. How to Use the ATAR Notes Forums And it shouldn't be, because CSSA papers are copyrighted. If you find a copy online (which may happen), it was illegally uploaded. Title: Re: Mathematics Question Thread Post by: Vinhtran on September 28, 2016, 06:38:35 pm I'm confused on how they found c in the tangent equation, please help. Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a Title: Re: Mathematics Question Thread Post by: MightyBeh on September 28, 2016, 06:59:20 pm I'm confused on how they found c in the tangent equation, please help. Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a Derivative of y at x=a will give you the gradient of the tangent. Subbing x=a back into y will give you a point, that the tangent will also pass through. Subbing this point into your tangent equation will give you c. Title: Re: Mathematics Question Thread Post by: RuiAce on September 28, 2016, 06:59:52 pm I'm confused on how they found c in the tangent equation, please help. Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a \begin{align*}y&=x^2e^x+1\\ \frac{dy}{dx}&=2xe^x+x^2e^x\end{align*}\\ \text{making use of the product rule} \text{At any arbitrary point }x=a\text{, we have}\\ \begin{align*}y&=a^2e^a+1\\ m_T=\frac{dy}{dx}&=2ae^a+a^2e^a\\ &=a(a+2)e^a\end{align*}\text{So by our point-gradient formula }y-y_1=m(x-x_1)\text{ we have}\\ \begin{align*}y-(a^2e^a+1)&=a(a+2)e^a(x-a)\end{align*}\\ \text{You may rearrange this however you wish. No form is tidy.} Title: Re: Mathematics Question Thread Post by: Vinhtran on September 28, 2016, 07:16:29 pm Didn't think about using gradient formula, thank you both Title: Re: Mathematics Question Thread Post by: samuels1999 on September 29, 2016, 09:21:25 pm Hi jamonwindeyer and jakeislove I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?) Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on September 29, 2016, 09:27:55 pm Hi jamonwindeyer and jakeislove I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?) Thanks Your textbook for mathematics doesn't matter at the end of the day, because you're going to be glued to past papers. Though with that out of the way, in general Cambridge is the recommended textbook. This is due to its wide range of coverage and its suitability in addressing all levels of student capability. It starts off nice and easy for people to start grasping the basics and still has a range of challenging problems for the very keen student. In general, I do advocate for it. But I didn't use it. I used Fitzpatrick for 2U and Maths in Focus (this textbook is actually strongly advised against under normal situations) for 3U. Note that every textbook has its pros and cons. An answer cannot be made definitively as it is entirely based off personal preference. Cambridge just happens to be the stronger option in general. With physics, Jacaranda and Physics in Focus are the two recommended. Not the place to be discussing it though. Jake's surname is "Silove" Title: Re: Mathematics Question Thread Post by: jakesilove on September 29, 2016, 09:29:03 pm Your textbook for mathematics doesn't matter at the end of the day, because you're going to be glued to past papers. Though with that out of the way, in general Cambridge is the recommended textbook. This is due to its wide range of coverage and its suitability in addressing all levels of student capability. It starts off nice and easy for people to start grasping the basics and still has a range of challenging problems for the very keen student. In general, I do advocate for it. But I didn't use it. I used Fitzpatrick for 2U and Maths in Focus (this textbook is actually strongly advised against under normal situations) for 3U. Note that every textbook has its pros and cons. An answer cannot be made definitively as it is entirely based off personal preference. Cambridge just happens to be the stronger option in general. With physics, Jacaranda and Physics in Focus are the two recommended. Not the place to be discussing it though. Jake's surname is "Silove" Agree completely with the above Title: Re: Mathematics Question Thread Post by: jamonwindeyer on September 29, 2016, 09:32:08 pm Hi jamonwindeyer and jakeislove I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?) Thanks To answer the question, Physics in Focus, and I used a very old textbook by Jones and Couchman for 3 Unit Mathematics (called, incidentally, 3 Unit Mathematics). In my opinion, actually better than Maths in Focus/Cambridge, based on the exposure I've had to the other two :) But yeah, really doesn't matter which you get, all have pros and cons :) :) Title: Re: Mathematics Question Thread Post by: pels on September 30, 2016, 11:44:46 am Hey all. Quick question here Evaluate lim θ -> 0 Of (sin2θ)/3θ Im not sure how to arrive at the final answer. Any help would be appreciated. cheers :^) Title: Mathematics Question Thread Post by: RuiAce on September 30, 2016, 11:58:54 am Hey all. Quick question here Evaluate lim θ -> 0 Of (sin2θ)/3θ Im not sure how to arrive at the final answer. Any help would be appreciated. cheers :^) \begin{align*}\lim_{\theta\to 0}\frac{\sin 2\theta}{3\theta}&=\frac{2}{3}\lim_{\theta\to 0}\frac{\sin 2\theta}{2\theta}\\&=\frac{2}{3}\times 1\\&=\frac32\end{align*} $\text{You need to be familiar with the result }\boxed{\lim_{x\to 0}\frac{\sin x}{x}=1}$ Title: Re: Mathematics Question Thread Post by: pels on September 30, 2016, 04:45:28 pm Can you explain how sin2θ/2θ = 1 Since sinx/x = 0, i still don't understand the above Title: Re: Mathematics Question Thread Post by: call.me.blueberry on September 30, 2016, 06:00:26 pm Can you explain how sin2θ/2θ = 1 Since sinx/x = 0, i still don't understand the above for the limit of sin(x)/(x), is x approaching 0? I check with wolframalpha the limit is =1 too: https://www.wolframalpha.com/input/?i=lim+x-%3E+0+sin(x)%2F(x) Title: Mathematics Question Thread Post by: RuiAce on September 30, 2016, 06:03:32 pm Can you explain how sin2θ/2θ = 1 Since sinx/x = 0, i still don't understand the above My apologies, I made a mistake in typing. That 0 is meant to be a 1. I will fix that now for the limit of sin(x)/(x), is x approaching 0? I check with wolframalpha the limit is =1 too: https://www.wolframalpha.com/input/?i=lim+x-%3E+0+sin(x)%2F(x) Yes. x goes to 0. Title: Re: Mathematics Question Thread Post by: lozil on September 30, 2016, 08:01:14 pm I'd appreciate if someone could help me with this one: Thanks! Title: Re: Mathematics Question Thread Post by: jakesilove on September 30, 2016, 09:15:11 pm I'd appreciate if someone could help me with this one: Thanks! Hey! So, essentially, we need to find a value for k such that the area under the curve from -6 to 4 equals the area under the curve from 4 to k (that way, the negative and positive areas cancel out! You can find the area under the first section in many ways; there's a formula, which I don't know, so I'll divide it into a triangle and a rectangle. The rectangle has area 10*2, and the triangle has area 10*(1/2)*3, so the total area will be 35 units squared. The area of the second section can be found in the same way; the rectangle will have area (k-4)(4) and the triangle will have area (1/2)(k-4)(2). So, the total area will be (k-4)(4)+(1/2)(k-4)(2). Now, we need these two areas to be the same, so $35=4(k-4)+(k-4)$ $35=5(k-4)$ $7=k-4$ $k=11$ As required. Jake Title: Re: Mathematics Question Thread Post by: lozil on September 30, 2016, 09:54:51 pm Hey! So, essentially, we need to find a value for k such that the area under the curve from -6 to 4 equals the area under the curve from 4 to k (that way, the negative and positive areas cancel out! You can find the area under the first section in many ways; there's a formula, which I don't know, so I'll divide it into a triangle and a rectangle. The rectangle has area 10*2, and the triangle has area 10*(1/2)*3, so the total area will be 35 units squared. The area of the second section can be found in the same way; the rectangle will have area (k-4)(4) and the triangle will have area (1/2)(k-4)(2). So, the total area will be (k-4)(4)+(1/2)(k-4)(2). Now, we need these two areas to be the same, so $35=4(k-4)+(k-4)$ $35=5(k-4)$ $7=k-4$ $k=11$ As required. Jake ok, thanks :) Title: Re: Mathematics Question Thread Post by: lozil on October 01, 2016, 12:28:20 pm Hi again :P So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks? Title: Re: Mathematics Question Thread Post by: RuiAce on October 01, 2016, 12:29:46 pm Hi again :P So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks? Yes that last step wouldn't be required. Still a rational denominator. Title: Re: Mathematics Question Thread Post by: lozil on October 01, 2016, 03:12:49 pm Yes that last step wouldn't be required. Still a rational denominator. ok, thanks! Title: Re: Mathematics Question Thread Post by: olivercutbill on October 03, 2016, 04:32:06 pm Hey guys, How does (d-6)^2 - 4 < 0 become 4 < d < 8 ? im confused with the individual steps Title: Re: Mathematics Question Thread Post by: RuiAce on October 03, 2016, 04:41:55 pm Hey guys, How does (d-6)^2 - 4 < 0 become 4 < d < 8 ? im confused with the individual steps \text{We are interested in a quadratic inequality here. The more straightforward approach is}\\ \begin{align*}&&(d-6)^2&< 4\\ &&-\sqrt{4} &< d-6 < \sqrt{4}&&\\ &&4 &< d < 8\end{align*} \text{Alternatively, expanding and refactorisation also suffices}\\ \begin{align*}d^2-12d+32&< 0\\ (d-4)(d-8)&< 0\\ 4&< d < 8\end{align*} $\text{For the second method, the last line }\textbf{IS NOT this mistake:}\\ d< 4\text{ and }d<8\\ \textbf{A quadratic inequality can NEVER be treated as though it was a linear inequality.}$ $\text{To observe why the second method works, consider }y=(d-4)(d-8)\\ \text{Sketch the graph of }y=(d-4)(d-8)$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-10-03%20at%204.40.03%20PM_zpsmotu2bgk.png) $\text{Because }y=(d-4)(d-8) \\ (d-4)(d-8)< 0\text{ corresponds to }y< 0\\ \text{From the graph, clearly this is when }4< d< 8$ Title: Re: Mathematics Question Thread Post by: jakesilove on October 03, 2016, 04:43:22 pm Hey guys, How does (d-6)^2 - 4 < 0 become 4 < d < 8 ? im confused with the individual steps Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps! $(d-6)^2-4<0$ Now, I usually work out inequalities by first solving an equation. As such, $(d-6)^2-4=0$ $(d-6)^2=4$ $d-6=\pm 2$ $d=8, d=4$ So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5. $(5-6)^2-4=(1)^2-4=-3<0$ Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is $4 < d < 8$ As required! Jake Title: Re: Mathematics Question Thread Post by: kevin217 on October 03, 2016, 04:45:09 pm Hey guys, How does (d-6)^2 - 4 < 0 become 4 < d < 8 ? im confused with the individual steps Rearrange the equation to get: (d-6)2 < 4 (d-6) < 2 or (d-6) > - 2 d < 8 or d > 4 therefore 4 < d < 8 I hope this helped Title: Re: Mathematics Question Thread Post by: Vinhtran on October 03, 2016, 09:29:54 pm Please help me on this question. So I managed to get the mean and standard deviation correct, but i didn't understand why they made certain choices. The probability of a seventeen year old boy having a part-time job while at school is 0.4 In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job. The answer key says the answer is 0.07 Title: Re: Mathematics Question Thread Post by: RuiAce on October 03, 2016, 09:43:12 pm Please help me on this question. So I managed to get the mean and standard deviation correct, but i didn't understand why they made certain choices. The probability of a seventeen year old boy having a part-time job while at school is 0.4 In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job. The answer key says the answer is 0.07 Statistics is not in the HSC mathematics course. $\text{Let }X\text{ be the number of 17 year old boys out of 80 with part time jobs}\\ \text{Then, }X \sim \mathcal{B}(80,0.4)\\ \text{i.e. }X\text{ follows a binomial distribution of sample size 80 with probability }0.4$ \text{Then we simply perform a ton of computations}\\ \begin{align*}&\> Pr(20 \le X \le 25) \\&=Pr(X=20)+Pr(X=21)+Pr(X=22)+Pr(X=23)+Pr(X=24)+Pr(X=25)\\ &= \binom{80}{20}0.6^{60}0.4^{20}+\binom{80}{21}0.6^{59}0.4^{21}+\binom{80}{22}0.6^{58}0.4^{22}+\binom{80}{23}0.6^{57}0.4^{23}+\binom{80}{24}0.6^{56}0.4^{24}+\binom{80}{25}0.6^{55}0.4^{25}\\ &= 0.06580277\dots \\ &\approx 0.07\end{align*} Title: Re: Mathematics Question Thread Post by: olivercutbill on October 04, 2016, 09:59:45 pm Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps! $(d-6)^2-4<0$ Now, I usually work out inequalities by first solving an equation. As such, $(d-6)^2-4=0$ $(d-6)^2=4$ $d-6=\pm 2$ $d=8, d=4$ So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5. $(5-6)^2-4=(1)^2-4=-3<0$ Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is $4 < d < 8$ As required! Jake This makes more sense to me than the solution posted above this. How are quadratic inequalities different from linear ones? In the first method posted by RuiAce, I don't understand the movement from the first line to the second. Thanks guys! Title: Re: Mathematics Question Thread Post by: RuiAce on October 04, 2016, 10:10:22 pm Ignore method 1 if it makes no sense and focus on method 2. Because it is the exact same as Jake's method ________________________ $\text{Although if you really need to know how method 1 works, the idea is that}\\ (d-6)^2 < 4\\ \text{becomes }\\ |d-6| < 2\\ \text{upon square rooting.}$ $\text{Then the absolute value inequality is evaluated.}\\ -2 < d-6 < 2$ $\textbf{An extremely common misconception}\text{ however not usually problematic in 2U.}\\ \text{People tend to think that }\sqrt{x^2} = x\\ \text{False. }\sqrt{x^2}=|x|$ ________________________ Quadratic inequalities (nothing's an identity here) are different from linear ones in that for starters, parabolas have a turning point. This messes things up. A linear function is just a straight line. Secondly, every quadratic function has two roots. (You just don't get taught that in 2U because you don't know what complex numbers are) Title: Re: Mathematics Question Thread Post by: IkeaandOfficeworks on October 04, 2016, 10:30:12 pm Hi guys! I'm a bit confused with this question: A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the rectangle and the square so that the total enclosed area is a minimum. I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 04, 2016, 10:48:45 pm Hi guys! I'm a bit confused with this question: A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the rectangle and the square so that the total enclosed area is a minimum. I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you :) $\text{Let the breadth of the triangle be }\ell\\ \text{Then, from the 3:1 ratio, the length of the triangle is }3\ell$ $\text{Additionally, let the length of the square be }x$ $\text{We know that all of the lengths were made from a 50cm wire.}\\ \text{This means, if we add the lengths back up, we'd get 50.}\\ \text{i.e. }50=\underbrace{6\ell+2\ell}_{\text{perimeter of rectangle}}+\underbrace{4x}_{\text{square}}$ $\text{So we have }\frac{25}{2}=2\ell+x\\ \text{Now we need to consider the area.}$ $\text{The area of the rectangle is }3\ell^2\\ \text{The area of the square is }x^2\\ \text{Hence the total area is given by }A=3\ell^2+x^2$ $\text{Substituting in the above result we get }\\ A=3\ell^2+\left(\frac{25}{2}-2\ell\right)^2\\ \text{This is now your regular max/min question, so expand, compute }\frac{dA}{d\ell}\text{ and carry through.}$ Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 05, 2016, 01:21:24 am A particle is moving in a straight life, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m s^-1. the displacement is given by x=2t - 3ln(t+1). Find the distance travelled by the particle in the first 2 seconds. (HSC Paper 2000). Can someone please help me out Title: Re: Mathematics Question Thread Post by: RuiAce on October 05, 2016, 08:40:22 am A particle is moving in a straight life, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m s^-1. the displacement is given by x=2t - 3ln(t+1). Find the distance travelled by the particle in the first 2 seconds. (HSC Paper 2000). Can someone please help me out $\text{We are interested in the }\textbf{distance travelled}\text{ by the particle}\\ \text{Not its }\textbf{displacement}\text{ after 2 seconds.}$ $\text{We need to address the fact that the particle may or may not turn around}\\ \text{within the first two seconds.}\\ v=2-\frac{3}{t+1}\\$ $\text{Solving }v=0\text{ gives }t=\frac{1}{2}$ $\text{By the second derivative test, }\ddot{x}=\frac{3}{(t+1)^2}> 0\text{ always.}\\ \text{Hence at }t=\frac{1}{2}\text{ there is a local minima.}\\ \text{So the particle DOES turn around at }t=\frac{1}{2}$ $\text{So to find the DISTANCE travelled, we must split up the cases.}\\ \text{Recall: }x=2t-3\ln (t+1) \\ \text{In the first two seconds, we have}\\ \text{Case 1: }0 < t < \frac12\\ d=|x_{t=\frac12}-x_{t=0}|=3\ln \frac{3}{2}-1\\ \text{Case 2: }\frac12 < t < 2\\ d=x_{t=2}-x_{t=\frac12}=(4-3\ln 3)-\left(1-3\ln \frac{3}{2}\right)=3-3\ln 2$ $\text{So the total distance travelled is}\\ \left(3\ln \frac{3}{2} - 1\right) + \left(3 - 3\ln 2\right) = 2 - 3 \ln \frac{4}{3}$ I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future. Title: Re: Mathematics Question Thread Post by: lha on October 05, 2016, 08:52:37 am What has been the main topic that they have assessed from the prelim course in the hsc exam? Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 05, 2016, 09:56:34 am What has been the main topic that they have assessed from the prelim course in the hsc exam? From what ive seen, locus & parabola. Also, just basic algebra and quadratics too. Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 05, 2016, 09:58:01 am $\text{We are interested in the }\textbf{distance travelled}\text{ by the particle}\\ \text{Not its }\textbf{displacement}\text{ after 2 seconds.}$ $\text{We need to address the fact that the particle may or may not turn around}\\ \text{within the first two seconds.}\\ v=2-\frac{3}{t+1}\\$ $\text{Solving }v=0\text{ gives }t=\frac{1}{2}$ $\text{By the second derivative test, }\ddot{x}=\frac{3}{(t+1)^2}> 0\text{ always.}\\ \text{Hence at }t=\frac{1}{2}\text{ there is a local minima.}\\ \text{So the particle DOES turn around at }t=\frac{1}{2}$ $\text{So to find the DISTANCE travelled, we must split up the cases.}\\ \text{Recall: }x=2t-3\ln (t+1) \\ \text{In the first two seconds, we have}\\ \text{Case 1: }0 < t < \frac12\\ d=|x_{t=\frac12}-x_{t=0}|=3\ln \frac{3}{2}-1\\ \text{Case 2: }\frac12 < t < 2\\ d=x_{t=2}-x_{t=\frac12}=(4-3\ln 3)-\left(1-3\ln \frac{3}{2}\right)=3-3\ln 2$ $\text{So the total distance travelled is}\\ \left(3\ln \frac{3}{2} - 1\right) + \left(3 - 3\ln 2\right) = 2 - 3 \ln \frac{4}{3}$ I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future. Oh yes sorry! It was a typo. Thank you! Title: Re: Mathematics Question Thread Post by: lha on October 05, 2016, 10:04:44 am From what ive seen, locus & parabola. Also, just basic algebra and quadratics too. Alright thank you! Do you know anywhere that I can find a whole bunch of locus questions from past papers (other than going through every past paper)? Title: Re: Mathematics Question Thread Post by: jakesilove on October 05, 2016, 11:10:50 am Alright thank you! Do you know anywhere that I can find a whole bunch of locus questions from past papers (other than going through every past paper)? Purchase a past paper book that goes by the dotpoint. Other than that, you'll have have to scroll through papers! (Which honestly won't take too long) :) Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 05, 2016, 11:33:16 am $\text{We are interested in the }\textbf{distance travelled}\text{ by the particle}\\ \text{Not its }\textbf{displacement}\text{ after 2 seconds.}$ $\text{We need to address the fact that the particle may or may not turn around}\\ \text{within the first two seconds.}\\ v=2-\frac{3}{t+1}\\$ $\text{Solving }v=0\text{ gives }t=\frac{1}{2}$ $\text{By the second derivative test, }\ddot{x}=\frac{3}{(t+1)^2}> 0\text{ always.}\\ \text{Hence at }t=\frac{1}{2}\text{ there is a local minima.}\\ \text{So the particle DOES turn around at }t=\frac{1}{2}$ $\text{So to find the DISTANCE travelled, we must split up the cases.}\\ \text{Recall: }x=2t-3\ln (t+1) \\ \text{In the first two seconds, we have}\\ \text{Case 1: }0 < t < \frac12\\ d=|x_{t=\frac12}-x_{t=0}|=3\ln \frac{3}{2}-1\\ \text{Case 2: }\frac12 < t < 2\\ d=x_{t=2}-x_{t=\frac12}=(4-3\ln 3)-\left(1-3\ln \frac{3}{2}\right)=3-3\ln 2$ $\text{So the total distance travelled is}\\ \left(3\ln \frac{3}{2} - 1\right) + \left(3 - 3\ln 2\right) = 2 - 3 \ln \frac{4}{3}$ I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future. in regards to this question, for case one, it is t(1/2) - t(0) & you got 3ln(3/2) - 1. how did you get that because im getting 1 - 3ln(3/2). Title: Re: Mathematics Question Thread Post by: RuiAce on October 05, 2016, 12:06:24 pm in regards to this question, for case one, it is t(1/2) - t(0) & you got 3ln(3/2) - 1. how did you get that because im getting 1 - 3ln(3/2). The absolute value brackets are there for a reason. Distance is positive, not negative Title: Re: Mathematics Question Thread Post by: Daliaradosevic on October 05, 2016, 07:59:56 pm Hey guys so i was just curious as to how to study for the finance (money) area of maths because i am terrible at understanding the question and answering them and i find that theres alot of finance based questions in the exams so please any advice would be grateful!! :) also i do so so so many practice questions and i still struggle in understanding them! Title: Re: Mathematics Question Thread Post by: RuiAce on October 05, 2016, 08:17:25 pm Hey guys so i was just curious as to how to study for the finance (money) area of maths because i am terrible at understanding the question and answering them and i find that theres alot of finance based questions in the exams so please any advice would be grateful!! :) also i do so so so many practice questions and i still struggle in understanding them! Effectively, every finance question is just the same thing in different words. 1. We always establish our interest rate first. 2. We start writing our recursive relationship. $\text{Consider an example with }r=0.05 p.a.\text{ and }500\text{ is paid back at the end of each year.}\\ \text{So if you don't understand what is going on you need to figure that out first.}$ $\text{Let }A_n\text{ be the amount owing after }n\text{ years}\\ \text{Let }P\text{ be the amount borrowed.}$ $\text{The money gains interest over the first year. And THEN a bundle of 500 gets paid off.}\\ \text{So we have }A_1=P(1.05)-500$ $\text{Whatever is still owing at the end of the first year carries forth into the second year.}\\ \text{Once again, it gains interest, and THEN a bundle of 500 gets paid off.}\\ A_2=A_1(1.05)-500\\ \text{But by just substituting in what we have earlier}\\ A_2=P(1.05)^2-500(1.05)-500$ $\text{Same thing in the third year: }A_3=A_2(1.05)-500$ 3. Make this go on for n years. There is ALWAYS a geometric progression that forms in these types of questions. \begin{align*}A_n&=P(1.05)^n-500-500(1.05)-500(1.05)^2-\dots-500(1.05)^{n-1}\\ &= P(1.05)^n-500(1+1.05+1.05^2+\dots+1.05^{n-1})\end{align*} 4. Use the GP formula Then whatever last is always a matter of understanding the wording. You will need to provide examples here, clearly indicating where you are stuck in the wording. And of course, they can hide the yearly/monthly/... repayment from you, the number of years from you or the interest rate from you. The process is still the exact same regardless of which is chosen. Title: Re: Mathematics Question Thread Post by: IkeaandOfficeworks on October 06, 2016, 12:55:24 pm Hey guys! I'm having trouble with this question. Thank you! :D Two dice with their faces numbered 1 to 6 are tossed. i. By listing the possible outcomes find the probability of a) one die showing a 6 with other showing at least 4 b) one die showing a 6 with the other showing 3 or less ii) Deduce from the results in part (i) the probability that neither die show s a 6. Title: Re: Mathematics Question Thread Post by: RuiAce on October 06, 2016, 01:05:39 pm Hey guys! I'm having trouble with this question. Thank you! :D Two dice with their faces numbered 1 to 6 are tossed. i. By listing the possible outcomes find the probability of a) one die showing a 6 with other showing at least 4 b) one die showing a 6 with the other showing 3 or less ii) Deduce from the results in part (i) the probability that neither die show s a 6. The question asks to list the possibilities, hence you should. However I'll structure my answers as though they weren't listed Let the rows correspond to the first dice, and the second correspond to the second dice. (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpso2y3tyak.png) The total possible outcomes is always 36 a) Suppose the first dice shows a 6. Then the second die shows either 4, 5 or 6. There are 3 possible choices (going down column 6) Suppose the second dice shows a 6. Then the first die shows either 4, 5 or 6. There are 3 possible choices (going across row 6) (6,6) is counted twice, so we deduct it once. Therefore there are 5 favourable outcomes, so the answer is 5/36 b) Suppose again the first dice shows a 6. Then the second shows either 1, 2 or 3 instead. There are 3 possible choices And similarly for the second dice However here, nothing gets counted twice. Therefore there are 6 favourable outcomes, so the answer is 6/36 = 1/6 For part ii), note that the sum of the above probabilities is 11/36. Notice that in the above parts, we've catered for all the cases where 6 must appear at least once. Hence the probability 6 doesn't appear at all is just the complement. This has a probability of 25/36 This can also be deduced from reading off the table, as we are basically asking for the probability that something not in row 6 and/or column 6 is chosen. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 06, 2016, 01:20:36 pm Guys I have no Idea how to answer this question at all. Its question 15 b) from the 2015 paper. (http://i63.tinypic.com/106mzac.png) I would say for: i) (triangle)ACB = (triangle)DCF AC=DC (Given) (angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given) - SAS somehow?? I dont know what else... i Title: Re: Mathematics Question Thread Post by: RuiAce on October 06, 2016, 01:28:15 pm Guys I have no Idea how to answer this question at all. Its question 15 b) from the 2015 paper. (http://i63.tinypic.com/106mzac.png) I would say for: i) (triangle)ACB = (triangle)DCF AC=DC (Given) (angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given) - SAS somehow?? I dont know what else... i $\text{Part (i) asks you to prove }\textbf{similarity.}\\ \text{Not }\textbf{congruence}.$ \text{There are an immediate pair of vertically opposite angles.}\\ \begin{align*}\angle FDC&=\angle ADE\tag{vert. opp.}\\ &= \angle EAD\tag{isos. triangle AED}\\ &= \angle BAC \end{align*} \text{So in }\triangle DCF\text{ and }\triangle ACB\\ \begin{align*}\angle FDC&=\angle BAC\tag{from above}\\ \angle DCF&=\angle ACB \tag{given}\\ \therefore \triangle DCF &||| \triangle ACD\tag{equiangular}\end{align*} AC is not equal to DC either. AD is equal to DC. And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 06, 2016, 11:22:25 pm $\text{Part (i) asks you to prove }\textbf{similarity.}\\ \text{Not }\textbf{congruence}.$ \text{There are an immediate pair of vertically opposite angles.}\\ \begin{align*}\angle FDC&=\angle ADE\tag{vert. opp.}\\ &= \angle EAD\tag{isos. triangle AED}\\ &= \angle BAC \end{align*} \text{So in }\triangle DCF\text{ and }\triangle ACB\\ \begin{align*}\angle FDC&=\angle BAC\tag{from above}\\ \angle DCF&=\angle ACB \tag{given}\\ \therefore \triangle DCF &||| \triangle ACD\tag{equiangular}\end{align*} AC is not equal to DC either. AD is equal to DC. And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it. Thanks So MUCH!!! Title: Re: Mathematics Question Thread Post by: BPunjabi on October 07, 2016, 12:05:26 am $\text{Part (i) asks you to prove }\textbf{similarity.}\\ \text{Not }\textbf{congruence}.$ \text{There are an immediate pair of vertically opposite angles.}\\ \begin{align*}\angle FDC&=\angle ADE\tag{vert. opp.}\\ &= \angle EAD\tag{isos. triangle AED}\\ &= \angle BAC \end{align*} \text{So in }\triangle DCF\text{ and }\triangle ACB\\ \begin{align*}\angle FDC&=\angle BAC\tag{from above}\\ \angle DCF&=\angle ACB \tag{given}\\ \therefore \triangle DCF &||| \triangle ACD\tag{equiangular}\end{align*} AC is not equal to DC either. AD is equal to DC. And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it. RUI im still lost... can you help out please! Title: Re: Mathematics Question Thread Post by: RuiAce on October 07, 2016, 12:19:07 am Guys I have no Idea how to answer this question at all. Its question 15 b) from the 2015 paper. (http://i63.tinypic.com/106mzac.png) I would say for: i) (triangle)ACB = (triangle)DCF AC=DC (Given) (angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given) - SAS somehow?? I dont know what else... i $\text{For part ii), from the pair of similar triangles}\\ \angle DFC = \angle ABC\\ \text{as they are matching angles on similar triangles.}$ $\text{This means }\angle EFB = \angle EBF\\ \text{Hence }\triangle EFB\text{ is isosceles as it has a pair of angles equal.}$ $\text{iii) gets a bit trickier.}\\ \text{It should make complete sense to use the triangle we just proven was isosceles:}\\ EB=EF$ $\text{Notice that }EF\text{ can be split into }ED+DF\\ \text{So }EB=ED+DF\\ \text{But then we have }\textit{another}\text{ pair of sides equal: }AE=ED\\ \text{Hence }\boxed{EB=AE+DF}$ $\text{Look at what we're trying to prove. }EB=3AE\\ \text{So somehow }DF = 2AE??$ $\text{Go back to part (i). You showed that }\triangle DCF ||| \triangle ACD\\ \text{Hence, using }\textit{proportional sides on similar triangles.}\\ \frac{DC}{AC}=\frac{DF}{AB}\\ \text{But }\frac{DC}{AC}\text{ is just }\frac{1}{2}$ $\text{So }\frac{DF}{AB}=\frac{1}{2}\iff 2DF = AB\\ \text{But then }\boxed{AB=AE+EB}\\ \text{So }\boxed{2DF = AE + EB}$ $\text{So sub the first box into the third:}\\ 2(EB-AE)=AE+EB \iff \boxed{EB = 3AE}$ $\textit{What to take out: Sometimes you have to split sides up into SMALLEr sides.}$ Title: Re: Mathematics Question Thread Post by: onepunchboy on October 07, 2016, 11:11:26 pm (http://uploads.tapatalk-cdn.com/20161007/58acb99cb53b5093233aba2f8a357806.jpg) Can someone help me with 15) ii. Im not sure about it Title: Re: Mathematics Question Thread Post by: jakesilove on October 07, 2016, 11:44:16 pm (http://uploads.tapatalk-cdn.com/20161007/58acb99cb53b5093233aba2f8a357806.jpg) Can someone help me with 15) ii. Im not sure about it Which ii) did you need help with? There are two in the image you've posted Title: Re: Mathematics Question Thread Post by: RuiAce on October 07, 2016, 11:45:31 pm (http://uploads.tapatalk-cdn.com/20161007/58acb99cb53b5093233aba2f8a357806.jpg) Can someone help me with 15) ii. Im not sure about it $\text{This question from my HSC is simply too hard without using a diagram.}\\\text{Use your sketch of }y=|2x+3|\text{ as a visual aid}\\ \text{Note that }y=mx+1\text{ is a line passing through }(0,1)\text{ with gradient }m$ $\text{There are three possibilities.}\\ \text{Case 1: The graph slants too far to the left and never touches it.}\\ \text{This happens when }m < -2$ $\text{Case 2: The graph slants too far to the right and no longer touches it twice.}\\ \text{This happens when }m \ge 2$ Essentially, 2 is the special number here, because $$y=|2x+3|$$ has gradient either 2 or -2. (Which depends on if $$x < -\frac32$$ or if $$x >\frac32$$, but that's not too important.) $\text{Case 3 (special case): What if the graphs intersect on the }x\text{-axis}\\ \text{This is a special case because }\left(\frac{3}{2},0\right)\text{ is the one critical point on }y=|2x-3|\\ \text{This occurs when }m=-\frac{2}{3}$ (Note: The critical point here is basically the corner of the absolute value function.) $\text{And they are all your cases.}$ BOSTES solutions has a diagram. GeoGebra simulation Edit: A remark for case 3 - the value of $$m = -\frac23$$ was computed by essentially plugging the points $$(0,1)$$ and $$\left( \frac32, 0\right)$$ into the usual gradient formula $$m = \frac{y_2-y_1}{x_2-x_1}$$ Which ii) did you need help with? There are two in the image you've posted I just noticed that lol Title: Re: Mathematics Question Thread Post by: onepunchboy on October 08, 2016, 12:11:04 am oh yea sorry about that, it was the graphing question lol. Thanks rui this really helped Title: Re: Mathematics Question Thread Post by: nibblez16 on October 08, 2016, 02:29:42 pm Hello, I am sort of having trouble with geometry, for similar triangles etc. How many proofs do we need for each?? ??? Title: Re: Mathematics Question Thread Post by: RuiAce on October 08, 2016, 02:32:39 pm Hello, I am sort of having trouble with geometry, for similar triangles etc. How many proofs do we need for each?? ??? What do you mean by "each"? You only prove two triangles are similar at a time, not three or four at a time. Instead, there are three tests that you need to know: - Equiangular: Two angles in both triangles are equal - All three sides in proprotion - Two sides in proportion, included angle equal (Note: There is a similar one for RHS but it is not needed) The similarity tests can be matched up with a congruence test Equiangular VS AAS 3 sides in prop VS SSS 2 sides in prop + inc. angle VS SAS Title: Re: Mathematics Question Thread Post by: nibblez16 on October 08, 2016, 02:36:31 pm What do you mean by "each"? You only prove two triangles are similar at a time, not three or four at a time. Instead, there are three tests that you need to know: - Equiangular: Two angles in both triangles are equal - All three sides in proprotion - Two sides in proportion, included angle equal (Note: There is a similar one for RHS but it is not needed) The similarity tests can be matched up with a congruence test Equiangular VS AAS 3 sides in prop VS SSS 2 sides in prop + inc. angle VS SAS Okay Thanks! What I meant by 'how many proofs for each' was similar triangles, congruent triangles, alternate etc :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 08, 2016, 02:40:26 pm Okay Thanks! What I meant by 'how many proofs for each' was similar triangles, congruent triangles, alternate etc :) These can be found in any textbook. They stem down from properties of the geometric figure we're interested in. Parallel lines have corresponding angles equal, alternate angles equal and cointerior angles corresponding, Hence, if one of the following are true: 1. If alternate angles equal 2. If corresponding angles equal 3. If cointerior angles add to 180deg Then they are parallel Congruent triangles look the exact same, so they must have the exact same sides and angles. Therefore 1. SSS works because you just showed all 3 sides match up 2. SAS works because imagine you have two lines equal in size. Join them together. The only thing you can change is the angle. If you fix that angle, you're stuck. 3. AAS works because imagine you have two angles. If two angles are equal, the third is equal because a triangle only has three angles. So what can you do? Make the triangle larger or smaller. If you fix a side, you end up fixing all three sides. 4. RHS - a special case, and only works because right angled triangles are nice Title: Re: Mathematics Question Thread Post by: :3 on October 08, 2016, 07:43:42 pm Hey Mathematicians, Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)? I often struggle to see the connections between what has been given and what is being asked. Much appreciated. :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 08, 2016, 08:37:44 pm Hey Mathematicians, Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)? I often struggle to see the connections between what has been given and what is being asked. Much appreciated. :P $\text{If it were that obvious, no way would it be Q10 material.}\\ \text{For starters, copy or trace the diagram so you can annotate it later on.}$ $\text{Find the side }QL\\ \text{Note that we want to look for }QL^2\text{. This immediately says }\textit{Pythagoras.}$ $\text{Since }PQ=12\text{ which you should've shown on your diagram}\\ \text{Clearly }KQ=12-(6+x)=6-x\\ \text{This sort of splitting sides apart is normal.}$ $\text{But here's where you need to read the question. What is }KL?\\ \text{The paper was }\textbf{folded}.\text{ Try folding your exam booklet.}\\\text{You will then see that }KL = PK = 6+x$ $\text{Now by Pythagoras' Theorem and perfect squares (or really clever diff. of 2 squares)}\\ QL^2=KL^2-KQ^2=(6+x)^2-(6-x)^2=24x$ ____________________________________________ $\text{The next part }\textbf{explicitly}\text{ says to prove similarity. So we will do so.}\\ \text{Find }\triangle QKL\text{ and }\triangle KLM.\\ \text{Perfect. They obviously both have a right angle. What else could be useful?}$ $\text{It looks too hard to prove two sides in proportion. So we will try another angle.}\\ \text{Let }\angle KLQ=\theta\\ \text{But then }\angle LMN=\theta\text{ as well!}$ $\textit{Do you see why?}\text{ Notice that when angles are literally side by side,}\\ \text{and ESPECIALLY given }\angle NLQ=180^\circ\\ \text{We should consider how angles add up.}\\ \text{(Alternatively just consider only the exterior angle of a triangle if you're good enough.)}$ $\text{Then we use angle sum of }\triangle NLQ\text{ and that is it.}\\ \text{You can tidy it up to prove similarity now via the equiangular test.}$ ____________________________________________ $\text{Now we just proved two triangles are similar, and we need to show something related to side lengths.}\\ \text{What would be common sense here? Proportional sides on similar triangles.}\\ \text{Note that }y\text{ must make its way in. This hints at }LM\text{ and consequently }LK.$ $\text{Now notice the peculiar square root in }\sqrt{x}\\ \text{But we just proved that }QL^2=24x\text{, so }QL=\sqrt{24x}=2\sqrt{6x}\\ \text{Well, that just gave us all we needed right?}$ \text{By proportional sides on similar triangles:}\\ \begin{align*}\frac{LM}{KL}&=\frac{NM}{QL}\\ \frac{y}{6+x}&=\frac{12}{2\sqrt{6x}}\\ y&=\frac{6(6+x)}{\sqrt{6}\sqrt{x}}\\ &=\frac{\sqrt{6}(6+x)}{\sqrt{x}}\end{align*} I will leave you to do the rest of the question. Answers may be found here. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 08, 2016, 11:40:46 pm Hey Mathematicians, Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)? I often struggle to see the connections between what has been given and what is being asked. Much appreciated. :P To add to Rui's answer, here are a few of my tips for Geometry questions: - Leave them until last. So this is more of a general thing, but if you struggle with a question type you need to leave it until last. Get the easy marks first! - Use some scrap paper or a doodling box or something to write everything you know, and everything you can find out about the situation. Don't try and find the solution, try and find everything. See if any of this helps you spot a solution. - Work backwards! If you need to prove a result, see if you can reverse engineer it to figure out what path you would need to take. If you need similarity, what angles do you need? What do you need to get those angles? Can you get that info? - When in doubt, assign pro numerals! It can help make the problem more logical/algebraic :) :) And practice, because I have to say it ;) Title: Re: Mathematics Question Thread Post by: Fadwa9 on October 09, 2016, 10:21:01 pm Hi, Can you please explain how the the solution to question 14 (d) is obtained and how exactly the graph is interpreted [2013 HSC] http://boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf Title: Re: Mathematics Question Thread Post by: RuiAce on October 09, 2016, 10:38:11 pm Hi, Can you please explain how the the solution to question 14 (d) is obtained and how exactly the graph is interpreted [2013 HSC] http://boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf $\text{Recall that the integral is just a measure of the area under the curve.}\\ \text{We use the geometrical interpretation to determine what the value of }a\text{ is.}\\ \text{Note that the boundaries were chosen so that one is the negative of the other.}$ \begin{align*}\int_{-1}^1f(x)dx&=2\\ \int_{-2}^2f(x)dx&= 3.5\\ \int_{-3}^3f(x)dx&=3\tag{3}\\ \int_{-4}^4f(x)dx&=1\\ \int_{-5}^5f(x)dx &= -1\\ \int_{-4.5}^{4.5}f(x)dx&=0\end{align*} $\text{The important observations that had to be made were:}\\ \int_{-3}^{-1}f(x)dx=0\text{ and }\int_2^3f(x)dx=0\\ \text{and that }\int_{-1}^2f(x)dx=3\\ \text{If you could make these observations}\\ \text{You could've skipped the first two lines of working out and started on line (3)}$ Material added later on $\text{Perhaps, a more formal way of approaching the question is this.}\\ \text{From the diagram, }\int_{-3}^{-1} f(x)\,dx = 0\text{ and }\int_2^3 f(x)\,dx = 0.$ $\text{Also, noting that the region is a rectangle,}\\ \int_{-1}^2 f(x)\,dx = 3\times 1 = 3.\\ \text{So we first gather that}\\ \int_{-3}^3 f(x)\,dx = \int_{-3}^{-1}f(x)\,dx + \int_{-1}^2 f(x)\,dx + \int_2^3 f(x)\,dx = 3$ \text{Therefore, for any }a \ge 3,\\ \begin{align*}\int_{-a}^a f(x)\,dx &= \int_{-a}^{-3} f(x)\,dx + \int_{-3}^3 f(x)\,dx + \int_{3}^a f(x)\,dx\\ &= -(a-3) + 3 - (a-3)\\ &= 9 - 2a \end{align*} having obtained the $$-(a-3)$$ by again using the formula for the area of a rectangle. $\text{So effectively speaking, we're just solving the equation}\\ 9-2a = 0.\\ \text{This gives }a = 4.5$ Title: Re: Mathematics Question Thread Post by: isaacdelatorre on October 10, 2016, 01:18:42 pm Hey guys, Could someone please help me with this question? I'm not sure where to start and have been blankly staring at it for a while now. Thanks in advance :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 10, 2016, 01:31:56 pm Hey guys, Could someone please help me with this question? I'm not sure where to start and have been blankly staring at it for a while now. Thanks in advance :) $\textit{Observe that this question says to EXPLAIN.}$ $\text{1}{4}\text{ is representative of 15 minutes according to the question.}\\ \text{Note that if Xuan or Yvette have waited for more than 15 minutes, they will leave.}\\ \text{Hence, the times that they arrive must not be further apart by more than 15 minutes.}$ $\text{The time taken for Xuan is represented by }x\\ \text{If Yvette arrives 15 minutes later, Xuan will go.}\\ \text{So }x-y \le \frac{1}{4}\\ \text{Similarly, if Yvette was earlier we'd have }y-x\le \frac{1}{4}$ ____________________________ $\text{This is simply too hard to visualise.}\\ \text{On your graph, sketch }x-y\le \frac14 \iff y\ge x-\frac14\\ \text{and also sketch }y-x\le \frac{1}{4}\iff y \le x+\frac{1}{4}$(http://i.imgur.com/DtqrOie.png) $\text{Observe that the brown area is reflective of the probability we need.}\\ \text{It is the region bound by the square and the two regions }x-y\le\frac14, y-x\le \frac14$ $\text{To determine the area, we can split the hexagon into two triangles and a rectangle.}\\ \text{The two triangles each have area }A=\frac{1}{2}\times \frac14\times \frac14\\ \text{The rectangle's area is harder.}$ $\text{For the rectangle, by either Pythagoras' theorem or trigonometry}\\ \ell = 0.75\sqrt{2}\\ b=0.25\sqrt{2}\\ \text{So the area is }2\times 0.75\times 0.25\\ \text{Add everything up, and you have your answer.}$ I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have. Title: Re: Mathematics Question Thread Post by: isaacdelatorre on October 10, 2016, 02:42:55 pm $\textit{Observe that this question says to EXPLAIN.}$ $\text{1}{4}\text{ is representative of 15 minutes according to the question.}\\ \text{Note that if Xuan or Yvette have waited for more than 15 minutes, they will leave.}\\ \text{Hence, the times that they arrive must not be further apart by more than 15 minutes.}$ $\text{The time taken for Xuan is represented by }x\\ \text{If Yvette arrives 15 minutes later, Xuan will go.}\\ \text{So }x-y \le \frac{1}{4}\\ \text{Similarly, if Yvette was earlier we'd have }y-x\le \frac{1}{4}$ ____________________________ $\text{This is simply too hard to visualise.}\\ \text{On your graph, sketch }x-y\le \frac14 \iff y\ge x-\frac14\\ \text{and also sketch }y-x\le \frac{1}{4}\iff y \le x+\frac{1}{4}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-10-10%20at%201.29.11%20PM_zpsnz6vxwit.png) $\text{Observe that the brown area is reflective of the probability we need.}\\ \text{It is the region bound by the square and the two regions }x-y\le\frac14, y-x\le \frac14$ $\text{To determine the area, we can split the hexagon into two triangles and a rectangle.}\\ \text{The two triangles each have area }A=\frac{1}{2}\times \frac14\times \frac14\\ \text{The rectangle's area is harder.}$ $\text{For the rectangle, by either Pythagoras' theorem or trigonometry}\\ \ell = 0.75\sqrt{2}\\ b=0.25\sqrt{2}\\ \text{So the area is }2\times 0.75\times 0.25\\ \text{Add everything up, and you have your answer.}$ I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have. Awesome!! Thanks for the help Rui. It took a few tries... and apparently a whole hour; but I got it in the end. Thanks :) Title: Re: Mathematics Question Thread Post by: asd987 on October 10, 2016, 10:10:56 pm Hi, I'm can some please complete this question for me. find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve Title: Re: Mathematics Question Thread Post by: jakesilove on October 10, 2016, 10:18:21 pm Hi, I'm can some please complete this question for me. find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve Hey! So, we can expand the graph out to $y=x^3-12x-16$ First, we find the derivative. $y'=3x^2-12$ We set it equal to zero, to find stationary point (where the gradient is zero) $3x^2-12=0$ $x=2, -2$ We find the second derivative to determine the nature of the stationary points. $y''=6x$ $x=2, y''=12>0$ Therefore, x=2 is a min $y''=6x$ $x=-2, y''=-12<0$ Therefore, x=-2 is a max. We know the x-intercepts (from the original graph, at x=4, and x=2). We can sub in our turning points to find y values. Finally, we know that it is a 'positive' cubic, is we get +x^3. Therefore, the graph will look like (http://i.imgur.com/o3wyjDH.png) Let me know if I can clarify anything! Title: Re: Mathematics Question Thread Post by: asd987 on October 10, 2016, 11:03:06 pm TYVM i understand. i just didn't know you had to expand the equation of the graph :-[ Title: Re: Mathematics Question Thread Post by: RuiAce on October 10, 2016, 11:07:27 pm TYVM i understand. i just didn't know you had to expand the equation of the graph :-[ $\text{It can be commenced without the expansion}\\ \text{The product rule and chain rule are both needed to differentiate.}\\ \frac{dy}{dx}=2(x-4)(x+2)+(x+2)^2$ $\text{Then to solve }\frac{dy}{dx}=0\\ \text{Factorise out }(x-2):\\ (x+2)[2(x-4)+(x+2)]=0 \iff (x-2)(3x-6)=0\\ \text{Which still solves to give }x=\pm 2$ Title: Re: Mathematics Question Thread Post by: anotherworld2b on October 10, 2016, 11:50:32 pm I was wondering how do we simply imaginary numbers? We just started doing them but im already confused. I was wondering if i could get help understanding how to graph cubics and tan graphs please Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:07:18 am I was wondering how do we simply imaginary numbers? We just started doing them but im already confused. I was wondering if i could get help understanding how to graph cubics and tan graphs please Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before. Title: Re: Mathematics Question Thread Post by: RuiAce on October 11, 2016, 12:08:58 am Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before. It's not in 2U. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:09:44 am It's not in 2U. Thank the heavens... Title: Re: 96 in 2U Maths: Ask me Anything! Post by: BPunjabi on October 11, 2016, 12:18:29 am Hey Liiz: HEY Im jake (obviously not haha). Im a year 12 student who completed my 2u HSC mathematics course in year 11 and lm just happy to help out here. Now, this type of question is amongst one of the most difficult ones in geometrical applications of calculus, and unfortunately in HSC exams there WILL be harder ones. But don't worry, once you have practised enough, you will begin to seize some tricks to approach these questions. Before I begin answering your question, just a few generally tips to help you answer questions like these where only one number value is provided: BEFORE YOU DO ANYTHING, DRAW A DIAGRAM WITH LABELS 1. Highlight all USEFUL INFORMATIONS (in this case, highlight rectangular box, square base, no top, 500cm^3 and least area) 2. Appoint two variables to the unknown sides (in this case, I named the side length of the square base as x, and the height of the rectangular box as L) 3. There will be at least one number quantity in every one of these questions in 2u mathematics, so the first equation you should construct, using your name variables to construct an equation that uses the numbers provided by the question 4. Draw out the relationship between the two variables through this equation that you have constructed 5. Construct another equation using your variables and the subject that is asked for in the question (In this case, for example, we constructed an Area equation which directly relates to what we are asked to find) 6. Substitute in the equivalent expression of a variable (In this case, for example, L = 500/x^2, so we substitute any L we see with 500/x^2) to reduce the total number of values down to one, so that we can construct an equation entire out of only one variable, which then allows us to perform differentiation 7. Clean up the equation after the substitution to make life easier 8. Differentiate the equation 9. Let this derivative = 0 to find any stationary points (In an Exam, YOU MUST STATE "LET dy/dx = 0 TO FIND ANY STATIONARY POINTS, OTHERWISE MARKS MAYBE DEDUCTED!!!) 10. Solve the derivative equation and find a value for your variable which will be your stationary point 11. Test both sides to show that a local minimum/maximum occurs at your stationary points 12. Substitute your minimum value back into the area equation (or maximum value if the question asks for maximum area) to find the minimum area (or the maximum area if you substitute in the maximum value) So here is my solution: (http://i.imgur.com/t6wtIk1.jpg) Hope you find my solutions clear and useful! If you are confused with anything, dont hesitate to ask!!! :D Best Regards Jacky He LOL WTF how were you guys so good at math? Did you constantly study! Title: Re: Mathematics Question Thread Post by: Alize on October 11, 2016, 11:47:21 am Hey, Could someone please explain to me how solve the following? 7sin3x=2x-1 Thank you in advance! ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 11, 2016, 11:55:41 am Hey, Could someone please explain to me how solve the following? 7sin3x=2x-1 Thank you in advance! ;D $\textbf{You can't.}\\ \text{There is no known algebraic method in the entire world of maths that can be used}\\ \text{to solve }\sin x = ax+b$ $\text{You can only }\textbf{approximate}\text{ the solution to }7\sin 3x=2x-1\\ \text{To do so, you need to sketch}\\ y=7\sin 3x\\ y=2x-1\\ \text{to a }\textbf{perfect scale}.$ $\text{Once you have your graph, you read the }x\text{-coordinates}\\ \text{of your point of intersection.}\\ \text{This will allow you to approximate a solution.}$ Your answers should be close to WolframAlpha's exact answers Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on October 11, 2016, 12:05:25 pm Hey! I feel like this would be really obvious but I'm struggling with a few questions like find the gradient of the tangent to the curve: y= square root of x at (4,2) and find the gradient of the normal to the curve y=x2 + 5 at (-2,9) Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:18:09 pm Well with gradient of tangent you differentiate. So you make the equation first x^1/2 then differentiate. Then it becomes 1/2x^-1/2. Sub in x=4 and then find the gradient. Grad become 1/4 . With normal same principals but switch it around so repeat first steps. Differentiate x^2+5 which is 2x. Sub in -2, making it =-4. Then put the reciprocal which is 1/4. Thats the grad of a normal Title: Re: Mathematics Question Thread Post by: RuiAce on October 11, 2016, 12:22:26 pm Hey! I feel like this would be really obvious but I'm struggling with a few questions like find the gradient of the tangent to the curve: y= square root of x at (4,2) and find the gradient of the normal to the curve y=x2 + 5 at (-2,9) $\text{The questions are just using definitions and you should get used to them.}\\ \text{For }y=\sqrt{x}, \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\\ \text{So when }x=4,\, m_T= \frac{dy}{dx}=\frac{1}{2\sqrt{4}}=\frac14$ $\textit{Recall: The derivative IS the gradient of the tangent at the point.}$ ___________ $\text{And as pointed above, for the gradient of the normal, recall that}\\ \text{perpendicular lines satisfy }m_1m_2=-1$ Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:33:23 pm $\text{The questions are just using definitions and you should get used to them.}\\ \text{For }y=\sqrt{x}, \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\\ \text{So when }x=4,\, m_T= \frac{dy}{dx}=\frac{1}{2\sqrt{4}}=\frac14$ $\textit{Recall: The derivative IS the gradient of the tangent at the point.}$ ___________ $\text{And as pointed above, for the gradient of the normal, recall that}\\ \text{perpendicular lines satisfy }m_1m_2=-1$ Rui I swear that the normal is the exact same method as tangent but you swap it to the reciprocal? Like in my method above^^ Title: Re: Mathematics Question Thread Post by: RuiAce on October 11, 2016, 12:38:51 pm Rui I swear that the normal is the exact same method as tangent but you swap it to the reciprocal? Like in my method above^^ *Negative reciprocal But yeah nothing I said really contradicts the main idea you posed Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:41:06 pm *Negative reciprocal But yeah nothing I said really contradicts the main idea you posed No I just meant the second part because I thought you were alluding that the answer was -1. Had no idea what you were trying to say Title: Re: Mathematics Question Thread Post by: RuiAce on October 11, 2016, 12:43:40 pm No I just meant the second part because I thought you were alluding that the answer was -1. Had no idea what you were trying to say I never said that. I said that two lines that are perpendicular have their gradients multiply to -1. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 11, 2016, 12:45:26 pm I never said that. I said that two lines that are perpendicular have their gradients multiply to -1. Oh I understand aha, Sorry for the confusion!! Title: Re: Mathematics Question Thread Post by: Alize on October 11, 2016, 01:26:46 pm $\textbf{You can't.}\\ \text{There is no known algebraic method in the entire world of maths that can be used}\\ \text{to solve }\sin x = ax+b$ $\text{You can only }\textbf{approximate}\text{ the solution to }7\sin 3x=2x-1\\ \text{To do so, you need to sketch}\\ y=7\sin 3x\\ y=2x-1\\ \text{to a }\textbf{perfect scale}.$ $\text{Once you have your graph, you read the }x\text{-coordinates}\\ \text{of your point of intersection.}\\ \text{This will allow you to approximate a solution.}$ Your answers should be close to WolframAlpha's exact answers Oh ok, that makes a lot more sense, thank you! ;D Title: Re: Mathematics Question Thread Post by: asd987 on October 11, 2016, 11:49:47 pm i need help with this question ty sketch y=√x+5 for −4≤x≤4 and find its max and min values. btw, the square root is over x+5, not just x Title: Re: Mathematics Question Thread Post by: RuiAce on October 12, 2016, 12:03:29 am i need help with this question ty sketch y=√x+5 for −4≤x≤4 and find its max and min values. btw, the square root is over x+5, not just x $\text{That's just sketching your typical square root curve. Only difference is we have a domain restriction.}$ (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps6wk2hleu.png) $\text{Then just looking at the graph, clearly}\\ \text{The minimum value occurs at }x=-4\text{ and equals }1\\ \text{The maximum value occurs at }x=4\text{ and equals }0$ Use brackets to show that you meant y=√(x+5) in the future. $\text{Was there something in particular that did not make sense}\\ \text{There was nothing technical in this question}$ Title: Re: Mathematics Question Thread Post by: wesadora on October 14, 2016, 04:56:04 pm This question. This was a 1998 2 unit paper... Was this part of the older syllabus or something? Cause nowadays they just ask us to do probability trees. I mean, tables of outcomes are pretty simple but do they even teach that....otherwise other people might be like wtf. Title: Mathematics Question Thread Post by: RuiAce on October 14, 2016, 04:57:19 pm This question. This was a 1998 2 unit paper and I know I didn't get taught what the hell a table of possible outcomes is. Was this part of the older syllabus or something? Cause nowadays they just ask us to do probability trees. It means to draw a table listing out every single outcome. An outcome is something that can happen. So for example you can get 4 on the blue die and 2 on the pink. It is to be interpreted literally. _________ And once you have a table you can literally just count the number of favourable outcomes for your probability. Title: Re: Mathematics Question Thread Post by: wesadora on October 14, 2016, 05:01:39 pm Thanks haha, yeah figured it out but like what the ._. Title: Mathematics Question Thread Post by: RuiAce on October 14, 2016, 05:03:09 pm You're not taught enumeration techniques in 2U so it is unreasonable to make you do part (ii) without the table. That being said I don't believe that this sort of thing is applicable for questions in 2U that are not about dice rolls Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 11:00:41 am I can't do this maths question! Help please! Differentiate p=2x+50/x with respect to x. Find any stationary points on the curve and determine their nature. Thanks Title: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 11:06:53 am $\text{The derivative is just }\frac{dp}{dx}=2-50x^{-2}$ $\text{This type of question is fairly standard in 2U and you need to get used to it.}\\ \text{Stationary points occur when }\frac{dp}{dx}=0\\ \text{Solving }2-50x^{-2}=0\text{ gives }x=\pm 5$ $\text{Then either test both sides or use the second derivative to find their nature.}$ Title: Re: Mathematics Question Thread Post by: asd987 on October 15, 2016, 11:14:10 am I'm stuck on part b) of the question, but this includes the whole thing: A farmer wants to make a rectangular paddock with an area of 4000m^2. However, fencing costs are high and she wants the paddock to have a minimum perimeter. a) Show that the perimeter is given by the equation P = 2x + 8000/x b) Find the dimensions of the rectangle that will give the minimum perimeter, correct to 1 decimal place. Title: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 11:18:04 am I'm stuck on part b) of the question, but this includes the whole thing: A farmer wants to make a rectangular paddock with an area of 4000m^2. However, fencing costs are high and she wants the paddock to have a minimum perimeter. a) Show that the perimeter is given by the equation P = 2x + 8000/x b) Find the dimensions of the rectangle that will give the minimum perimeter, correct to 1 decimal place. $\text{Functionally similar to the question above.}\\ \text{We must use the first derivative to minimise.}\\ \frac{dP}{dx}=2-8000x^{-2}$ $\text{Solving }\frac{dP}{dx}=2-8000x^{-2}=0\text{ gives }x=\pm \sqrt{4000}\\ \text{Ignore the negative case (length can't be negative) and we have }x=\sqrt{4000}$ $\text{Then check that it is a minimum and hence find the dimensions}$ $\text{Convert your final answer to 1 d.p.}$ Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 11:29:42 am (Sorry new to this - don't know how to reply to one comment on its own...) In reference to my differentiation and nation of SP question, I tried testing points on either side but it doesn't match up to the answers or what the actual graph looks like. Could you solve it and let me see if my answers match. The 2 SP I got were (5, 20) and (-5,-20). I thought the first was the minimum and the second a maximum. Thanks in advance. Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 11:34:32 am (Sorry new to this - don't know how to reply to one comment on its own...) In reference to my differentiation and nation of SP question, I tried testing points on either side but it doesn't match up to the answers or what the actual graph looks like. Could you solve it and let me see if my answers match. The 2 SP I got were (5, 20) and (-5,-20). I thought the first was the minimum and the second a maximum. Thanks in advance. Can you please show me your working out? There's guides to how you can upload images onto this forum if you don't want to type it. Or you can upload through the ATARnotes app. Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 11:59:01 am Can you please show me your working out? There's guides to how you can upload images onto this forum if you don't want to type it. Or you can upload through the ATARnotes app. Ok, so my differentiation is in my working. I used the table in the picture to figure out if it was the maximum or minimum. For (5,20), because it is negative before and positive after, I thought that would make it a minimum but the answers have maximum. Turns out my diagram isn't right according to geogebra so ignore that btw. I haven't learnt using second derivatives, but am keen to learn it if helps me understand the question. Thanks again Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 12:04:41 pm Ok, so my differentiation is in my working. I used the table in the picture to figure out if it was the maximum or minimum. For (5,20), because it is negative before and positive after, I thought that would make it a minimum but the answers have maximum. Turns out my diagram isn't right according to geogebra so ignore that btw. I haven't learnt using second derivatives, but am keen to learn it if helps me understand the question. Thanks again Its 2x+50/x right? I'm pretty sure that (5,20) is a minimum and (-5,-20) is a maximum. GeoGebra should confirm that. One thing to note: the graph has an asymptote at x=0. You have to analyse this separately. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 15, 2016, 01:17:42 pm Hey Rui! Need help for the whole question please! (http://i63.tinypic.com/260d1xl.png) Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 01:32:37 pm $\textbf{2014 HSC 2U}$ Hey Rui! Need help for the whole question please! (http://i63.tinypic.com/260d1xl.png) $\text{The triangles clearly look similar and immediately there's a common angle.}\\ \text{Which reduces the workload considerably. Just gotta look for something else.}\\ \text{And here it is: The parallel lines will help us.}$ \text{In }\triangle DEF\text{ and }\triangle DSR\\ \begin{align*}\angle EDF& = \angle SDR\tag{common angle}\\ \angle DFE &= \angle DRS \tag{corresponding, RS||FE}\\ \therefore \triangle DEF&||| \triangle DER \tag{equiangular}\end{align*}\\ \text{Make sure in similarity proofs you know which triangles you are proving to have a focus}\\ \text{as to what to stare at.} _________________________ $\text{Part (ii) is a free mark.}\\ \text{It should be obvious that you need proportional sides to similar triangles.}$ _________________________ $\text{Here's where it gets interesting. We are suddenly interested in areas.}\\ \text{Think a bit more outside the box here.}$ $A_1\text{ is the area of }\triangle DSR\\ A\text{ is the area of }\triangle DEF\\ \text{These are, coincidentally, the same triangles we proved similarity in part (i)}$ $\text{Part (ii) drops a hint - we may very likely need that ratio.}$ $\text{Now, reidentify the geometrical figure. We are interested in triangles.}\\ \text{Indeed, we have TWO formulae for the area of a triangle.}\\ \text{However, it looks to be that }Area=\frac{1}{2}bh\text{ will be useless here.}$ \text{Let's try the other one: }Area=\frac{1}{2}ab\sin C:\\ \begin{align*}A_1=\frac{1}{2}\times DR \times DS \times \sin \angle D\\ A=\frac{1}{2} \times DF \times DE \times \sin \angle D\end{align*} $\text{How interesting. If we divide one over the other, some nice things cancel out!}\\ \frac{A_1}{A}=\frac{DR}{DF}\times \frac{DS}{DE}\\ \text{And now we just use the exact same working for part (ii)}\\ \frac{A_1}{A}=\frac{x}{x+y}\times \frac{x}{x+y}\\ \text{Take positive square roots and we're done.}$ _________________________ $\text{I'll let you try part (iv) first by yourself. Hint though:}\\ \sqrt{\frac{A_2}{A}}=\frac{y}{x+y}\\ \text{You need not prove this result, but you need to see how you could get there.}$ Title: Re: Mathematics Question Thread Post by: justdoit on October 15, 2016, 01:39:40 pm Hi, can someone please help me with the 2015 q.16 c. Looked at the working out and cant understand it :/ Thank youuu :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 01:47:43 pm Hi, can someone please help me with the 2015 q.16 c. Looked at the working out and cant understand it :/ Thank youuu :) In the future please provide a link to the paper or the question $\text{This is one of the classics in the harder end of 2U.}\\ \text{The proof always begins with similar triangles.}\\ \textit{Similar triangles are very common in difficult 2U problems.}$ $\text{Due to the messiness of the given diagram}\\ \textbf{Draw a 2D diagram clearly showing the front view of the cylinder.}\\ \text{From your diagram, it should become somewhat clearer which pair of triangles we need.}$ $\text{Using similar triangles}\\ \frac{R-x}{R}=\frac{y}{H}\\ \text{So }y=\frac{H}{R}(R-x)\\ \text{The purpose of doing this was because what we had to prove had }(R-x)\text{ in it.}$ \text{And now we just apply the volume formula they gave.}\\ \begin{align*}V&=\pi r^2 h\\ &= \pi \times x^2 \times \left(\frac{H}{R}(R-x)\right)\end{align*} ____________________________________ $\text{Now, I will assume that you had no trouble deducing the value}\\ \text{of }x\text{ which maximises the volume.}\\ \text{This value is }x=\frac{2R}{3}$ \text{So what is the maximum volume? Subbing }x=\frac{2R}{3}\text{ back in for }V:\\ \begin{align*}V&=\frac{H}{R}\pi\left(\frac{2R}{3}\right)^2\left(R-\frac{2R}{3}\right)\\ &= \frac{H}{R}\pi \left(\frac{4R^2}{9}\right)\left(\frac R3\right)\\ &=\frac{4R^2H\pi}{27}\end{align*} $\text{We've listened to their instruction. We've found the maximum. Look back at the question.}\\ \text{It says to compare it to the volume of the }\textbf{cone}.\\ \text{What is the volume of the cone? }V_{Cone}=\frac{\pi R^2H}{3}$ $\text{What do you notice?}\\ {V_{Cylinder}}_{Max}=\frac{4}{9}\times \frac{\pi R^2H}{3} = \frac{4}{9}V_{Cone}\\ \text{Hence since that's the MAXIMUM volume of the cylinder}\\ \text{The volume of the cylinder never exceeds }\frac{4}{9}\text{ the volume of the cone.}$ Title: Re: Mathematics Question Thread Post by: kevin217 on October 15, 2016, 02:28:02 pm What are the steps to graphing this sine curve? Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 02:36:43 pm What are the steps to graphing this sine curve? $\text{We want to sketch }y=\sin \left[2\left(x+\frac{\pi}{6}\right)\right]$ $\text{Method 1:}\\ \text{1. Sketch }y=\sin x\\ \text{2. Double the period by sketching }y=\sin (2x)\\ \text{3. Shift the graph by }\frac{\pi}{6}\text{ to the left by sketching }y=\sin \left[2\left(x+\frac{\pi}{6}\right)\right]$ $\text{Method 2:} \\ \text{1. Sketch }y=\sin x\\ \text{2. Shift the graph by }\frac{\pi}{6}\text{ to the left by sketching }y=\sin \left(x+\frac{\pi}{6}\right)\\ \text{3. Using }x=-\frac{\pi}{6},\text{ double the period by sketching }y=\sin \left[2\left(x+\frac{\pi}{6}\right)\right]$ As an exercise, you should use Desmos to graph all of the mentioned curves for you. You should observe what happens - both how the graphs stretch, and how the x-intercepts change. GeoGebra is a good substitute for Desmos. Title: Re: Mathematics Question Thread Post by: justdoit on October 15, 2016, 02:42:04 pm Hi, thanks for the help but how did u get 2R/3 Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 02:50:25 pm Hi, thanks for the help but how did u get 2R/3 \text{Like stated, this is the standard method to find a maxima/minima.}\\ \begin{align*}V&=\frac{H}{R}\pi x^2(R-x)\\ &=\frac{H}{R}\pi (Rx^2-x^3)\\ \frac{dV}{dx}&=\frac{H}{R}\pi (2Rx-x^2)\end{align*}\\ \text{Set }\frac{dV}{dx}=0\text{, find your maxima and proceed as with every other similar question.}\\ \text{Note that }x=0\text{ is obviously discarded because that implies no radius.} Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 03:33:52 pm I have to differentiate: 3x/√3x-4+√3x-4 I end up getting (9√2)/8 but the answer is (3√2)8 I'm so close yet so far... help! Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 03:40:08 pm I have to differentiate: 3x/√3x-4+√3x-4 I end up getting (9√2)/8 but the answer is (3√2)8 I'm so close yet so far... help! $\text{You said you need to differentiate }\frac{3x}{\sqrt{3x-4}}+\sqrt{3x-4}\\ \text{Using a messy combination of quotient and chain rule this becomes }\frac{9(x-2)}{(3x-4)^\frac{3}{2}}$ $\text{So I am not sure where the numbers you arrived at came from.}\\ \text{Differentiating does not produce a number here.}$ Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 03:43:27 pm $\text{You said you need to differentiate }\frac{3x}{\sqrt{3x-4}}+\sqrt{3x-4}\\ \text{Using a messy combination of quotient and chain rule this becomes }\frac{9(x-2)}{(3x-4)^\frac{3}{2}}$ $\text{So I am not sure where the numbers you arrived at came from.}\\ \text{Differentiating does not produce a number here.}$ Oh I'm so sorry, I meant to add that f'(x) = 2 Title: Re: Mathematics Question Thread Post by: jakesilove on October 15, 2016, 03:50:24 pm Oh I'm so sorry, I meant to add that f'(x) = 2 Using Rui's answer, and setting it equal to two, you find that x is a complex number. Potentially you haven't fully described the question properly; can you just post a screenshot? Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 03:52:04 pm Oh I'm so sorry, I meant to add that f'(x) = 2 Doesn't mean anything because that has no (real) solutions. Are you sure you haven't differentiated already and were trying to say that what you gave us WAS the derivative? Title: Re: Mathematics Question Thread Post by: jakesilove on October 15, 2016, 03:53:42 pm Doesn't mean anything because that has no (real) solutions. Are you sure you haven't differentiated already and were trying to say that what you gave us WAS the derivative? Nup, that still produces a complex solution Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 03:55:14 pm Nup, that still produces a complex solution I have no clue either then. We need the actual question. Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 04:07:51 pm I have no clue either then. We need the actual question. Here you go :) Sorry if I stuffed up the question Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 04:10:52 pm Here you go :) Sorry if I stuffed up the question $\text{That is just }\textbf{evaluating }f^{\prime\prime}(2)\\ \text{We aren't solving anything here.}$ \text{Here's the correction:}\\ \begin{align*}f^\prime(x)&=\frac{3x}{2\sqrt{3x-4}}+\sqrt{3x-4}\end{align*} Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 04:17:07 pm $\text{That is just }\textbf{evaluating }f^{\prime\prime}(2)\\ \text{We aren't solving anything here.}$ \text{Here's the correction:}\\ \begin{align*}f^\prime(x)&=\frac{3x}{2\sqrt{3x-4}}+\sqrt{3x-4}\end{align*} I understand how to get f'(x) but don't know how to get f''(x) which will let me finish the question as I can sub in 2 then. Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 04:21:14 pm I understand how to get f'(x) but don't know how to get f''(x) which will let me finish the question as I can sub in 2 then. \text{Inevitably, you need the quotient rule on top of the chain rule.}\\ \text{Clever algebra makes it neater for us.}\\ \begin{align*}f^\prime(x)&=\frac{3x}{2\sqrt{3x-4}}+\sqrt{3x-4}\\ &= \frac{3x}{2\sqrt{3x-4}}+\frac{2(3x-4)}{2\sqrt{3x-4}}\\&=\frac{9x-8}{2\sqrt{3x-4}}\end{align*} $\text{To which you'd use the quotient rule now.}\\ f^{\prime\prime}(x)=\frac{2\sqrt{3x-4}(9)-(9x-8)\frac{3}{\sqrt{3x-4}}}{(2\sqrt{3x-4})^2}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 04:30:18 pm Sorry, your post came up with random symbols thrown into it making it very difficult to read... If you're in an app, go into web view. LaTeX does not get produced on the app; you need a proper web browser. Otherwise I'm not sure at all what the problem is and you will need to provide sufficient evidence including screenshots for technical help. Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 04:38:44 pm \text{Inevitably, you need the quotient rule on top of the chain rule.}\\ \text{Clever algebra makes it neater for us.}\\ \begin{align*}f^\prime(x)&=\frac{3x}{2\sqrt{3x-4}}+\sqrt{3x-4}\\ &= \frac{3x}{2\sqrt{3x-4}}+\frac{2(3x-4)}{2\sqrt{3x-4}}\\&=\frac{9x-8}{\sqrt{3x-4}}\end{align*} $\text{To which you'd use the quotient rule now.}\\ f^{\prime\prime}(x)=\frac{\sqrt{3x-4}(9)-(9x-8)\frac{3}{2\sqrt{3x-4}}}{(\sqrt{3x-4})^2}$ Sorry for being so naggy, but where did the 2 go at the end of the first differentiation? Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 04:40:51 pm Sorry for being so naggy, but where did the 2 go at the end of the first differentiation? Fixed, my bad on the accident there Title: Re: Mathematics Question Thread Post by: julzzz on October 15, 2016, 04:56:06 pm Fixed, my bad on the accident there Cheers Really appreciate all your help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Title: Re: Mathematics Question Thread Post by: BPunjabi on October 15, 2016, 05:52:33 pm How do you do these questions? How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ? Title: Re: Mathematics Question Thread Post by: jakesilove on October 15, 2016, 05:57:06 pm How do you do these questions? How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ? Hey! For that equation to hold true, either $sin(x)-1=0$ or $tan(x)+2=0$ Looking to the first equation, $sin(x)=1$ $x=\frac{\pi}{2}$ Looking to the second equation $tan(x)=-2$ $x=2.03, 5.18$ So those should be our three solutions! Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:14:28 pm How do you do these questions? How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ? $\text{Jake walked into a trap that I was hoping he would walk into.}$ $\text{Upon glancing, the solutions are}\\ x=\frac{\pi}{2}, 2.03, 5.18$ $\text{But wait. What is }\tan \frac{\pi}{2}?\\ \text{Well fuck. We just broke maths.}$ $\text{In actuality, we have to }\textbf{discard the solution }x=\frac{\pi}{2}\\ \text{Because otherwise }(\sin x - 1)(\tan x - 2)\text{ is undefined!}$ $\textit{Hence there are only TWO solutions.}$ Title: Re: Mathematics Question Thread Post by: BPunjabi on October 15, 2016, 06:15:15 pm Hey! For that equation to hold true, either $sin(x)-1=0$ or $tan(x)+2=0$ Looking to the first equation, $sin(x)=1$ $x=\frac{\pi}{2}$ Looking to the second equation $tan(x)=-2$ $x=2.03, 5.18$ So those should be our three solutions! Sorry man the answers says 2. Its multiple choice question 7 for the 2014 paper. Also could you explain to me how you quickly worked out the fractions for the solutions like: $sin(x)=1$ $x=\frac{\pi}{2}$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 15, 2016, 06:16:54 pm Hello, can you please help me with HSC 2007 question 7a(i)... Its a little confusing.... Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:17:23 pm Sorry man the answers says 2. Its multiple choice question 7 for the 2014 paper. Also could you explain to me how you quickly worked out the fractions for the solutions like: $sin(x)=1$ $x=\frac{\pi}{2}$ $\text{If you don't know the boundary values then you need to study your graphs again.}\\ \sin 0 = 0, \sin \frac{\pi}{2}=1, \sin \pi = 0, \sin \frac{3\pi}{2} = -1, \sin 2\pi = 0$ Anyway, I explained his mistake. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 15, 2016, 06:19:22 pm $\text{If you don't know the boundary values then you need to study your graphs again.}\\ \sin 0 = 0, \sin \frac{\pi}{2}=1, \sin \pi = 0, \sin \frac{3\pi}{2} = -1, \sin 2\pi = 0$ Anyway, I explained his mistake. Sorry I quoted and said the wrong thing... I meant the answers for the second part. When I put those values into my calc it does not equal exactly zero, and is there a way to figuring it out fast? Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:20:22 pm Hello, can you please help me with HSC 2007 question 7a(i)... Its a little confusing.... In the future, please provide a link to the paper. $\text{The parabola we're given is just }x^2=y-4\\ \text{Recall the standard form }(x-h)^2=4a(y-k)$ $\text{So here, the vertex is }V(0,4)\\ \text{And the focal length is }4a=1 \implies a=\frac{1}{4}$ $\text{So using the vertex and focal length, our focus is }S\left(0, 4+\frac{1}{4}\right)\\ \text{i.e. }\left(0, \frac{17}{4}\right)$ $\text{All formula work.}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:21:10 pm Sorry I quoted and said the wrong thing... I meant the answers for the second part. When I put those values into my calc it does not equal exactly zero, and is there a way to figuring it out fast? 2.03 and 5.18 are obviously rounded. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 15, 2016, 06:23:53 pm 2.03 and 5.18 are obviously rounded. Yes but can you please tell me how you get that answer quickly... you dont just randomly test those points in your calculator.. there must be some method to getting those values. Thanks, Bobby Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:25:13 pm Yes but can you please tell me how you get that answer quickly... you dont just randomly test those points in your calculator.. there must be some method to getting those values. Thanks, Bobby $\text{He just solved the trigonometric equation.}$ $\tan x + 2 = 0\\ \tan x = -2$ $\text{Using ASTC, tan is negative in the second and fourth quadrants}\\ \text{So }x=\pi - \tan^{-1}2, x = 2\pi - \tan^{-1}2$ $\text{This is a method you learnt in prelim.}$ Title: Re: Mathematics Question Thread Post by: asd987 on October 15, 2016, 06:30:15 pm help plz A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced Title: Re: Mathematics Question Thread Post by: nibblez16 on October 15, 2016, 06:34:40 pm In the future, please provide a link to the paper. $\text{The parabola we're given is just }x^2=y-4\\ \text{Recall the standard form }(x-h)^2=4a(y-k)$ $\text{So here, the vertex is }V(0,4)\\ \text{And the focal length is }4a=1 \implies a=\frac{1}{4}$ $\text{So using the vertex and focal length, our focus is }S\left(0, 4+\frac{1}{4}\right)\\ \text{i.e. }\left(0, \frac{17}{4}\right)$ $\text{All formula work.}$ Thank You so much! Sorry, I'll provide a link next time... Thanks :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 06:45:59 pm help plz A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsffwxytg1.png) $\text{The trick is that the diagonal of the rectangle is the same as}\\ \text{the diameter of the circle.}\\ \text{Note that the radius is 6, so the diameter will be 12.}$ $\text{Let the length of the rectangle be }x\\ \text{Then by Pythagoras' Theorem, the breadth has length }\sqrt{144-x^2}$ $\text{Hence the area is given by }A=\ell b = x\sqrt{144-x^2}\\ \text{The problem now falls to maximising this area.}$ $\frac{dA}{dx}=\sqrt{144-x^2} - \frac{x^2}{\sqrt{144-x^2}}\\ \text{Setting }\frac{dA}{dx}=0\text{ gives }x=6\sqrt{2}$ $\text{You should be able to take it from there.}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 07:58:15 pm hi, i always get stuck when it comes to drawing the derivative function of a graph, has anyone got any tips to draw these? Be more specific please? If you're given the function, say, f(x)=x3+3x2 then all you have to do is differentiate it and then continue. Do you mean drawing the graph of the derivative, given the graph of the original thing? Title: Re: Mathematics Question Thread Post by: nimasha.w on October 15, 2016, 11:23:36 pm Be more specific please? If you're given the function, say, f(x)=x3+3x2 then all you have to do is differentiate it and then continue. Do you mean drawing the graph of the derivative, given the graph of the original thing? sorry for the confusion! like with (iii) i could get it all except from t1 to t3 Title: Re: Mathematics Question Thread Post by: asd987 on October 15, 2016, 11:29:43 pm I need help with this question please A surfboard is in the shape of a rectangle and semicircle. The perimeter is to be 4m. Find the maximum area of the surfboard, correct to 2 decimal places. Title: Re: Mathematics Question Thread Post by: RuiAce on October 15, 2016, 11:43:34 pm I need help with this question please A surfboard is in the shape of a rectangle and semicircle. The perimeter is to be 4m. Find the maximum area of the surfboard, correct to 2 decimal places. Draw a diagram and show us your working out thus far, like how I used a diagram for your earlier question. You cannot simply do those questions without a diagram. sorry for the confusion! like with (iii) i could get it all except from t1 to t3 $\text{Think about what the derivative is.}\\ \text{The derivative measures the gradient of the tangent at that point.}$ $\text{If you have a function that is constant (and thus a horizontal line), the gradient of the tangent is always 0}\\ \text{This is matched by the fact that when you differentiate a constant you get 0.}\\ \frac{d}{dx} c = 0$ $\text{So wherever that curve is completely flat, its derivative is just 0.}$ ______________________________________________________$\text{Then you just use what the derivative is.}$ $\text{If }y\text{ is increasing, then }\frac{dy}{dx}> 0\\ \text{If }y\text{ is decreasing, then }\frac{dy}{dx}< 0$ $\text{So for your graph, if you look at }1 \le t \le 3, \frac{dx}{dt}\text{ is decreasing.}\\ \text{So }\frac{d^2x}{dt^2}< 0\text{, i.e. }\frac{d^2x}{dt^2}\text{ will be below the }t\text{-axis.}$ $\text{Similarly for }5\le t\le 6, \, \frac{dx}{dt}\text{ will be above the }x\text{-axis.}$ $\textit{In fact, if you draw it correctly, the P.O.I. at }t=2\textit{ on }\frac{dx}{dt}\\ \textit{becomes a stationary point on }\frac{d^2x}{dt^2}$ $\text{Now try sketching the curve. Feel free to post up your attempt for feedback.}$ Title: Re: Mathematics Question Thread Post by: kevin217 on October 16, 2016, 10:36:28 am Depending on whether a function is concave up or down, how does Simpsons rule over/underestimate the exact area? Title: Re: Mathematics Question Thread Post by: RuiAce on October 16, 2016, 10:43:03 am Depending on whether a function is concave up or down, how does Simpsons rule over/underestimate the exact area? Recall that Simpson's rule uses a parabolic arc to estimate the area. Given any three points, sketch a parabola through them. Look at whether the area under the parabola is greater than, or less than the original area. Note that the concavity of the function itself isn't enough for Simpson's rule. For the trapezoidal rule it is all that's needed because you're estimating with just trapeziums (and hence you draw straight lines). For Simpson's rule, you need to physically consider what the parabola through the three points looks like to make a conclusion. An example of where this was needed was in the 2013 HSC. Title: Re: Mathematics Question Thread Post by: kevin217 on October 16, 2016, 11:13:16 am Recall that Simpson's rule uses a parabolic arc to estimate the area. Given any three points, sketch a parabola through them. Look at whether the area under the parabola is greater than, or less than the original area. Note that the concavity of the function itself isn't enough for Simpson's rule. For the trapezoidal rule it is all that's needed because you're estimating with just trapeziums (and hence you draw straight lines). For Simpson's rule, you need to physically consider what the parabola through the three points looks like to make a conclusion. An example of where this was needed was in the 2013 HSC. I see, thanks Title: Re: Mathematics Question Thread Post by: MysteryMarker on October 16, 2016, 06:55:22 pm For question 15c, could someone explain to me how to do it both graphically and algebraically. Isn't that if I equate the two expressions, square them and solve for their discriminant to = 0, i should get a value for m? (I tried and its not working, just curious as to why). Thanks. Title: Re: Mathematics Question Thread Post by: RuiAce on October 16, 2016, 07:39:50 pm For question 15c, could someone explain to me how to do it both graphically and algebraically. Isn't that if I equate the two expressions, square them and solve for their discriminant to = 0, i should get a value for m? (I tried and its not working, just curious as to why). Thanks. The graphical method was done in post #454 ______________________ $\text{Now, when we square both sides the absolute value disappears.}\\ \text{We're left with a quadratic. So we should be able to use the discriminant, right?}\\ \textbf{Wrong!}\text{ (Unfortunately.)}$ $\text{The problems lie in exactly that; the absolute values.}\\ \text{The LHS has absolute values, but the RHS does not. This means that}\\ \text{When we sketch }y=|2x-3|\text{, we have NONE of the graph below the }x\text{-axis}\\ \text{This happens because if }x< \frac{2}{3}\text{, we flip it back up to get the V shape,}$ $\text{This does not happen for }y=mx+1\\ \text{That just keeps going.}$ $\text{So on one hand, we have both graphs going the same way.}\\ \text{On the other hand, we don't.}\\ \textbf{Squaring demolishes this problem when it's not supposed to.}\\ \text{If both graphs DO happen to be above the }x\text{-axis, the whole thing breaks down.}$ $\text{In general, squaring is unsafe and used only because we have to.}\\ \text{Here, we would much rather avoid it}$ ___________________________________ $\text{There is no convenient algebraic method that can possibly be asked out of a 2U student.}\\ \text{Far too many things must be analysed all at once.}$ Title: Re: Mathematics Question Thread Post by: epherbertson on October 16, 2016, 07:56:53 pm Help This is from question 10 of the 1995 paper and am having trouble with the working of the question. I got part I out but i am lost from there. any help would be largely appreciated... Thanks Emily :-\ Title: Re: Mathematics Question Thread Post by: imtrying on October 16, 2016, 08:13:59 pm Hey:) Just need a hand with this question from the 2012 paper. I've tried looking at the solutions but they just arent making sense to me. Thanks! Title: Re: Mathematics Question Thread Post by: RuiAce on October 16, 2016, 08:23:40 pm Hey:) Just need a hand with this question from the 2012 paper. I've tried looking at the solutions but they just arent making sense to me. Thanks! $\text{You know nothing about line }PT\text{ whatsoever.}\\ \text{Yet somehow }\theta\text{ makes its way into the equation of the tangent.}$ $\text{You have to look at what }\theta\text{ is.}\\ \theta\text{ is the angle }OP\text{ makes with the }x\text{-axis.}$ $\text{Hence, recall the formula}\\ \boxed{m=\tan \theta}$ $\text{The gradient of }OP\text{ IS }\tan \theta\\ \text{Conveniently, }OP\perp PT\\ \text{Hence using the formula }m_1m_2=-1\text{ we have}\\ \boxed{m_{PT}=-\cot \theta}$ $\text{And now we use the classic point-gradient formula.}\\ \text{Except, what are the coordinates of }T?$ $\text{Draw a line from }T\text{ perpendicular to the }x\text{-axis.}\\ \text{Use trigonometry to figure out why }x=\cos \theta, y=\sin \theta\\ \text{i.e. }T\text{ has coordinates }\boxed{(\cos \theta, \sin \theta)}$ $\text{Then it's just}\\ y-\sin \theta = -\cot \theta(x-\cos \theta)\\ y-\sin \theta = -\frac{\cos \theta}{\sin \theta}(x-\cos \theta)$ Title: Re: Mathematics Question Thread Post by: imtrying on October 16, 2016, 08:28:38 pm $\text{You know nothing about line }PT\text{ whatsoever.}\\ \text{Yet somehow }\theta\text{ makes its way into the equation of the tangent.}$ $\text{You have to look at what }\theta\text{ is.}\\ \theta\text{ is the angle }OP\text{ makes with the }x\text{-axis.}$ $\text{Hence, recall the formula}\\ \boxed{m=\tan \theta}$ $\text{The gradient of }OP\text{ IS }\tan \theta\\ \text{Conveniently, }OP\perp PT\\ \text{Hence using the formula }m_1m_2=-1\text{ we have}\\ \boxed{m_{PT}=-\cot \theta}$ $\text{And now we use the classic point-gradient formula.}\\ \text{Except, what are the coordinates of }T?$ $\text{Draw a line from }T\text{ perpendicular to the }x\text{-axis.}\\ \text{Use trigonometry to figure out why }x=\cos \theta, y=\sin \theta\\ \text{i.e. }T\text{ has coordinates }\boxed{(\cos \theta, \sin \theta)}$ $\text{Then it's just}\\ y-\sin \theta = -\cot \theta(x-\cos \theta)\\ y-\sin \theta = -\frac{\cos \theta}{\sin \theta}(x-\cos \theta)$ Thanks so much, your explanation really helped:) Title: Re: Mathematics Question Thread Post by: RuiAce on October 16, 2016, 08:41:43 pm Help This is from question 10 of the 1995 paper and am having trouble with the working of the question. I got part I out but i am lost from there. any help would be largely appreciated... Thanks Emily :-\ The question says to show us part a) so please link part a) next time. $\text{Note that time is positive, i.e. }t > 0\\ \text{The speeds are the same if }4\cos t = 2 - t\\ \text{These are the exact same curves you sketched in part a).}$ $\text{Looking at your graph, for }x > 0\\ \text{There were only two points of intersections.}\\ \text{Hence here, there are only two times }t_1\text{ and }t_2\text{ where their speeds equal.}\\ \text{Note that we only know that the points of intersection exist, not where they are.}$ _____________________________ $\text{Inspect the graph you drew in a) once again.}\\ \text{You will notice that at }t=t_1, v>0\\ \text{whereas at }t=t_2, v< 0\\ \text{This implies that the particle turned around somewhere.}$ $\text{From formulae work or calculus methods, we know that the turning point on }\\ x=-6+2t-\frac{1}{2}t^2\text{ is at }t=2$ $\text{This means that the particle travels to the right in the first 2 seconds}\\ \text{but then starts travelling to the left.}\\ \text{Hence, we must analyse the two separate components of distance.}$ $\textbf{Remember: Displacement takes into account direction. Distance does NOT.}$ \text{Between times }t=t_1\text{ and }t=2:\\ \begin{align*}d_1 &= \left(-6+2(2)-\frac{1}{2}(2)^2\right)-\left(-6+2t_1-\frac{1}{2}t_1^2\right)\\ &= 2-2t_1+\frac{1}{2}t_1^2\end{align*} \text{Between times }t=2\text{ and }t=t_2\\ \begin{align*}d_2&= \left|\left(-6+2(2)-\frac{1}{2}(2)^2\right)-\left(-6+2t_1-\frac{1}{2}t_1^2\right)\right|\\ &= \left| -2 + 2t_2^2 - \frac{1}{2}t_2^2\right|\\ &= 2-2t_2+\frac{1}{2}t_2^2\end{align*} $\text{Adding the distances gives}\\ d=4-2(t_1+t_2)+\frac{1}{2}(t_1^2+t_2^2)$ ____________________ Part (iv) cannot be done because you don't know how to solve -6+2t-t2/2 = 4sin(t) Hence, try a graphical approach. You may find the solutions here Title: Re: Mathematics Question Thread Post by: cajama on October 17, 2016, 09:24:10 am Hello, I'm not sure if this question has been answered before and I know it's been talked about in one of the ATARNotes Maths lectures but are we still allowed to use abbreviations in the reasonings (e.g. isosceles Δ abbreviated to isos.Δ / supplementary to supp.)? A friend of mine said that this year, we wouldn't be allowed to use them and I'm just a little confused. Title: Re: Mathematics Question Thread Post by: RuiAce on October 17, 2016, 09:37:13 am Hello, I'm not sure if this question has been answered before and I know it's been talked about in one of the ATARNotes Maths lectures but are we still allowed to use abbreviations in the reasonings (e.g. isosceles Δ abbreviated to isos.Δ / supplementary to supp.)? A friend of mine said that this year, we wouldn't be allowed to use them and I'm just a little confused. I haven't heard anything about this (maybe the two other mods will input further). Personally I always chose when to abbreviate. At times, abbreviating is a convenient shorthand, but at other times it reduces the formality of your work and begins to read like laziness. Title: Re: Mathematics Question Thread Post by: isaacdelatorre on October 17, 2016, 11:30:37 am Hey guys, Just a question, regarding rounding. I know that you shouldn't round any numbers until the last part of the question. But when it says round to 4dp for example, do you continue to use the whole decimal or just 4 decimal places? E.g. rates of change when finding k - continue to use the whole thing or whatever decimal places they initially say Title: Re: Mathematics Question Thread Post by: imtrying on October 17, 2016, 12:02:16 pm Just needing some help with part ii of this question from the 2011 HSC. I've done part i, the answer is (n+1)1/2 - n1/2. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 17, 2016, 12:13:15 pm Hey guys, Just a question, regarding rounding. I know that you shouldn't round any numbers until the last part of the question. But when it says round to 4dp for example, do you continue to use the whole decimal or just 4 decimal places? E.g. rates of change when finding k - continue to use the whole thing or whatever decimal places they initially say Hey Isaac! If they say find to 4dp, and then the question continues, you can use that value!! :) but they would probably pay either or ;D Title: Mathematics Question Thread Post by: jakesilove on October 17, 2016, 12:53:19 pm Just needing some help with part ii of this question from the 2011 HSC. I've done part i, the answer is (n+1)1/2 - n1/2. Hey! Cool question, check out my solution below and let me know if I can clarify anything! (http://i.imgur.com/94nRCvj.png) Jake Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator Annotation by Jake of Annotation by Rui: Cheers buddy Title: Re: Mathematics Question Thread Post by: imtrying on October 17, 2016, 01:49:22 pm Hey! Cool question, check out my solution below and let me know if I can clarify anything! (http://i.imgur.com/94nRCvj.png) Jake Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator Thanks so much! :) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 17, 2016, 03:37:18 pm Hello, I'm not sure if this question has been answered before and I know it's been talked about in one of the ATARNotes Maths lectures but are we still allowed to use abbreviations in the reasonings (e.g. isosceles Δ abbreviated to isos.Δ / supplementary to supp.)? A friend of mine said that this year, we wouldn't be allowed to use them and I'm just a little confused. Sorry cajama I missed this! Why does your friend say specifically this year? What do they think has changed? That's the curious bit to me :) In general, you want to make your work as readable for the marker as is humanly possible. I am reasonably sure that abbreviations are fine, and that they would pay you for those that you've provided, but I was always hesitant to use them. Not because they wouldn't recognise them and not pay you for it intentionally, but more along Rui's line of thinking. It's about clarity and formality of proof, and basically, I wanted to take no chances that what I'd written had been misinterpreted. My thinking is this. The BOSTES sample solutions don't use those abbreviations, so if you want to be 100% sure that you are okay, then take the 1 second to write the full word. Of course, symbols like Δ and // are universal and definitely acceptable :) Title: Re: Mathematics Question Thread Post by: jakesilove on October 17, 2016, 03:47:06 pm Sorry cajama I missed this! Why does your friend say specifically this year? What do they think has changed? That's the curious bit to me :) In general, you want to make your work as readable for the marker as is humanly possible. I am reasonably sure that abbreviations are fine, and that they would pay you for those that you've provided, but I was always hesitant to use them. Not because they wouldn't recognise them and not pay you for it intentionally, but more along Rui's line of thinking. It's about clarity and formality of proof, and basically, I wanted to take no chances that what I'd written had been misinterpreted. My thinking is this. The BOSTES sample solutions don't use those abbreviations, so if you want to be 100% sure that you are okay, then take the 1 second to write the full word. Of course, symbols like Δ and // are universal and definitely acceptable :) I definitely used abbreviations for anything that was absolutely obvious (alt. angles on || are =, for instance), however if I thought there was a chance of confusion I would avoid abbreviations all together. I wouldn't say this makes much of a difference though! Title: Re: Mathematics Question Thread Post by: olivercutbill on October 17, 2016, 04:43:53 pm I've seen questions like this before but they've given a time constraint. How do I do 12? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 17, 2016, 05:01:18 pm I've seen questions like this before but they've given a time constraint. How do I do 12? Are you an Extension student Oliver? Because I think this has strayed into that territory of difficulty :P So we have this expression for cost in terms of speed, per hour: $$c=150+\frac{v^2}{80}$$ What we want is the total cost for a 500km trip. To find how long THIS takes, we use our speed formula: $$\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\v=\frac{500}{t}\\\therefore t=\frac{500}{v}$$ So our trip takes that many hours. To find our total cost, we multiply the cost per hour by the amount of hours taken that we just found. $$C_\text{Total}=ct=\frac{500}{v}\left(50+\frac{v^2}{80}\right)$$ And now we're back to familiar territory, differentiate and find your minima! Does that help? :) Title: Re: Mathematics Question Thread Post by: olivercutbill on October 17, 2016, 05:45:32 pm Are you an Extension student Oliver? Because I think this has strayed into that territory of difficulty :P So we have this expression for cost in terms of speed, per hour: $$c=150+\frac{v^2}{80}$$ What we want is the total cost for a 500km trip. To find how long THIS takes, we use our speed formula: $$\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\v=\frac{500}{t}\\\therefore t=\frac{500}{v}$$ So our trip takes that many hours. To find our total cost, we multiply the cost per hour by the amount of hours taken that we just found. $$C_\text{Total}=ct=\frac{500}{v}\left(50+\frac{v^2}{80}\right)$$ And now we're back to familiar territory, differentiate and find your minima! Does that help? :) No 2u, this was in the textbook :) I get 63km/h as a minimum - textbook answer says 110km/h. What am I doing wrong? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 17, 2016, 05:51:04 pm No 2u, this was in the textbook :) I get 63km/h as a minimum - textbook answer says 110km/h. What am I doing wrong? Oh I mean like, it's strayed into that level of difficulty, definitely 2U content. I mean like, don't stress that it's hard for you, because it is a hard question, aha ;D Snap a picture of your working? I'll see if I can spot ;D Title: Re: Mathematics Question Thread Post by: olivercutbill on October 17, 2016, 06:25:28 pm Oh I mean like, it's strayed into that level of difficulty, definitely 2U content. I mean like, don't stress that it's hard for you, because it is a hard question, aha ;D Snap a picture of your working? I'll see if I can spot ;D Figured it out, and did another similar question and got that too. Thanks :) Title: Re: Mathematics Question Thread Post by: Rikahs on October 17, 2016, 07:37:01 pm Does anyone know how to answer the attached question (all parts please). Im super confused :( Title: Re: Mathematics Question Thread Post by: RuiAce on October 17, 2016, 07:54:44 pm Does anyone know how to answer the attached question (all parts please). Im super confused :( Part (iii) was already addressed in post #533 $\text{For the previous parts of the question, understand what the question is asking.}\\ \text{You are given the velocity, which tells you information about the displacement.}$ $\text{If the velocity is positive, then the particle is }\textbf{continuously moving to the right of the origin.}\\ \text{Note how in the first 2 seconds, this is what the particle does.}$ $\text{Is there anything more to identify about }t=2?\\ \text{Yes. Note that at }t=2\text{ the velocity becomes 0}\\ \text{Once }t> 2\text{ the velocity is negative.}\textbf{ The particle turns around at }t=2$ $\text{Once the particle turns around, its displacement starts going back down.}\\ \text{Hence the maximum displacement is at }t=2.$ $\textit{This should not be surprising. The velocity is the derivative of the displacement.}\\ \textit{Stationary points on }x\textit{ occur when }\frac{dx}{dt}=0\\ \textit{Indeed, at }t=2\textit{, we have }v=0$ _________________________________ $\text{Part ii) needs a slightly more careful analysis.}\\ \text{The particle turns around at }t=2\text{ as established above.}\\ \text{When does it go back enough so that the displacement perfectly resets to }x=0?$ $\text{Recall that }x=\int v\, dt\\ \text{The displacement is just the antiderivative of the velocity.}$ $\text{Effectively, we want the displacement to be 0, ultimately.}\\ \text{This means that we want }\int_0^T v\, dt = 0\\ \text{where }T\text{ is the answer we're after.}$ $\textit{Note that we want when it returns to the origin}\\ \textit{So the initial time }T=0\textit{ is useless.}$ $\textbf{Recall that the integral measures the area under the curve.}\\ \text{Look at time }t=4\\ \text{By comparing areas. it should be clear that }\int_0^4 v \, dt = 0$ $\text{Hence }t=4$ _______ $\text{This can also be achieved analytically.}\\ \text{The particle travels at some varying speeds in the first 2 seconds.}\\ \text{But then it travels at the same speeds in the next 2, just in }\textbf{the opposite direction}.$ $\text{Because the same amount of time was taken}\\ \text{Since the velocities, in a way, 'collapse' on themselves}\\ t=4\text{ is the answer.}$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 17, 2016, 09:59:50 pm Hello, can you please explain the last question... I always get confused on those types... Thanks :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 17, 2016, 10:01:39 pm Hello, can you please explain the last question... I always get confused on those types... Thanks :) Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant. Title: Re: Mathematics Question Thread Post by: nibblez16 on October 17, 2016, 10:11:38 pm Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant. Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 12:43:31 am $\text{Observe that }P\text{ and }r\text{ remain, however }\theta\text{ vanishes in what we need to prove.}$ $\text{The intuition to this question is not nice.}\\ \text{Notice that we are proving a restriction}\\ \text{(Or in particular, a restriction on }r)$ $\text{Since }\theta\text{ vanishes, we must identify if there are any restrictions on }\theta\\ \text{This is now targetting the very upper end of 2U capabilities.}$ $\text{Fundamentally, }\boxed{0 < \theta < 2\pi}\\ \text{Hence }2 < \theta+2 < 2\pi+2\\ \text{So }2r < r(\theta+2) < 2r(\pi + 1)$ $\text{Thus }2r < P < 2r(\pi +1)$ $\text{Split the inequality up at this point:}\\ 2r < P \implies r < \frac{P}{2}\\ P < 2r(\pi + 1) \implies r > \frac{P}{2(\pi+1)}\\ \text{Recombine and done.}$ $\textit{The main question was where was the inspiration in all of this?}\\ \text{Once again, notice how there was only a restriction on }\theta\\ \text{Since }\theta\text{ vanishes, starting on it is probably a good idea.}$ $\text{You may have chosen to work backwards in pencil.}\\ \text{For some algebraic proofs, working backwards gives you a better sense of direction.}$ $\text{Working backwards, }P < 2r(\pi + 1) < P(\pi + 1)\\ \text{But we know from part (i) that }P=r(\theta + 2)\\ \text{It becomes harder to see what's going on here.}\\ \text{This last bit just assesses your analytical skills.}\\ \text{Ultimately you have to watch out for small details. That's how to get 100 percent in 2U.}$ Title: Re: Mathematics Question Thread Post by: FallonXay on October 18, 2016, 09:04:14 am Hello, with this question, why does the minimum of D occur at the same time as the minimum of D^2? Also, why do you use the negative root in the volume question? And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D Thanks! Title: Re: Mathematics Question Thread Post by: kimmie on October 18, 2016, 09:49:54 am Hey can you please help me with this question. It's from the 2013 hsc paper. I did part (i) but I don't understand how to do part (ii) Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:49:57 am Hello, with this question, why does the minimum of D occur at the same time as the minimum of D^2? Also, why do you use the negative root in the volume question? And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D Thanks! $\text{In Extension 2, a trick known as implicit differentiation is taught.}\\ \frac{d(D^2)}{dt}=2D\frac{dD}{dt}\\ \text{So seeting }\frac{d(D^2)}{dt}=0\text{ has the same effect as setting }\frac{dD}{dt}=0$ $\text{This can also be interpreted graphically.}\\ \text{Open up GeoGebra. Sketch some functions and their squares. Compare where the stationary points are.}\\ \text{e.g. }(x-3)(x-2)(x-1)\text{ and }[(x-3)(x-2)(x-1)]^2, \sin x\text{ and }\sin^2 x$ $\text{To prove the result that }D^2\text{ has a S.P. when }D\text{ has a S.P.}\\ \text{Let }y=[f(x)]^2\\\text{so that }y\text{ is just the square of }f(x) \\\text{Then }\frac{dy}{dx}=2f(x)f^\prime(x)$ $\text{For a stationary point, }\frac{dy}{dx}=0\\ \text{So }f(x)=0 \text{ or }f^\prime(x)=0$ $\text{The }f(x)=0\text{ case tells us something extra, but all we are interested is that}\\ \frac{dy}{dx}=0\text{ when }f^\prime(x)=0\\ \text{i.e. the derivative of the square is 0, when the derivative of the original thing is 0.}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:52:08 am Hello, with this question, why does the minimum of D occur at the same time as the minimum of D^2? Also, why do you use the negative root in the volume question? And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D Thanks! $\text{For the volumes question, consider what is going on.}$ $\text{You have the equation of the parabola }y=(x-2)^2\\ \text{Taking roots, }x-2 = \pm \sqrt{y} \implies x = 2\pm \sqrt{y}$ $\text{But what is the difference? This is an important feature.}\\ \text{The positive root corresponds to the }\textbf{right branch}\text{ of the parabola}\\ \text{The negative root corresponds to the }\textbf{left branch}\text{ of the parabola}$ $\text{Again, you should graph }x-2=\sqrt{y}\text{ and }x-2=-\sqrt{y}\text{ separately on GeoGebra or Desmos.}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:54:31 am Hey can you please help me with this question. It's from the 2013 hsc paper. I did part (i) but I don't understand how to do part (ii) Already done in post #454 Title: Re: Mathematics Question Thread Post by: FallonXay on October 18, 2016, 10:00:47 am $\text{For the volumes question, consider what is going on.}$ $\text{You have the equation of the parabola }y=(x-2)^2\\ \text{Taking roots, }x-2 = \pm \sqrt{y} \implies x = 2\pm \sqrt{y}$ $\text{But what is the difference? This is an important feature.}\\ \text{The positive root corresponds to the }\textbf{right branch}\text{ of the parabola}\\ \text{The negative root corresponds to the }\textbf{left branch}\text{ of the parabola}$ $\text{Again, you should graph }x-2=\sqrt{y}\text{ and }x-2=-\sqrt{y}\text{ separately on GeoGebra or Desmos.}$ Why do we specifically need to use the left branch though? Sorry - I still don't understand :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 10:08:26 am Why do we specifically need to use the left branch though? Sorry - I still don't understand :P $\text{Sketch it firstly so that you can convince yourself.}\\ \text{If you need to, also draw up a table of values.}$ $\text{Geometrically, why this occurs is}\\ \text{Recall that }y=(x-2)^2\text{ is saying that }y\text{ is a function of }x\\ \text{But }x=-2\pm \sqrt{y}\text{ fails a 'horizontal line test'.}\\ \text{So }x\text{ is not a function of }y$ $\text{Indeed, }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\text{ split up are functions}\\ \text{In both cases (split up), }x\text{ is a function of }y.$ $\textit{Take care that the analysis slightly differs now that we're considering }x\textit{ in terms of }y\\ \textit{Not the usual vice-versa}$ $x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections of each other.}\\ \text{Note that a square root curve is just a parabola, with half of it chopped off.}\\ \text{Recall that }x^2=y\text{ is just }y=\sqrt{x}\text{ and its reflection }y=-\sqrt{x}.\text{ This works similarly here.}$ $\text{The reason why }y=\sqrt{x}\text{ and }y=-\sqrt{x}\text{ are reflections is because}\\ \textbf{they are the negative of each other.}\\ \text{Recall that the graph of the negative is just the same graph flipped about the }x\text{-axis.}$ $\text{Instead, here, we shift our analysis to the }y\text{-axis instead.}\\ x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections about the }y\text{-axis.}\\ x=\sqrt{y}\text{ represents what's to the right of the }y\text{-axis}\\ \text{whereas }x=-\sqrt{y}\text{ represents wha's to the left of the }y\text{-axis}$ $\text{The only difference is that we have }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\\ \text{i.e. an extra -2 floats around.}\\ \text{This just means that we use this analogy along }x=2\text{ and not }x=0(y\text{-axis})$ Title: Re: Mathematics Question Thread Post by: FallonXay on October 18, 2016, 10:20:43 am $\text{Sketch it firstly so that you can convince yourself.}\\ \text{If you need to, also draw up a table of values.}$ $\text{Geometrically, why this occurs is}\\ \text{Recall that }y=(x-2)^2\text{ is saying that }y\text{ is a function of }x\\ \text{But }x=-2\pm \sqrt{y}\text{ fails a 'horizontal line test'.}\\ \text{So }x\text{ is not a function of }y$ $\text{Indeed, }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\text{ split up are functions}\\ \text{In both cases (split up), }x\text{ is a function of }y.$ $\textit{Take care that the analysis slightly differs now that we're considering }x\textit{ in terms of }y\\ \textit{Not the usual vice-versa}$ $x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections of each other.}\\ \text{Note that a square root curve is just a parabola, with half of it chopped off.}\\ \text{Recall that }x^2=y\text{ is just }y=\sqrt{x}\text{ and its reflection }y=-\sqrt{x}.\text{ This works similarly here.}$ $\text{The reason why }y=\sqrt{x}\text{ and }y=-\sqrt{x}\text{ are reflections is because}\\ \textbf{they are the negative of each other.}\\ \text{Recall that the graph of the negative is just the same graph flipped about the }x\text{-axis.}$ $\text{Instead, here, we shift our analysis to the }y\text{-axis instead.}\\ x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections about the }y\text{-axis.}\\ x=\sqrt{y}\text{ represents what's to the right of the }y\text{-axis}\\ \text{whereas }x=-\sqrt{y}\text{ represents wha's to the left of the }y\text{-axis}$ $\text{The only difference is that we have }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\\ \text{i.e. an extra -2 floats around.}\\ \text{This just means that we use this analogy along }x=2\text{ and not }x=0(y\text{-axis})$ Ahh k, I see. thx :) Title: Re: Mathematics Question Thread Post by: kimmie on October 18, 2016, 10:28:28 am Already done in post #454 I checked it and I'm still confused why would we use 2 for the gradient thing? Title: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 10:33:07 am I checked it and I'm still confused why would we use 2 for the gradient thing? $\text{Like mentioned, 2 is a crucial number here.}\\ \text{2 and -2 are the gradients of }y=|2x-3|$ $\text{This implies that 2 and -2 are where the gradients become equal}\\ \text{This is critical because we are analysing when the lines are parallel.}\\\text{The instant lines become parallel and stop being parallel tells us a lot about the points of intersection.}$ $\text{Hence 2 and -2 become the basis for comparison,}\text{Utilise the GeoGebra simulation to have a better idea}$ Title: Re: Mathematics Question Thread Post by: kimmie on October 18, 2016, 10:46:13 am Oh I get it now thank you so much $\text{Like mentioned, 2 is a crucial number here.}\\ \text{2 and -2 are the gradients of }y=|2x-3|$ $\text{This implies that 2 and -2 are where the gradients become equal}\\ \text{This is critical because we are analysing when the lines are parallel.}\\\text{The instant lines become parallel and stop being parallel tells us a lot about the points of intersection.}$ $\text{Hence 2 and -2 become the basis for comparison,}\text{Utilise the GeoGebra simulation to have a better idea}$ Title: Re: Mathematics Question Thread Post by: kimmie on October 18, 2016, 10:48:01 am Can you help me with part (iv) and (v) please Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 11:36:30 am Can you help me with part (iv) and (v) please Hey! Cool question, I think this was from my year's paper :) We know that the equation for Carp population will be $P=10e^{0.2t}$ From the information given. This means that the rate of increase in Carp will be $\frac{dP}{dt}=0.2e^{0.02t}$ The rate of decrease in Trout is $\frac{dN}{dt}=-0.04e^{0.04t}$ Now, for part iv), we want these two rates to be equal! We absolute value them, because one is decreasing and one is increasing. So, $0.2e^{0.02t}=0.04e^{0.04t}$ $e^{0.02t-0.04t}=\frac{0.04}{0.2}$ $e^{-0.02t}=0.2$ $-0.02t=-1.6094$ $t=80.47 months$ For the next part, we just set the numbers equal to each other $10e^{0.2t}=375-e^{0.04t}$ This is a tricky solution. BOSTES give fairly straight forward answers to this part. (http://i.imgur.com/9pkfwAr.png) Let me know if I can clarify anything! Title: Re: Mathematics Question Thread Post by: kevin217 on October 18, 2016, 11:52:46 am Hi, could someone provide me a solution for the second derivative in i). Thanks Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 12:06:20 pm Hi, could someone provide me a solution for the second derivative in i). Thanks Hey! So we've got $x=t-(1+t)^{-1}$ To find the first derivative, we can just apply the chain rule to the second part $v=1-(-1)(1+t)^{-2}=1+(1+t)^{-2}$ To find the second derivative, we can just apply the chain rule again. $a=(-2)(1+t)^{-3}=-2*(1+t)^{-3}$ As t is always positive, (1+t) will always be positive, and so (1+t)^(-3) will always be positive. Therefore, -2*positive will always be negative, as required. Title: Re: Mathematics Question Thread Post by: kevin217 on October 18, 2016, 12:10:56 pm Hey! So we've got $x=t-(1+t)^{-1}$ To find the first derivative, we can just apply the chain rule to the second part $v=1-(-1)(1+t)^{-2}=1+(1+t)^{-2}$ To find the second derivative, we can just apply the chain rule again. $a=(-2)(1+t)^{-3}=-2*(1+t)^{-3}$ As t is always positive, (1+t) will always be positive, and so (1+t)^(-3) will always be positive. Therefore, -2*positive will always be negative, as required. Sweet, thanks for the solution. Title: Re: Mathematics Question Thread Post by: nimasha.w on October 18, 2016, 03:04:34 pm hey! could someone please explain how you would do this question :-) Title: Re: Mathematics Question Thread Post by: mfjw on October 18, 2016, 03:33:50 pm Can someone help me with this question, please? Title: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 03:37:52 pm Can someone help me with this question, please? $a^{-6}=\frac{1}{a^6}$ $5x^{-3}=\frac{5}{x^3}$ There. The indices are all positive now. Title: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 03:39:08 pm hey! could someone please explain how you would do this question :-) Im on a phone right now so I can't link but part i is already done. Edit: Page 37 post 549 or something Title: Re: Mathematics Question Thread Post by: mfjw on October 18, 2016, 03:41:43 pm I'm a bit stuck on this question right now. Could someone please explain how to do Q4g and Q4h? Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 03:43:04 pm I'm a bit stuck on this question right now. Could someone please explain how to do Q4g and Q4h? The answer is the question itself. It's called a trick question; it is already answered for you Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 03:43:24 pm I'm a bit stuck on this question right now. Could someone please explain how to do Q4g and Q4h? Remember that $x^{-a}=\frac{1}{x^a}$ So, $\frac{x^{-3}}{2}=\frac{1}{2x^3}$ I'll let you try h) Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 03:46:34 pm Remember that $x^{-a}=\frac{1}{x^a}$ So, $\frac{x^{-3}}{2}=\frac{1}{2x^3}$ I'll let you try h) I nearly walked into that. Jake, that index was already positive Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 03:49:18 pm I nearly walked into that. Jake, that index was already positive I know it looks positive, but I assumed the picture was just bad quality. Surely they wouldn't give an already-positive indice in numerous questions, as the image appears. Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 03:50:38 pm I know it looks positive, but I assumed the picture was just bad quality. Surely they wouldn't give an already-positive indice in numerous questions, as the image appears. Yeah it makes sense because there was a "space" there. Oh well. Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 03:51:52 pm Yeah it makes sense because there was a "space" there. Oh well. Either way, questioner get's a comprehensive response! Title: Re: Mathematics Question Thread Post by: zachary99 on October 18, 2016, 06:28:03 pm just did last years paper and scored 100% ;D Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100) Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 06:30:46 pm just did last years paper and scored 100% ;D Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100) Yes. Whilst it's extremely hard, it is possible to score 100. 1st place in 2U mathematics always gets 100. Title: Re: Mathematics Question Thread Post by: zachary99 on October 18, 2016, 06:43:09 pm Cheers! yeah it will be very hard Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 18, 2016, 07:33:25 pm just did last years paper and scored 100% ;D Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100) In case anyone is curious, the reason for this is because an ATAR of 100 means you beat 100% of students in the state. Including yourself ;) Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 07:36:57 pm Yep ditto with Jamon. Reminder that the ATAR is a rank, as opposed to the HSC giving you a set of marks. Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 18, 2016, 07:43:20 pm In case anyone is curious, the reason for this is because an ATAR of 100 means you beat 100% of students in the state. Including yourself ;) That's a fair point but in that case why stop at 99.95? why not go to 99.99? Title: Re: Mathematics Question Thread Post by: jakesilove on October 18, 2016, 08:52:06 pm That's a fair point but in that case why stop at 99.95? why not go to 99.99? Policy reasons; an ATAR is one a tool to limit university intake on a supply/demand basis. There's really no point scaling people about 99.95, since only like 40 people in the state get that mark anyway! Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 18, 2016, 08:53:00 pm Policy reasons; an ATAR is one a tool to limit university intake on a supply/demand basis. There's really no point scaling people about 99.95, since only like 40 people in the state get that mark anyway! ahh okay. Fair enough Title: Re: Mathematics Question Thread Post by: FallonXay on October 18, 2016, 09:01:36 pm Hiya all! I'm having some trouble understanding how to go about doing part v of this question. ??? Thanks for the help btw! Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:04:15 pm Hiya all! I'm having some trouble understanding how to go about doing part v of this question. ??? Thanks for the help btw! This was answered somewhere in the last two pages (on my phone right now so can't link) Title: Re: Mathematics Question Thread Post by: FallonXay on October 18, 2016, 09:05:34 pm This was answered somewhere in the last two pages (on my phone right now so can't link) ahh k, found it. cheers ;D Title: Re: Mathematics Question Thread Post by: bethjomay on October 18, 2016, 09:07:16 pm (http://uploads.tapatalk-cdn.com/20161018/ae3ca657354ac4d4ef18ca34ea27cb96.jpg) I have a couple of questions! For the first one, I've done part I) but I'm having trouble with the transformation to reach the answer in part ii) and a bit lost after that. In the second question I got all but part three. The answers mentioned 'favourable outcomes' which seems quite simple but I'm not familiar, could someone enlighten me? Thank you! (http://uploads.tapatalk-cdn.com/20161018/5809966f3ef60c285d37eff0bfca1a82.jpg)(http://uploads.tapatalk-cdn.com/20161018/cb463a433b62ebdeb078691f6b64554e.jpg) Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:42:38 pm (http://uploads.tapatalk-cdn.com/20161018/ae3ca657354ac4d4ef18ca34ea27cb96.jpg) I have a couple of questions! For the first one, I've done part I) but I'm having trouble with the transformation to reach the answer in part ii) and a bit lost after that. In the second question I got all but part three. The answers mentioned 'favourable outcomes' which seems quite simple but I'm not familiar, could someone enlighten me? Thank you! (http://uploads.tapatalk-cdn.com/20161018/5809966f3ef60c285d37eff0bfca1a82.jpg)(http://uploads.tapatalk-cdn.com/20161018/cb463a433b62ebdeb078691f6b64554e.jpg) $\text{The distance }AB\text{ is just}\\ \sqrt{(\alpha-\beta)^2+(\alpha^2-\beta^2)^2}\\ \text{So using the identity they gave us}\\ AB=\sqrt{(\alpha-\beta)^2[1+(\alpha+\beta)^2]]}$ $\text{At this point it should be very clear that }\alpha+\beta=m\text{ can be substituted in.}$ \text{It falls to transforming }(\alpha-\beta)^2\text{ into }(m^2+4b)\text{. Check this out.}\\ \text{By expanding and completing the square:}\\ \begin{align*}(\alpha-\beta)^2&=\alpha^2-2\alpha\beta+\beta^2\\ &= (\alpha^2+2\alpha\beta+\beta)^2-4\alpha\beta\\ &= (\alpha+\beta)^2-4\alpha\beta\\ &=m^2-4b\end{align*}\\ \text{Always look at what you're trying to prove to have a sense of direction.} Some hints now: For part (iii), recall your usual coordinate methods in geometry questions. You're given a length, and you want to find the area of a triangle with it. Usually, to use A=1/2 bh, we consider an associated perpendicular distance. (iv) is just your brute force max/min question Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 09:44:24 pm $\text{For that probability question. recall the standard probability formula.}\\ \text{Probability of an event} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$ $\text{The total outcomes is what you can get, without restrictions altogether.}\\ \text{The favourable outcomes are what you want.}\\ \text{Here, you want one winning ticket of each colour. It's how many possible ways this can be achieved.}$ Note that in this context, however, it's used to list the amount of favourable combinations of tickets. Title: Re: Mathematics Question Thread Post by: :3 on October 18, 2016, 10:33:20 pm 2007, Q10: having trouble understanding the answer. (https://i.gyazo.com/801fbb3e27932847d4000021a9557c6d.png) Thanks :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 18, 2016, 11:19:25 pm 2007, Q10: having trouble understanding the answer. (https://i.gyazo.com/801fbb3e27932847d4000021a9557c6d.png) Thanks :P $\text{Simpson's rule is just chucking it into formulae}\\ d\approx \frac{2}{3}(0+4(1)+5)=6$ ______________ $\text{Think about what decreasing and increasing mean.}\\ \text{If the particle's displacement is decreasing, then the first derivative (the velocity) must be }\textbf{negative}\\ \text{If the particle's displacement is increasing, then the first derivative must be }\textbf{positive}.$ $\text{Clearly, }v< 0\text{ when }t> 5\\ \text{That's when the displacement decreases.}$ ______________ $\textit{It becomes tricker here.}\\ \text{Note that just like the velocity is the first derivative of the displacement}\\ \text{The displacement is the }\textbf{integral}\text{ of the velocity.}$ $\text{Think about when }4 < t < 6\\ \text{When }4 < t < 5\text{, it's going to travel some distance away from the origin. Again.}\\ \text{But when }5 < t < 6\text{ it literally travels that exact same distance in the other direction.}$ $\text{Do you see why? This is because clearly the areas cancel out}\\ \int_4^6 v\, dt = 0\\ \text{Hence, displacement wise, overall the particle has gone nowhere at all in those 2 seconds.}$ $\text{It falls to do a similar analogy to the other area we haven't addressed.}\\ \text{i.e. }0< t < 4$ $\text{You estimated in (i), by Simpson's rule, that the area is 6.}\\ \text{So we want the area under the }t\text{-axis, ignoring the triangle we already used, to also be 6}\\ \text{That way those areas will cancel out as well.}$ $Q: \int_6^{???}v\, dt = -\int_0^4 v\, dt$ $\text{Well since that's just a rectangle, we have a formula. }A=\ell b\\ \text{Here, }b=5\\ \text{So }5T=6\implies T=1.2\\ \text{So recalling that we start at }t=6, 6+1.2=7.2\\ \text{Hence at around }7.2s$ Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 19, 2016, 12:18:09 am How do we solve these sort of questions?? Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 19, 2016, 12:22:35 am How do we solve these sort of questions?? Hey Fizzy! This one is a bit of intuition, a bit of Trial and Error, you are looking for a value of x on either side of the origin that puts as much area above the x-axis as below! This way, the areas cancel each other out and you get zero ;D 1 to -1: 2 square units above, 0 square units below 2 to -2: 3.5 square units above, 0 square units below 3 to -3: 3.75 square units above, 0.75 square units below 4 to 0 -4: 3.75 square units above, 2.75 square units below And I'll let you think about the value we'd need to have them match ;) note that I am finding these areas using basic area formula for rectangles and triangles! ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 12:26:07 am How do we solve these sort of questions?? Covered in post #464 or alternatively just use what Jamon said Title: Re: Mathematics Question Thread Post by: fizzy.123 on October 19, 2016, 01:00:39 am thank you both! Also, another question. how many solutions of the equation (sinx -1)(tanx +2) = 0 lie between 0 & 2pi How do we solve these sort of questions? Title: Re: Mathematics Question Thread Post by: ml125 on October 19, 2016, 04:06:25 am thank you both! Also, another question. how many solutions of the equation (sinx -1)(tanx +2) = 0 lie between 0 & 2pi How do we solve these sort of questions? $(\sin x-1)(\tan x+2)=0$ $\text{We first need to solve the two parts of the equation separately.}$ \begin{align*}\sin x-1&=0\\\sin x&=1\\x&=\frac{\pi}{2}\end{align*} $\text{This has one solution.}$ \begin{align*}\tan x+2&=0\\\tan x&=-2\end{align*} $\text{This has two solutions.}$ $\text{However, }x=\frac{\pi}{2}\text{ cannot be taken as a solution as }\tan\frac{\pi}{2}\text{ is undefined.}$ $\text{Therefore, there can only be two solutions! :)}$ Rui edit: Just fixed up some LaTeX for ya Title: Re: Mathematics Question Thread Post by: smile123 on October 19, 2016, 09:07:26 am Hey :) I was wondering if some one can help with this question part (iii) Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 09:18:45 am Hey :) I was wondering if some one can help with this question part (iii) $\text{This expression was found in part (ii)}\\ A_n=360000(1.005)^n - M\left(\frac{1.005^n-1}{0.005}\right)\\ \text{And we just found that }M=2319.50$ $\text{So we are trying to solve}\\ 180000=360000(1.005)^n - 2319.50\left(\frac{1.005^n-1}{0.005}\right)\\ \text{This is probably the harder of the bunch but still standard and you need to be able to do it.}$ \begin{align*}180000&=360000(1.005)^n-463900(1.005^n - 1)\\ 180000 &= 360000(1.005)^n -463900(1.005)^n +463900\\ -283900 &= -103900(1.005)^n\\ \frac{2839}{1039}&=1.005^n\\ \log \frac{2839}{1039}&=\log 1.005^n\\ n\log 1.005 &= \log \frac{2839}{1039}\\ n &= \frac{\frac{2839}{1039}}{\log 1.005}\end{align*} Title: Re: Mathematics Question Thread Post by: smile123 on October 19, 2016, 09:33:18 am thanks RuiAce :) that made things easier Title: Re: Mathematics Question Thread Post by: kimmie on October 19, 2016, 12:24:26 pm soz for barging in but can you explain why tanx= -2 has two solutions? im confused about that $(sinx-1)(tanx+2)=0$ $\text{We first need to solve the two parts of the equation separately.}$ \begin{align*}sinx-1&=0\\sinx&=1\\x&=\frac{\pi}{2}\end{align*} $\text{This has one solution.}$ \begin{align*}tanx+2&=0\\tanx&=-2\end{align*} $\text{This has two solutions.}$ $\text{However, }x=\frac{\pi}{2}\text{ cannot be taken as a solution as }tan\frac{\pi}{2}\text{ is undefined.}$ $\text{Therefore, there can only be two solutions! :)}$ Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 19, 2016, 12:31:28 pm soz for barging in but can you explain why tanx= -2 has two solutions? im confused about that No worries if you look at a graph of tanx like the one here, we can see that from 0 to $$2\pi$$, there are two locations where $$\tan{x}=-2$$ ;D it's a graphical approach that we take here (or you can use your ASTC rules to find the values exactly, if you want to) ;D Title: Re: Mathematics Question Thread Post by: rinagee12 on October 19, 2016, 03:54:49 pm Would someone please be able to help me with this question? :) Just with the last part, (iv). Title: Re: Mathematics Question Thread Post by: jakesilove on October 19, 2016, 04:01:41 pm Would someone please be able to help me with this question? :) Just with the last part, (iv). Hey! To find the total distance traveled, you would usually just think to integrate, right? However, you need to think about different sections separately. We already found that velocity equals zero at Pi/3. Before that, the velocity is negative, and after that, the velocity is positive. Therefore, it is travelling in opposite directions in these two sectors. To answer the question, just integrate between 0 and Pi/3, and the Pi/3 to Pi. Absolute value both solutions, and add them together! Let me know if you want a fully worked solution. Jake Title: Re: Mathematics Question Thread Post by: FallonXay on October 19, 2016, 04:14:33 pm 2014 Q7: How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ? The answers say 2, but shouldn't it be 3??? Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 04:35:03 pm 2014 Q7: How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ? The answers say 2, but shouldn't it be 3??? $\text{No. Because when }x=\frac{\pi}{2}\\ \sin x - 1 =0\\ \text{but }\tan x + 2 = \tan \frac{\pi}{2}+ 2 = \text{ you just broke maths}$ That was the trap in that question. Title: Re: Mathematics Question Thread Post by: FallonXay on October 19, 2016, 04:46:39 pm Hiya. How do you do b) part (ii) of this question? Thanks. :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 06:22:44 pm Hiya. How do you do b) part (ii) of this question? Thanks. :) $\text{You're just finding }A_{60}\\ \text{But the pattern is weird. Because the payments themselves also increase.}$ \begin{align*}A_1&= 500(1.003)\\ A_2&=[A_1+500(1.01)](1.003)\\ &=500(1.003)^2+500(1.01)(1.003)\\ A_3&=[A_2+500(1.01)^2](1.003)\\ &= 500(1.003)^3+500(1.01)(1.003)^2+500(1.01)^2(1.003)\\ A_4&=[A_3+500(1.01)^3](1.003)\\ &= 500(1.003)^4+500(1.01)(1.003)^3+500(1.01)^2(1.003)^2+500(1.003)(1.01)^3\end{align*} $\text{Try to continue the sequence from here.}\\ \text{The trick for the GP is that the common ratio is }(1.003)^{-1}(1.01)$ Title: Re: Mathematics Question Thread Post by: FallonXay on October 19, 2016, 07:33:28 pm With this question, how come the little 'triangle' section under the semi circle is shaded? Doesn't the region of the semicircle stop at the x axis? Thankyou! Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 07:45:24 pm With this question, how come the little 'triangle' section under the semi circle is shaded? Doesn't the region of the semicircle stop at the x axis? Thankyou! $\text{Note that the region }y\le \sqrt{9-x^2}\text{ is this ENTIRE region below the semi-circle.}$ (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsjszycfb4.png) $\text{Try to see why.}\\ \text{Hint: That's a positive root there. A negative number is always less than a positive number.}$ And then you're combining regions. $\text{Check a random point for confirmation, say, }(0,-2)\\ -2 \le \sqrt{9-0^2}\text{ TRUE}$ $\textit{Note that a circle isn't like a semi-circle. A circle is closed.}$ Title: Re: Mathematics Question Thread Post by: SimplyNikhil on October 19, 2016, 07:49:59 pm Hey, can someone please show how to do this question? Thanks in advance! Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 07:52:20 pm Hey, can someone please show how to do this question? Thanks in advance! \begin{align*}e^{1-\ln 2}&=e^1 \times e^{-\ln 2}\\ &= \frac{e}{e^{\ln 2}}\\ &= \frac{e}{2}\end{align*} $\text{Note that }e^{\ln x}=x\text{ is an identity which holds true for all }x> 0$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 07:54:34 pm Makes sense, but then why is the range for semicircles always stated between 0 and some number (i.e in this case stated as between 0<=y<=3 instead of just y<=3?) $\text{Range deals with what values can y }\textbf{equal}\\ \text{Not be less than}$ Title: Re: Mathematics Question Thread Post by: FallonXay on October 19, 2016, 07:56:51 pm $\text{Range deals with what values can y }\textbf{equal}\\ \text{Not be less than}$ Yeah, just realised. Thanks :P Title: Re: Mathematics Question Thread Post by: Rikahs on October 19, 2016, 08:11:51 pm Hi everyone, I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way. Thanks Title: Re: Mathematics Question Thread Post by: jakesilove on October 19, 2016, 08:17:55 pm Hi everyone, I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way. Thanks Personally, I would suggest that that you prove it another way, because it isn't a 'classic' 2U proof. As such, markers may not be happy with you 'assuming' things beyond the curriculum. Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 08:22:34 pm Hi everyone, I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way. Thanks Definitely not. The midpoint on a hypotenuse theorem is not assumed and you need to prove it from what you're given Title: Re: Mathematics Question Thread Post by: rahaf on October 19, 2016, 09:02:11 pm Hi! Can somebody help me with this please. Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 09:16:44 pm Hi! Can somebody help me with this please. Addressed in post #620 Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 10:09:28 pm Saw that but was wondering how is there 2 solutions??? I understand sinx-1=0 but how do you solve the tan one. How do you know it has two solutions? $\tan x = -2$ $\text{That is now just a trig equation to which you apply the rule of ASTC.}\\ \text{According to ASTC, tan is negative in the second and fourth quadrants.}$ $\text{So using methods you learnt in preliminary}\\ x = \pi - \tan^{-1}2, x = 2\pi - \tan^{-1}2\\ \text{or whatever your calculator says }\tan^{-1}2\text{ is.}$ Title: Re: Mathematics Question Thread Post by: Deng on October 19, 2016, 11:32:15 pm Hey guys, i was wondering if anyone could help give me iron out some stuff: e.g locus and parabola for y^2=4ax, is the directrix y=-a and the focus (x+ or - a, y). For some reason i cant grasp the y^2 side despite knowing how the x^2 side works This may be hard to explain but how would i graph velocity graphs ? I can work out the intercepts and roots, but i get stumped when wondering how to draw them and i dont necessarily have a question on hand The sum and product of roots and the variations outside of alpha + beta, alpha beta, there is one of them which i cant exactly remember ( which i know isnt helpful in trying to get help for) but its a more 'complex' one that requires a bit more algebraic manipulation Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on October 19, 2016, 11:41:33 pm Hey guys, i was wondering if anyone could help give me iron out some stuff: e.g locus and parabola for y^2=4ax, is the directrix y=-a and the focus (x+ or - a, y). For some reason i cant grasp the y^2 side despite knowing how the x^2 side works This may be hard to explain but how would i graph velocity graphs ? I can work out the intercepts and roots, but i get stumped when wondering how to draw them and i dont necessarily have a question on hand The sum and product of roots and the variations outside of alpha + beta, alpha beta, there is one of them which i cant exactly remember ( which i know isnt helpful in trying to get help for) but its a more 'complex' one that requires a bit more algebraic manipulation Thanks Recall that the parabola x^2=4ay has directrix y=-a and focus (0,a) Hence, the parabola y^2=4ax has directrix x=-a and focus (a,0) _________ Sorry but yes you will need to find a question. What you're saying isn't coming through. If you just meant derivative curve sketching that's somewhere in this thread already. ________ $\text{The weirdest one of the sum of roots is}\\ \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$ Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 12:02:55 am This example isnt the best example because i know how to graph this one, essentially the motion questions where i have to graph the velocity as part of it, also for part V for questions like that when would i know i would have to integrate or i can use t = 0;x=2 , t=1;x=4 etc Thanks again ! Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:12:29 am This example isnt the best example because i know how to graph this one, essentially the motion questions where i have to graph the velocity as part of it, also for part V for questions like that when would i know i would have to integrate or i can use t = 0;x=2 , t=1;x=4 etc Thanks again ! Still not sure what the problem is. If you're given the equation then like you said, you basically know how to draw it. You need to further clarify exactly what you mean by "you don't know how to sketch it". If you're given the actual equation of the velocity, you will virtually always know when and how to sketch it. That one is basically sketching y=3x2-9x+6 except with y switched for v and x switched for t _________________________ Part (v) can be made slightly easier with the answer to part (i). The idea is that the displacement takes into account direction, but the distance does not. So you need to watch out for whenever the particle turns around. Of course, you CAN use integration for part (v). You just don't need to. _________________________ Also, if that is a past HSC question, consider checking worked solutions. Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 12:42:18 am For the equation (y-k)^2= 4a(x-h)^2 Would the focus be (k + a, h) and the directix y= k -a So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght? Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:10:47 am For the equation (y-k)^2= 4a(x-h)^2 Would the focus be (k + a, h) and the directix y= k -a So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght? Thanks Recall that for the parabola (x-h)^2 = 4a(y-k) The vertex is at (h,k). Since the parabola concaves UP, the focus is a units ABOVE the vertex, and the directrix is a units BELOW the vertex. i.e. (h, k+a) and y=k-a I never memorised these formulae because there was no point. I used common sense So for the parabola (y-k)^2 = 4a(x-h) The vertex is still at (h,k). The parabola concaves to the RIGHT now. So the focus is a units TO THE RIGHT of the vertex, and the directrix is a units to the LEFT of the vertex i.e. (h+a, k) and x=h-a Title: Re: Mathematics Question Thread Post by: epherbertson on October 20, 2016, 09:19:21 am This is a question from the 2005 paper that no one in my class has figured out yet so just looking for a bit of help to conquer this mess. Thanks, Emily :-\ Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 10:47:53 am This is a question from the 2005 paper that no one in my class has figured out yet so just looking for a bit of help to conquer this mess. Thanks, Emily :-\ Already done in post #466 Title: Re: Mathematics Question Thread Post by: SimplyNikhil on October 20, 2016, 11:32:34 am ??? Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 11:48:15 am Could you please include each working step http://pasteboard.co/h2n5xrD5Q.png (https://cdn.pbrd.co/images/h2n5xrD5Q.png) Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 11:49:18 am @SimplyNikhil ( not sure how to tag your post ) But to find the area it would be top curve - bottom curve ln2x-lnx By index laws that would become ln2 Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve Hope i answered your question Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 11:50:34 am @SimplyNikhil ( not sure how to tag your post ) But to find the area it would be top curve - bottom curve ln2x-lnx By index laws that would become ln2 Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve Hope i answered your question Lol I was thinking that too but then I choked and was like what the fuck? do we integrate ln2 or divide now?? Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 11:52:11 am @BPunjabi How i solve these ( which may not be correct, Rui may correct me after if im wrong) Equation is Sin(2x+pi/3) To find the where it crosses the x axis let 2x+pi/3=0 2x = -pi/3 x = -pi/6 Therefore when we inspect A,B,C,D A and B are eliminated And between C and D , D is -pi/6 Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 11:56:03 am Hey guys, i was wondering on how i would identify the second derivative for this q The first derivative would be f'(a) < 0 since the gradient is negative I just cant picture in my head what the second derivate would look like ( some reason i picture a horizontal line since f(x) looks like a parabola to me Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:00:15 pm ??? This is partly wrong. Please wait for a correction. This question is actually much harder than what it looks like. $\text{You do not how to integrate }\ln x\\ \text{Hence, analyse the scenario with respect to the }y\text{-axis.}$ $\text{Firstly, rearranging the given equations:}\\ y=\ln x \implies x = e^y\\ y=\ln (2x) \implies x = \frac{1}{2}e^y$ $\text{Now, start considering the areas with respect to the }y\text{-axis}$\ Everything that's wrong starts from about here. But I'm not going to delete this post. I'm going to KEEP this mistake here, but also post up a new solution. (http://i1164.photobucket.com/albums/q566/Rui_Tong/tediousness_zpscmj0pqbu.png) $\text{Note that we want the area of the violent region ONLY.}\\ \text{The area of the violent PLUS the pink region given is}\\ A_1=\int_0^{\ln (2e)}\left(e^y-\frac{1}{2}e^y\right)dy\\ \text{which is left as your exercise.}$ $\textbf{But then we need to bust out that pink area that we did not want.}\\ \textbf{Note that this is a FOUR mark question. It is NOT meant to be simple.}$ $\text{The pink area is given by the area of the rectangle, deduct the unnecessary stuff.}\\ \text{Which is given by }A_2=\ell b - \int_0^{\ln 2}\frac12e^y dy\\ A_2=1\times \ln 2 - \int_0^{\ln 2} \frac{1}{2}e^y dy$ $\text{Your required answer is }A_2-A_1$ Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 12:05:41 pm @BPunjabi How i solve these ( which may not be correct, Rui may correct me after if im wrong) Equation is Sin(2x+pi/3) To find the where it crosses the x axis let 2x+pi/3=0 2x = -pi/3 x = -pi/6 Therefore when we inspect A,B,C,D A and B are eliminated And between C and D , D is -pi/6 Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me You are a fucking Beast, probs to you man!! Thanks so much, btw to reply click quote on the top right of every post if you want to reply to that message! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 20, 2016, 12:08:05 pm @BPunjabi How i solve these ( which may not be correct, Rui may correct me after if im wrong) Equation is Sin(2x+pi/3) To find the where it crosses the x axis let 2x+pi/3=0 2x = -pi/3 x = -pi/6 Therefore when we inspect A,B,C,D A and B are eliminated And between C and D , D is -pi/6 Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me Good method Deng! I'd personally suggest a more intuitive approach based on rules that you can find here, once you get the hang of doing it this way, purely for speed! ;D @SimplyNikhil ( not sure how to tag your post ) But to find the area it would be top curve - bottom curve ln2x-lnx By index laws that would become ln2 Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve Hope i answered your question This is also correct, and extremely clever, well done Deng ;D Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 12:12:28 pm At Rui for the area under the curve integral question Would this not give the correct working Area = integrate between e and 1 for ln2x-ln2 = ln 2 + ln x - ln x = ln2 ( constant there is no x, therefore integratable ) Therefore, it becomes xln2 between e and 1 eln2 - ln2 = ln2 ( e - 1 ) I'm not sure if the method you provided gave same answer because i dont understand it Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 12:13:39 pm Good method Deng! I'd personally suggest a more intuitive approach based on rules that you can find here, once you get the hang of doing it this way, purely for speed! ;D So Jamon, we are basically looking at (Bx+C) Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:14:15 pm Following from my mistake (http://i1164.photobucket.com/albums/q566/Rui_Tong/tediousness_zpsxuxhax8e.png) I oversimplified it. You actually have to analyse THREE regions, not just TWO. I have labelled them in the diagram for reference. $\text{For regions }A_1\text{ and }A_3\text{ you must analyse them using rectangle minus integral.}\\ \text{For region }A_2\text{ you must analyse it like an area between curves.}$ $\text{Effectively, we want }A=A_1+A_2+A_3\\ \text{Here are the relevant expressions.}$ \begin{align*}A_1&=\ell b - \int = (e-1)(\ln (2e)-1) - \int_1^{\ln (2e)}e^ydy\\ A_2&= \int_{\ln 2}^1\left(e^y-\frac12e^y\right)dy\\ A_3&=\int - \ell b = \int_0^{\ln 2}\frac{1}{2}e^ydy-(1)(\ln 2)\end{align*} Jamon please double check Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 20, 2016, 12:15:03 pm At Rui for the area under the curve integral question Would this not give the correct working ... I'm not sure if the method you provided gave same answer because i dont understand it Jamon please double check Yeah to reiterate above, Rui's method is overkill in this case (though I think correct), Deng your answer is definitely correct ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:15:42 pm At Rui for the area under the curve integral question Would this not give the correct working Area = integrate between e and 1 for ln2x-ln2 = ln 2 + ln x - ln x = ln2 ( constant there is no x, therefore integratable ) Therefore, it becomes xln2 between e and 1 eln2 - ln2 = ln2 ( e - 1 ) I'm not sure if the method you provided gave same answer because i dont understand it Ah. I missed the logarithm law. Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 12:16:21 pm anyone know how to do this? (https://cdn.pbrd.co/images/h2KkrmZyj.png) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 20, 2016, 12:16:34 pm So Jamon, we are basically looking at (Bx+C) Yeah precisely! ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:17:45 pm Hey guys, i was wondering on how i would identify the second derivative for this q The first derivative would be f'(a) < 0 since the gradient is negative I just cant picture in my head what the second derivate would look like ( some reason i picture a horizontal line since f(x) looks like a parabola to me Do not picture the second derivative in your head. Use definitions. The first derivative tells us whether it is increasing or decreasing. Since it is decreasing, f'(a) < 0 The second derivative tells us its concavity. Since it is concave up, f"(a) > 0 Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 20, 2016, 12:18:55 pm anyone know how to do this? (https://cdn.pbrd.co/images/h2KkrmZyj.png) Edit: Woops, dictation error The denominator of a function cannot be zero, so: $\sqrt{x+3}\ne0\implies x\ne-3$ But also we can't square root a negative number: $x+3\ge0\implies x\ge-3$ Put these together to obtain $$x>-3$$ ;D Title: Re: Mathematics Question Thread Post by: Deng on October 20, 2016, 12:37:07 pm For this question, if i did the maths right, the x value where it is increasing is 2/3 and 2, but how would i know if its 2/3<x<2 or if its split I've been told to draw a parabola but i dont particularly understand this method and just want more clarification Thanks Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:39:05 pm For this question, if i did the maths right, the x value where it is increasing is 2/3 and 2, but how would i know if its 2/3<x<2 or if its split I've been told to draw a parabola but i dont particularly understand this method and just want more clarification Thanks $\text{Note that }f^\prime (x) = (x-2)(3x-2)$ $\text{We are just solving the quadratic inequality }f^\prime(x) > 0\\ \text{i.e. }(x-2)(3x-2) > 0$ $\text{There are heaps of ways of doing this.}\\ \text{The most common way is to graph }y=f^\prime(x)\text{ and find when }y> 0\\ \text{You choose what's best for you.}$ Title: Re: Mathematics Question Thread Post by: BPunjabi on October 20, 2016, 12:42:23 pm Edit: Woops, dictation error The denominator of a function cannot be zero, so: $\sqrt{x+3}\ne0\implies x\ne-3$ But also we can't square root a negative number: $x+3\ge0\implies x\ge-3$ Put these together to obtain $$x>-3$$ ;D Thanks Jamon but the answer says a? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:43:46 pm Thanks Jamon but the answer says a? That's what he got. He said in the last line that hence the answer is x > -3 Title: Re: Mathematics Question Thread Post by: rinagee12 on October 20, 2016, 12:44:59 pm Hey I'm stuck on this question: For what values of k will the equation 2x^2 + 8kx - (1-k) = 0 have roots that differ by 2? My first thought was to use the formula α+β=-b/a and αβ=c/a I tried to replace 'β' with (α-2) but I'm not sure if I'm doing the right thing, I can't seem to get the answer Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:48:12 pm Hey I'm stuck on this question: For what values of k will the equation 2x^2 + 8kx - (1-k) = 0 have roots that differ by 2? My first thought was to use the formula α+β=-b/a and αβ=c/a I tried to replace 'β' with (α-2) but I'm not sure if I'm doing the right thing, I can't seem to get the answer $\text{By letting }\beta = \alpha - 2\text{ we have, sum of roots:}\\ \alpha + (\alpha-2)=-4k \implies \alpha = -2k+1$ $\text{Hence }\beta = -2k-1$ \text{So from product of roots:}\\ \begin{align*}(-2k+1)(-2k-1)&=\frac{k-1}{2}\end{align*} That should give the answer. Please show us your working if something happened from here onwards Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 12:49:54 pm $\text{By letting }\beta = \alpha - 2\text{ we have, sum of roots:}\\ \alpha + (\alpha-2)=-4k \implies \alpha = -2k+1$ $\text{Hence }\beta = -2k-1$ \text{So from product of roots:}\\ \begin{align*}(-2k+1)(-2k-1)&=\frac{k-1}{2}\end{align*} That should give the answer. Please show us your working if something happened from here onwards Would you also need to do a case where beta = alpha + 2? As the question only asks for 'differs'? Like I dunno if you'd get different answers, but I suppose you would. Conceptually, it shouldn't make a difference, but since I think it would change the answer... Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 12:52:06 pm Would you also need to do a case where beta = alpha + 2? As the question only asks for 'differs'? Like I dunno if you'd get different answers, but I suppose you would. Conceptually, it shouldn't make a difference, but since I think it would change the answer... Those cases should all be addressed at once though... If we do beta = alpha + 2 we just get beta = -2k+1 and alpha = -2k-1 Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 12:54:40 pm Those cases should all be addressed at once though... If we do beta = alpha + 2 we just get beta = -2k+1 and alpha = -2k-1 Yeah cool cool, just couldn't be bothered actually checking for consistency aha cheers Title: Re: Mathematics Question Thread Post by: onepunchboy on October 20, 2016, 03:16:54 pm (http://uploads.tapatalk-cdn.com/20161020/f16fe5c9672d37242bfd4693afcab75d.jpg) Hey guys can someone help me with part iii) and iv).. Thanks in adv !! Title: Re: Mathematics Question Thread Post by: imtrying on October 20, 2016, 03:18:01 pm Hey:) I'm sure this is fairly simple but i cant remember how to do it: A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced Thanks in advance:) Title: Re: Mathematics Question Thread Post by: MysteryMarker on October 20, 2016, 03:23:30 pm Hey guys, just a question from the 2005 HSC mathematics exam. I don't really understand this question, so i'd appreciate it if someone could explain how to do each part. Cheers :P Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 03:33:34 pm (http://uploads.tapatalk-cdn.com/20161020/f16fe5c9672d37242bfd4693afcab75d.jpg) Hey guys can someone help me with part iii) and iv).. Thanks in adv !! Hey! For this question, you need to think about the area under the curve. The area under the curve, when positive, is movement (displacement) in the positive direction. The area under the curve, when negative, is movement (displacement) in the negative direction. You need to find the point at which the area in the NEGATIVE portion is equal to the area in the POSITIVE portion. For the last part, just consider how integration works. When the gradient equals zero, (ie. dy/dx =0), what does that mean? Have a go, and if you still need help let me know! Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 03:41:00 pm Hey:) I'm sure this is fairly simple but i cant remember how to do it: A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced Thanks in advance:) Hey! We basically need to set up some sort of equation, and then differentiate to find a maximum value! Not a simple question at all. There are actually heaps of ways to do this. I would probably draw a circle, and throw in the angles to your rectangle. You can break the problem into four triangles, and four minor segments. You can easily find the area of the four segments, and then find the MINIMUM area of the four minor segments using calculus. Then, when you find this MINIMUM external area, you can subtract that from the area of the circle to find the MAXIMUM area of the rectangle! Give it a go, and let me know if you can't get a solution out. Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 03:42:06 pm Hey:) I'm sure this is fairly simple but i cant remember how to do it: A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced Thanks in advance:) Done in post #538 Hey guys, just a question from the 2005 HSC mathematics exam. I don't really understand this question, so i'd appreciate it if someone could explain how to do each part. Cheers :P Done in post #466 Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 20, 2016, 04:34:34 pm Done in post #538Done in post #466 How the hell do you remember these? ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 04:39:39 pm How the hell do you remember these? ;D (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsfzvianbw.png) I just have to remember a few words I typed and the fact I answered it before :P Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 05:53:00 pm Hello, can you please help me with the last question. I cannot understand properly... Thanks Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 05:55:20 pm Hello, can you please help me with the last question. I cannot understand properly... Thanks Check out my answer HERE. Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 06:04:08 pm Hey! Cool question, check out my solution below and let me know if I can clarify anything! (http://i.imgur.com/94nRCvj.png) Jake Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator Annotation by Jake of Annotation by Rui: Cheers buddy Thank You so muchh! Much more easier to understand :D Quick question... Why does root 2 go infront and root 1 becomes negative? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 06:09:10 pm Thank You so muchh! Much more easier to understand :D Quick question... Why does root 2 go infront and root 1 becomes negative? $\text{Look at the formula}\\ \frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}$ $\text{The formula he used puts the larger one in front of the smaller one.}$ $\text{Of course, if you insist on writing it as }-\sqrt{n}+\sqrt{n+1}\text{ that is not penalisable.}$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 06:13:41 pm $\text{Look at the formula}\\ \frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}$ $\text{The formula he used puts the larger one in front of the smaller one.}$ $\text{Of course, if you insist on writing it as }-\sqrt{n}+\sqrt{n+1}\text{ that is not penalisable.}$ Ohhh! Ok cool! :D Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 06:43:46 pm Can you please help me with the last question Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 06:49:26 pm Can you please help me with the last question Hey! So we can say that Alex's salary grows like this $33,000 \rightarrow 33000+1500 \rightarrow 33000+1500+1500$ If they save a third of their funds, then they will save $\frac{1}{3}(33,000+33000+1500+33000+1500+1500+....)$ $\frac{1}{3}(33,000+34,500+36,000+37,500+...)$ etc. Now, we want this value to equal87,500. You can use the sum of an arithmetic sequence (a formula you should definitely know!) to simplify the terms in brackets. Then, set that equal to 87,500 and solve for n! Let me know if you need a worked solution for this.

Jake
Post by: yasmineturner on October 20, 2016, 06:54:17 pm
Hi everyone,
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).
Post by: RuiAce on October 20, 2016, 06:56:46 pm
Hi everyone,
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).
Refer to post #522
Post by: jakesilove on October 20, 2016, 06:57:41 pm
Hi everyone,
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).

Hey! Welcome to the forums! You can find velocity by differentiating displacement, so

$\frac{dx}{dt}=2-t$
$\frac{dx}{dt}=-4cost$

To show that there are only two times where these velocities are the same, I would graph the two functions. There will only be two intercepts, thus showing that there are only two times in which the particles have the same velocity!

Annnnd I'm going to stop answering, because Rui is finding a solution he already wrote up. Bear with us! Good luck tomorrow :)
Post by: nibblez16 on October 20, 2016, 07:01:17 pm
Hey! So we can say that Alex's salary grows like this

$33,000 \rightarrow 33000+1500 \rightarrow 33000+1500+1500$

If they save a third of their funds, then they will save

$\frac{1}{3}(33,000+33000+1500+33000+1500+1500+....)$
$\frac{1}{3}(33,000+34,500+36,000+37,500+...)$

etc. Now, we want this value to equal 87,500. You can use the sum of an arithmetic sequence (a formula you should definitely know!) to simplify the terms in brackets. Then, set that equal to 87,500 and solve for n! Let me know if you need a worked solution for this. Jake Okay understandable! But what do we do with the 1/3? Thats where I am confused, how do we place it into formula? Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 07:11:09 pm Okay understandable! But what do we do with the 1/3? Thats where I am confused, how do we place it into formula? We'll end up getting something that looks like $\frac{1}{3}\frac{n(2a+d(n-1))}{2}=87,500$ Where we know d, and we know a. Therefore, it should be fairly straight-forward to solve for n! Title: Re: Mathematics Question Thread Post by: yasmineturner on October 20, 2016, 07:43:54 pm Hi again, I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 07:45:54 pm Hi again, I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou Both ∆ABC and ∆ACD are isosceles. This implies that we can use base angles on isosceles triangles twice. < ACD = < ABC (= common < A) Title: Re: Mathematics Question Thread Post by: yasmineturner on October 20, 2016, 07:48:32 pm Both ∆ABC and ∆ACD are isosceles. This implies that we can use base angles on isosceles triangles twice. < ACD = < ABC (= common < A) How do you do part iv? Thankyou :) Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 07:52:08 pm Hi again, I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou Hey! Firstly, note that y must be positive. We know that cos(theta) can be between -1 and 1, so actually, the maximum value of 2cos(theta) is 2. However, (1-2cos(theta))=1-2=-1 is negative, which is not allowed for the size of a line. If you let cos(theta)=-1, you get 2cos(theta)=-2. So, the maximum value is a(1-2cos(theta))=a(1-2(-1))=3a. y will always have to be less than that! This was really badly explained. Hope it made sense! Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 07:53:11 pm Oh apologies on misreading $\text{Part iv) is the very cliche but easy to miss:}\\ -1 \le \cos \theta \le 1\\ \text{Yes: This is just the range of cos.}$ $\text{Which implies }-1 \le -\cos \theta \le 1 \text{ as well.}$ \text{So by substituting that into our result from part (iii):}\\ \begin{align*}y&=a(1-2\cos \theta)\\ \therefore y&\le a(1+2)\\ y&\le 3a\end{align*} ...Ok don't worry Jake beat me to it Title: Re: Mathematics Question Thread Post by: MysteryMarker on October 20, 2016, 07:57:57 pm Hey, I came across this question and realised that I have absolutely no idea what simple interest is. Could someone explain to me the last part of this question, the parts before are fine just the last part. The answers used some PRT/100 = I formula, which i have never used lol. Cheers :P Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 08:05:57 pm Help please Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 08:09:16 pm Help please Hey! So, we need to set up a geometric series. We can easily see that $T_n=3x(-2x)^{n-1}$ So, which of the terms fit into this pattern? It's clearly looking at the 10th and 11th term (as that will get us a power of 10 and 11 for x). $T_{10}=(3x)(-2x)^9=-1536x^{10}$ Okay, not helpful $T_{11}=(3x)(-2x)^10=3072x^{11}$ So, the answer is C Title: Re: Mathematics Question Thread Post by: olivercutbill on October 20, 2016, 08:11:02 pm Hey, I came across this question and realised that I have absolutely no idea what simple interest is. Could someone explain to me the last part of this question, the parts before are fine just the last part. The answers used some PRT/100 = I formula, which i have never used lol. Cheers :P what year is this from? :) If i check I can help Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:14:55 pm Hey, I came across this question and realised that I have absolutely no idea what simple interest is. Could someone explain to me the last part of this question, the parts before are fine just the last part. The answers used some PRT/100 = I formula, which i have never used lol. Cheers :P $\text{That's just the formula for simple interest.}\\ I=PrT = \frac{PRT}{100}\\ \text{Note: }\frac{R}{100}=r$ To illustrate the r and R, for a rate of 6% p.a. R = 6 r = 0.06 $\text{The proof of the compound interest formula works}\\ \text{in that }T=1\text{ in the simple interest formula.}$ $\text{Consider a principal value }P\\ \text{It gains interest }I=PrT = Pr\\ \text{Hence, the amount after one period is }A = P+Pr = P(1+r)$ $\text{Now, the }P(1+r)\text{ gains interest }I=[P(1+r)]r\\ \text{Hence the amount after two periods is }A=P(1+r)+[P(1+r)]r = P(1+r)^2$ $\text{Following the pattern, after applying the formula }n\text{ times:}\\ A=P(1+r)^n$ _____________ That being said: To do that question, just deduct off the original principal and plug it into the formula. I = PrT = PRT/100 is assumed knowledge from Year 10. Title: Re: Mathematics Question Thread Post by: caninesandy on October 20, 2016, 08:15:42 pm Hey guys! :D I just have a quick question about an easy question but I always get confused... Solve 3 tan(x) + 1= 0 for 0° ≤ x ≤ 360° Are the answers: 150 and 210??? Thank you! :D Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 08:15:50 pm Hey! So, we need to set up a geometric series. We can easily see that $T_n=3x(-2x)^{n-1}$ So, which of the terms fit into this pattern? It's clearly looking at the 10th and 11th term (as that will get us a power of 10 and 11 for x). $T_10=(3x)(-2x)^9=-1536x^{10}$ Okay, not helpful $T_10=(3x)(-2x)^10=3072x^{11}$ So, the answer is C Alright! Thank You! Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ??? Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 08:17:56 pm Alright! Thank You! Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ??? Yeah, ratio should be between -1 and 1. Sounds like a weird question! Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:18:29 pm Alright! Thank You! Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ??? $\text{Yes. The limiting sum formula does not hold if }|r| >1\\ \text{That implies the question did something wrong otherwise.}$ Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 08:19:30 pm $\text{Yes. The limiting sum formula does not hold if }|r| >1\\ \text{That implies the question did something wrong otherwise.}$ This would imply that you can have a complex series with a limiting sum? Interesting.... Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:20:08 pm Hey guys! :D I just have a quick question about an easy question but I always get confused... Solve 3 tan(x) + 1= 0 for 0° ≤ x ≤ 360° Are the answers: 150 and 210??? Thank you! :D $\text{Unless you meant }\sqrt{3}\text{, that one has no tidy solution.}\\ \text{Because we're solving }\tan x = -\frac{1}{\sqrt 3}$ Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:21:00 pm This would imply that you can have a complex series with a limiting sum? Interesting.... What are you trying to pull here Jake lol I don't get it Title: Re: Mathematics Question Thread Post by: caninesandy on October 20, 2016, 08:31:41 pm $\text{Unless you meant }\sqrt{3}\text{, that one has no tidy solution.}\\ \text{Because we're solving }\tan x = -\frac{1}{\sqrt 3}$ Oh, yeah sorry I meant the root of 3 :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:34:32 pm Oh, yeah sorry I meant the root of 3 :) $\text{If it's }\sqrt{3}\text{ we have}\\ \tan x = -\frac{1}{\sqrt 3}$ $\text{Now, recall the rule of ASTC.}\\ \text{tan is positive in the first and THIRD quadrants.}\\ \text{Hence tan is negative in the second and FOURTH quadrants.}$ $\text{Working in degrees (more common in prelim, radians far more common in HSC)}\\ \text{We know that }\tan 30^\circ = \frac{1}{\sqrt3}\text{ as you correctly deduced.}$ $\text{The second and fourth quadrant angles are }180^\circ-\theta, 360^\circ-\theta\\ \text{Hence our answers are }150^\circ, 330^\circ$ Title: Re: Mathematics Question Thread Post by: caninesandy on October 20, 2016, 08:36:39 pm OHHH!! Thank you so much!!! :D ;D Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 08:41:09 pm What are you trying to pull here Jake lol I don't get it I suck at this part of series :P Help on last q Title: Re: Mathematics Question Thread Post by: olivercutbill on October 20, 2016, 08:43:17 pm $\text{That's just the formula for simple interest.}\\ I=PrT = \frac{PRT}{100}\\ \text{Note: }\frac{R}{100}=r$ To illustrate the r and R, for a rate of 6% p.a. R = 6 r = 0.06 $\text{The proof of the compound interest formula works}\\ \text{in that }T=1\text{ in the simple interest formula.}$ $\text{Consider a principal value }P\\ \text{It gains interest }I=PrT = Pr\\ \text{Hence, the amount after one period is }A = P+Pr = P(1+r)$ $\text{Now, the }P(1+r)\text{ gains interest }I=[P(1+r)]r\\ \text{Hence the amount after two periods is }A=P(1+r)+[P(1+r)]r = P(1+r)^2$ $\text{Following the pattern, after applying the formula }n\text{ times:}\\ A=P(1+r)^n$ _____________ That being said: To do that question, just deduct off the original principal and plug it into the formula. I = PrT = PRT/100 is assumed knowledge from Year 10. Can you do the solution? I think I was taught a different way... Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:46:52 pm I suck at this part of series :P Help on last q Refer to post #631 Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 08:47:19 pm Can you do the solution? I think I was taught a different way... I don't have all the answers to the previous parts Title: Re: Mathematics Question Thread Post by: imtrying on October 20, 2016, 09:03:10 pm Hi, just wondering if someone would be able to explain to me how to do part ii) of this question? Thank you! Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 09:06:02 pm Hi, just wondering if someone would be able to explain to me how to do part ii) of this question? Thank you! $\text{It's 1 mark because the maximum value of }-2\cos \theta \text{ is }2.\\ \text{Do you see why? This is because the range of }-\cos \theta\text{ is }-1 \le -\cos \theta \le 1$ $\text{So the max is }1+2 = 3$ Title: Re: Mathematics Question Thread Post by: jakesilove on October 20, 2016, 09:06:31 pm Hi, just wondering if someone would be able to explain to me how to do part ii) of this question? Thank you! This is just something that you need to be comfortable using $-1 The maximum value of velocity will clearly occur when cos(theta)=-1, thus $v=1-2(-1)=3$ Title: Re: Mathematics Question Thread Post by: imtrying on October 20, 2016, 09:17:07 pm This is just something that you need to be comfortable using $-1 The maximum value of velocity will clearly occur when cos(theta)=-1, thus $v=1-2(-1)=3$ $\text{It's 1 mark because the maximum value of }-2\cos \theta \text{ is }2.\\ \text{Do you see why? This is because the range of }-\cos \theta\text{ is }-1 \le -\cos \theta \le 1$ $\text{So the max is }1+2 = 3$ Thanks, that helps:) Definitely wasn't thinking that one through properly! Title: Re: Mathematics Question Thread Post by: Rikahs on October 20, 2016, 09:33:55 pm Thanks, that helps:) Definitely wasn't thinking that one through properly! Btw, there is an easy method typically known by 3u or 4u students, maximum velocity occurs when the acceleration is zero; if you just differentiat velocity then make it equal to zero to find t then just sub back into velocity you get max velocity. May seem a bit more confusing or may make you go like "oh yeahhhh I see" but it can sometimes make life easier especially if you get ugly equations. Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 09:36:10 pm Btw, there is an easy method typically known by 3u or 4u students, maximum velocity occurs when the acceleration is zero; if you just differentiat velocity then make it equal to zero to find t then just sub back into velocity you get max velocity. May seem a bit more confusing or may make you go like "oh yeahhhh I see" but it can sometimes make life easier especially if you get ugly equations. That works in general. 2U are taught this. It just so happens that this is a 1 mark question and thus that method is not necessary because of the trick. Title: Re: Mathematics Question Thread Post by: Rikahs on October 20, 2016, 09:40:42 pm Wait I just realized something, did you do ur 2u hsc in year 10? Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 09:46:48 pm The whole reference sheet is in the exam right? Coz ive been hearing rumours that theyre taking out integrals part... Just confirming that the rumour is a rumour lol Title: Re: Mathematics Question Thread Post by: Rikahs on October 20, 2016, 09:48:05 pm Pretty damn sure that is a rumour, they can't just take out the integral section without telling the students. Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 09:55:09 pm Pretty damn sure that is a rumour, they can't just take out the integral section without telling the students. Yea! I guess people are making up stuff, I mean HSC is already stressful, y give more surprises lol Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 09:56:47 pm Wait I just realized something, did you do ur 2u hsc in year 10? Yes The whole reference sheet is in the exam right? Coz ive been hearing rumours that theyre taking out integrals part... Just confirming that the rumour is a rumour lol Pretty damn sure that is a rumour, they can't just take out the integral section without telling the students. Yea! I guess people are making up stuff, I mean HSC is already stressful, y give more surprises lol Only 2 or 3 of the standard integrals are not on the reference sheet. Every other standard integral has been transferred The 2 (or 3) that have been taken out are also useless for 2U AND 3U. Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 10:12:00 pm Heyy! How do we know the procedure to solve this? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 10:21:28 pm Heyy! How do we know the procedure to solve this? $\text{When you see "at least", you should be thinking the complement}\\ \text{That is, the probability of what we DON'T want.}$ $\text{What we don't want is for neither of them to win.}\\ \text{The probability of this is }\frac{8}{9}\times \frac{15}{16}=\frac{5}{6}$ $\text{Hence the probability at least one of them wins is }\frac16$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 10:24:12 pm $\text{When you see "at least", you should be thinking the complement}\\ \text{That is, the probability of what we DON'T want.}$ $\text{What we don't want is for neither of them to win.}\\ \text{The probability of this is }\frac{8}{9}\times \frac{15}{16}=\frac{5}{6}$ $\text{Hence the probability at least one of them wins is }\frac16$ The answer is different, its 91/216 Title: Re: Mathematics Question Thread Post by: emilyf on October 20, 2016, 10:27:26 pm Sorry if this has been answered previously but could someone help me out with this question please? The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i). For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead?? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 10:31:07 pm The answer is different, its 91/216 $\text{I was wondering what they meant when they said 3 weeks.}\\ \text{The question is a bit stupid in that it doesn't say that those are the probabilities for ONE week.}$ $\text{Since the probability of no wins in 1 week is }\frac{5}{6}\\ \text{The probability of no wins in 3 weeks is }\frac{125}{216}$ $\text{So our required answer is }1-\frac{125}{216}=\frac{91}{216}$ Title: Re: Mathematics Question Thread Post by: :3 on October 20, 2016, 10:33:24 pm Sorry if this has been answered previously but could someone help me out with this question please? The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i). For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead?? tan(x) is positive in the first and 3rd quadrant. Therefore, the respective values for x are 60 (π/3) and 180 +60 (240, 4π/3; not 2π/3). Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 10:33:31 pm Sorry if this has been answered previously but could someone help me out with this question please? The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i). For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead?? You're looking at the wrong quadrants. Tan is positive in the first and THIRD quadrants, not first and second. (Rule of ASTC.) So the valuse are π/3, and π + π/3, which is 4π/3 Not π/3 with π - π/3 , which is 2π/3 Edit: Done above Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 10:36:08 pm $\text{I was wondering what they meant when they said 3 weeks.}\\ \text{The question is a bit stupid in that it doesn't say that those are the probabilities for ONE week.}$ $\text{Since the probability of no wins in 1 week is }\frac{5}{6}\\ \text{The probability of no wins in 3 weeks is }\frac{125}{216}$ $\text{So our required answer is }1-\frac{125}{216}=\frac{91}{216}$ Okay! But y do we cube it? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 10:37:21 pm Okay! But y do we cube it? They give you the probabilities for 1 week. You want the probability for a 3 week period. Not just 1. So you want the same event to happen 3 times. It's like saying I lost, and then I lost again, and then yet I lost one more time. So if the probability I lose is, say, 1/2 The probability I lose 3 times is 1/2 times 1/2 times 1/2, which is 1/8 Successive events, if you will Title: Re: Mathematics Question Thread Post by: nibblez16 on October 20, 2016, 11:18:24 pm Does anyone have prelim formula sheet? Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 11:25:18 pm Does anyone have prelim formula sheet? What's a prelim formula sheet? You're going to be given that same reference sheet as you have been tomorrow Title: Re: Mathematics Question Thread Post by: :3 on October 20, 2016, 11:47:31 pm For iii), I'm unsure as to why AP = AC (according to the answer, its because of part II). (https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png) Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 20, 2016, 11:54:39 pm For iii), I'm unsure as to why AP = AC (according to the answer, its because of part II). (https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png) Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P AP is not equal to AC. AP/AC = 1/2 according to (ii) Title: Re: Mathematics Question Thread Post by: :3 on October 21, 2016, 12:01:55 am AP is not equal to AC. AP/AC = 1/2 according to (ii) Here are the answers from BOSTE 2009 HSC exam. (https://i.gyazo.com/f78e5a0f152aba594d4c4502e6e9cf27.png) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 12:02:36 am Here are the answers from BOSTE 2009 HSC exam. (https://i.gyazo.com/f78e5a0f152aba594d4c4502e6e9cf27.png) That's CP. Not AC. Title: Re: Mathematics Question Thread Post by: Rikahs on October 21, 2016, 12:04:16 am For iii), I'm unsure as to why AP = AC (according to the answer, its because of part II). (https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png) Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P First of all AC= 2AP Then for iii) I would use congruent triangles, my proof: In triangle AMP and traingle MPC MP is common AC = 2AP ( from ii)) Since AC = AP + PC therefore AP = PC Angle MPA = Angle MPC = 90 (MP bisects AC) Therefore Triangle AMP is congruent to Triangle MPC (SAS) Therefore Angle MAP = Angle MCP ( corresponding angles in congruent triangles) Therefore Triangle AMC is Isos ( base angles are equal) Title: Re: Mathematics Question Thread Post by: :3 on October 21, 2016, 12:11:40 am Thanks and sorry Rui and Rikahs, My mind totally blanked out (probably because its geometry + late at night). Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 12:17:35 am It's past midnight. Time to get a good night sleep Title: Re: Mathematics Question Thread Post by: Hua Fei on October 21, 2016, 12:27:10 am Hi there! I don't understand how to do these questions: 1) the answer is 4.5. I don't even know how to approach this... 2) the answer is 2. Wouldn't it be 3? Sin gives us one solution and tan gives us the 2nd? Thank you so much in advance! Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 12:39:30 am Hi there! I don't understand how to do these questions: 1) the answer is 4.5. I don't even know how to approach this... 2) the answer is 2. Wouldn't it be 3? Sin gives us one solution and tan gives us the 2nd? Thank you so much in advance! Addressed in post #464 Addressed in post #620 $\sin x -1 = 0\text{ appears to give }x=\frac{\pi}{2}\text{ but plugging back in, }\tan \frac{\pi}{2}\text{ is undefined.}$ Title: Re: Mathematics Question Thread Post by: Hua Fei on October 21, 2016, 12:44:40 am Addressed in post #464 Addressed in post #620 $\sin x -1 = 0\text{ appears to give }x=\frac{\pi}{2}\text{ but plugging back in, }\tan \frac{\pi}{2}\text{ is undefined.}$ Thank you! You've really helped me out :) Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 01:04:51 am Help please last part Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 01:08:53 am Help please last part $\text{The gradient of }AB\text{ is }-2\\ \text{So using the rule }m_1m_2=-1\text{ for perpendicular lines}\\ \text{You want the gradient of }CN\text{ to be }\frac{1}{2}$ $\text{Then since you know the coordinates of }C\\ \text{Just use your }\frac{\text{rise}}{\text{run}}\text{ formula}$ $\text{Let the coordinates of }N\text{ be }(x,y)\\ \text{Then }\frac{y-4}{x-7}=\frac{1}{2}\\ 2y-1=x$ $\text{So we have the equation of }CN:\, x=2y-1\\ \text{And the equation of }AB: \, 2x+y+8\\ \text{Since }N\text{ is the point of intersection, just solve these equations simultaneously.}$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 06:37:03 am Good Morning :) For a question like (iii) we have to differentiate, so can be times the power 'k' with the whole number? In this case k= 0.0347 (4dp) but the whole number is 0.03465735903 Can we times it with the whole number? because the answer multplies it with only 4dp.... Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 08:03:19 am Good Morning :) For a question like (iii) we have to differentiate, so can be times the power 'k' with the whole number? In this case k= 0.0347 (4dp) but the whole number is 0.03465735903 Can we times it with the whole number? because the answer multplies it with only 4dp.... Morning! You would be paid for either, but as a rule of thumb, if you find a rounded value in an earlier part of the question, carry it through the rest of the questions! It is cleaner and still yields the required level of accuracy ;D Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 08:34:58 am Morning! You would be paid for either, but as a rule of thumb, if you find a rounded value in an earlier part of the question, carry it through the rest of the questions! It is cleaner and still yields the required level of accuracy ;D Okay Thanks! Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 08:36:35 am How do we know when to use a discriminant and which one to use (>, <, =) Title: Re: Mathematics Question Thread Post by: zemilyx on October 21, 2016, 09:07:26 am hi with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why. Thanks :)) Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 09:09:43 am hi with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why. Thanks :)) Its too late now, dont worry just wing it! Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 09:13:21 am How do we know when to use a discriminant and which one to use (>, <, =) Its too late now just wing it ;D. But here: 1. if triangle is <0, roots are real, rational and equal 2. If triangle is =0 roots are unreal and unrational 3. If triangle is >0 roots are unreal, irrational and unequal Title: Re: Mathematics Question Thread Post by: WLalex on October 21, 2016, 09:13:32 am hi with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why. Thanks :)) Do not count 2007 for interest as the 5000 is deposited at the beginning therefore the P is only gaining interest until 2006 e.g. 7 times not 8 (8 would be at the end of 2007) Title: Re: Mathematics Question Thread Post by: WLalex on October 21, 2016, 09:15:06 am Its too late now just wing it ;D. But here: 1. if triangle is <0, roots are real, rational and equal 2. If triangle is =0 roots are unreal and unrational 3. If triangle is >0 roots are unreal, irrational and unequal If the discriminate is less than 0 there is no real roots... = 0 is 1 real root > 0 is 2 real, unequal roots > and equal to means real roots Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 09:25:51 am Its too late now, dont worry just wing it! 5 hours is heaps of time to get some things clarified mate :) If the discriminate is less than 0 there is no real roots... = 0 is 1 real root > 0 is 2 real, unequal roots > and equal to means real roots This is the correct breakdown. Thanks WLalex! ;D and for your response above, awesome stuff ;D Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:26:36 am How do we know when to use a discriminant and which one to use (>, <, =) Note that f'(x) ≥ 0 means monotonic increasing f'(x) ≤ 0 means monotonic decreasing f'(x) = 0 for stationary points f'(x) > 0 means strictly increasing f'(x) < 0 means strictly decreasing When they don't say monotonic, assume strictly Title: Re: Mathematics Question Thread Post by: WLalex on October 21, 2016, 09:28:04 am Note that f'(x) ≥ 0 means monotonic increasing f'(x) ≤ 0 means monotonic decreasing f'(x) = 0 for stationary points f'(x) > 0 means strictly increasing f'(x) < 0 means strictly decreasing When they don't say monotonic, assume strictly so on top of this for clarity you will have to differentiate f(x) and make the discriminate less than 0 (as it will have no roots) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:29:18 am Hi, my answers say that d.) is correct but how did they get to this point? \text{This is just a geometric series.}\\ \begin{align*}\sum_{k=3}^{15}2^k&=8+16+32+64+\dots+2^{15}\\ &= \frac{8(2^{13}-1)}{2-1}\\ &= 8(2^{13}-1)\\ &= 2^{16}-8\end{align*} Title: Re: Mathematics Question Thread Post by: WLalex on October 21, 2016, 09:30:56 am Hi, my answers say that d.) is correct but how did they get to this point? I would do the calculation the normal way (just add them all together) and then see which answer matches your answers via calculation Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:31:24 am so on top of this for clarity you will have to differentiate f(x) and make the discriminate less than 0 (as it will have no roots) In theory, if we were to fully complete the question Not only do we want ∆ < 0, we want the leading coefficient to be positive. We basically want f'(x) to be positive definite. Because if the leading coefficient were negative, and ∆ < 0, we would be strictly DECREASING. However, it's easy to check that 1 > 0 and you are certainly right in using the discriminant here. Because f'(x) is indeed quadratic. Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:32:03 am I would do the calculation the normal way (just add them all together) and then see which answer matches your answers via calculation Takes far too long. That's what you do after you've finished the paper and it's time to check your answers, not during. There are 13 terms to add up in the sum. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 09:36:04 am Takes far too long. That's what you do after you've finished the paper and it's time to check your answers, not during. There are 13 terms to add up in the sum. Eh, when in doubt go back to basics; the techniques are obviously more effective. But if you have a choice between spending 1 minute getting an answer you are 50% confident with, and 3 minutes getting an answer you KNOW is correct, then why not (assuming you know you've got the time). Plus, doing the calculator work for this with 13 terms isn't actually that bad :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:39:26 am Eh, when in doubt go back to basics; the techniques are obviously more effective. But if you have a choice between spending 1 minute getting an answer you are 50% confident with, and 3 minutes getting an answer you KNOW is correct, then why not (assuming you know you've got the time). Plus, doing the calculator work for this with 13 terms isn't actually that bad :P Even at the fastest typing speed on the calculator that still takes me a full 45 seconds to get through all the options whereas I could've evaluated it in 15-25 :P They say to check your answers for reasons - get an answer on the paper first, and then calm the paranoia down when you find time to. Of course, if you have a stress attack though then yeah that's when you should just go back to basics or you'll be overly paranoid Title: Re: Mathematics Question Thread Post by: WLalex on October 21, 2016, 09:49:58 am Even at the fastest typing speed on the calculator that still takes me a full 45 seconds to get through all the options whereas I could've evaluated it in 15-25 :P They say to check your answers for reasons - get an answer on the paper first, and then calm the paranoia down when you find time to. Of course, if you have a stress attack though then yeah that's when you should just go back to basics or you'll be overly paranoid Yep valid point Rui but since it was a simple calculation thats just where my mind went :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 09:51:24 am Yep valid point Rui but since it was a simple calculation thats just where my mind went :) At least it's 100% not penalisable Title: Re: Mathematics Question Thread Post by: lha on October 21, 2016, 09:58:31 am You know those velocity graphs or acceleration graphs and it says describe the motion of the particle at t=whatever? Can someone give me a run down on how to do that please! Title: Re: Mathematics Question Thread Post by: kavinila on October 21, 2016, 09:59:58 am hey guys! i was just wondering if someone could please quickly run through locus? i don't think i understand it and i just noticed it in the syllabus 😳 please and thankyouu!! Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:05:05 am hey guys! i was just wondering if someone could please quickly run through locus? i don't think i understand it and i just noticed it in the syllabus 😳 please and thankyouu!! Find the section on locus Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:09:04 am You know those velocity graphs or acceleration graphs and it says describe the motion of the particle at t=whatever? Can someone give me a run down on how to do that please! Well if you need to describe the motion based off what you have. Suppose you have the graph of the displacement. Velocity is the first derivative of the displacement Therefore if the graph is increasing, velocity is positive If graph is decreasing, velocity is negative If graph is at a stationary point, particle is at rest Acceleration is the second derivative of the displacement Therefore if the graph is concave up, acceleration is positive If the graph is concave down, acceleration is positive If we are at a point of inflexion, acceleration is constant (possibly 0 but we don't know). And say we were given the graph of the velocity. Velocity is the first derivative Hence v > 0 means particle is travelling away from the origin (+'ve direction, to the right, however you memorised it) v < 0 means particle is travelling in the other direction from the origin (-'ve direction, to the left etc.) And those are just some examples. If you know what the derivatives actually MEAN then this is more intuitive than trying to rote learn a few dot points. Title: Re: Mathematics Question Thread Post by: onepunchboy on October 21, 2016, 10:10:38 am Hey anyone know how to find the asymptotes for tan graphs? For normal tanx graph asymptote is at pi/2 and 3pi/2 but what if it was tan(1/2 x) or tan (2x) how to find new asymptote.. thanks Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:12:16 am Hey anyone know how to find the asymptotes for tan graphs? For normal tanx graph asymptote is at pi/2 and 3pi/2 but what if it was tan(1/2 x) or tan (2x) how to find new asymptote.. thanks $\text{Analyse it.}\\ \tan \frac{\pi}{2}\text{ is undefined.}\\ \text{Hence, what value of }x\text{ will turn }\tan \frac{x}{2}\text{ into }\tan \frac{\pi}{2}?\\ \text{The answer is }\pi.\text{ When }x=\pi, \tan \frac{x}{2}=\tan \frac{\pi}{2}\\ \text{So }\tan\frac{x}{2}\text{ has an asymptote at }x=\pi$ $\text{Similarly, it can be shown that }\tan 2x\text{ has an asymptote at }x=\frac{\pi}{4}$ Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 10:13:23 am How do we know how to solve the last 2 questions Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 10:16:51 am How do we know how to solve the last 2 questions Hey! For the second part, you need to think about the cases. Pat can win on the first turn (probability same as i) ), or Pat can win on the second turn. If he wins on the second turn, it will be (1-P(win on first turn))*P(win). Then, we add up these two cases! You need to create some sort of 'series' which you can add up in a sum-to-infinity type equation for the last part. Can you see how? Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 10:20:31 am Guys have confidence, we will smash it! Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 21, 2016, 10:21:41 am Guys have confidence, we will smash it! Really good attitude to have :) :) Title: Re: Mathematics Question Thread Post by: Rikahs on October 21, 2016, 10:26:01 am Hi, I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero. Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it. Thanks!!! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 10:27:12 am Hi, I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero. Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it. Thanks!!! Confirming, monotonic means $$f'(x)\ge0$$ ;D Title: Re: Mathematics Question Thread Post by: Rikahs on October 21, 2016, 10:27:27 am Hi again, What kind of question would want you to make the discriminant greater and equal to zero? Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:27:35 am Hi, I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero. Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it. Thanks!!! Ditto to Jamon. Monotonic means it can equal. STRICTLY means it can't equal. Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:28:19 am Hi again, What kind of question would want you to make the discriminant greater and equal to zero? The discriminant is positive OR zero when ALL you care about is real roots. When it's strictly positive (greater than zero), i.e. >, it means your two real roots must be DISTINCT. Title: Re: Mathematics Question Thread Post by: nibblez16 on October 21, 2016, 10:28:37 am Hey! For the second part, you need to think about the cases. Pat can win on the first turn (probability same as i) ), or Pat can win on the second turn. If he wins on the second turn, it will be (1-P(win on first turn))*P(win). Then, we add up these two cases! You need to create some sort of 'series' which you can add up in a sum-to-infinity type equation for the last part. Can you see how? Its still confusing... Title: Re: Mathematics Question Thread Post by: g98 on October 21, 2016, 10:33:53 am Hi, I have a couple of questions: 1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either? 2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way? 3. What are acceptable abbreviations for proving geometrical questions? 4. do we leave our answer in improper fraction or mixed numeral - if not specified in question? 5. If not specified in question how many decimals should you round to? Thank You! Title: Re: Mathematics Question Thread Post by: Belindka on October 21, 2016, 10:35:22 am Hey! Just wondering, would changing our calculators to radians by default change our calculations for other answers that don't need radians? I always have mine in degrees and change it to radians whenever i need it ??? Thanks! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 10:35:47 am Its still confusing... All good! So: $P(\text{Pat Wins First Throw})=\frac{1}{36}$ For Pat to win on the second throw, he has to not win on his first throw, and then win on his second: $P(\text{Pat Wins Second Throw}) = \frac{35}{36}\times\frac{1}{36}$ So the answer to B is: $\frac{1}{36}+\frac{35}{36}\frac{1}{36}$ If we went to three throws, it would be similar; he needs to lose TWO throws and then win on his third: $\frac{1}{36}+\frac{35}{36}\frac{1}{36}+\left(\frac{35}{36}\right)^2\frac{1}{36}$ See the pattern? We have a geometric series forming, $$a=\frac{1}{36}$$ and $$r=\frac{35}{36}$$ ;D Now this value of r lets us take an infinite series and get a value; so use that formula for infinite series from your reference sheet with those values, and that is how you answer Part C ;D Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 21, 2016, 10:38:43 am Hi, I have a couple of questions: 1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either? 2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way? 3. What are acceptable abbreviations for proving geometrical questions? 4. do we leave our answer in improper fraction or mixed numeral - if not specified in question? 5. If not specified in question how many decimals should you round to? Thank You! From what I have been told by my teachers if the question doesn't specify you should always leave it in exact form but they probably won't mark you down for putting it into decimal form Again with the second question, if its not specifies then they won't take marks of for how you have left it but they usually always want to simplified answer which mean taking out common factors and factorizing. For the third question I've heard mixed fractions are better but again if not specified either works And for your last question I would personally round to about 3 or 4 to keep accuracy. Hope this helps Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 10:38:57 am Hi, I have a couple of questions: 1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either? 2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way? 3. What are acceptable abbreviations for proving geometrical questions? 4. do we leave our answer in improper fraction or mixed numeral - if not specified in question? 5. If not specified in question how many decimals should you round to? Thank You! 1. If it doesn't specify and you have the capability, exact form. 2. Factorising your final answer is something you can do, but it is unnecessary; you won't get marked down as long as you factorise don't divide 3. $$\Delta$$ for a triangle, \\ for parallel lines, and the congruency and similarity symbols to name a few. Try not to abbreviate words like "isosceles" unless you are short on time (you will probably be fine even if you do) 4. Improper Fraction is nicer imo, but BOSTES does use mixed numeral at times. Either or :) 5. If not specified, pick a sensible number. General rule is the most amount of decimal places used in the question (but then I usually just add one to that just to be safe) :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 10:40:32 am Some chaos occurs with mixed numerals. Don't smack me Jamon but I need to emphasise your point on this one :P preferably improper fractions please. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 10:40:47 am Hey! Just wondering, would changing our calculators to radians by default change our calculations for other answers that don't need radians? I always have mine in degrees and change it to radians whenever i need it ??? Thanks! Yes, you need to be careful about this because it does affect your answers for, say, geometry questions that use degrees. Leave your calculator in radians, then swap to degrees as soon as you see the degree symbol, then swap back as soon as that question is done. That's what I recommend since radians is the default unit of angular measure ;D Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 10:41:43 am Some chaos occurs with mixed numerals. Don't smack me Jamon but I need to emphasise your point on this one :P preferably improper fractions please. Happy to be emphasised; it does create room for more error! ;D (although BOSTES uses them in sample solutions, I never did, and was fine. Don't create more work for yourself) :) Title: Re: Mathematics Question Thread Post by: rinagee12 on October 21, 2016, 10:58:02 am Good luck to everyone who is sitting the exam today ;D (although I'm probably the one that's going to need it ;) ) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 11:00:46 am Good luck to everyone who is sitting the exam today ;D (although I'm probably the one that's going to need it ;) ) Have faith, don't imply that about yourself Title: Re: Mathematics Question Thread Post by: isaacdelatorre on October 21, 2016, 11:20:51 am Hey guys, Just a question with rates of change. If t equals something like 152.65 months, what is your final answer if it says when does P = x, occur. Do we leave it as a decimal? Or round to a month? Title: Re: Mathematics Question Thread Post by: MysteryMarker on October 21, 2016, 11:21:17 am Hey guys, this is more just a rule that i'm asking, but I remember that when there are parallel lines and two transversal intersect them, the ration between the two transversals are equal or something? Could someone just rephrase that properly or enlighten me on whether that is the correct rule? Cheers, ALSO GOOD LUCK PEEPS and a MASSIVE shoutout to all the moderators and members that make revising for 2U that much easier. :P Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 11:22:51 am Hey guys, Just a question with rates of change. If t equals something like 152.65 months, what is your final answer if it says when does P = x, occur. Do we leave it as a decimal? Or round to a month? I recommend rounding so that your units are consistent, but I think unless they explicitly say how many years and months it isn't that much of a concern to use a decimal answer. If they say explicitly how many years and months then you have to answer the question. I'll let others comment on this one as well Title: Re: Mathematics Question Thread Post by: g98 on October 21, 2016, 11:24:25 am is using the @ symbol ok? Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 11:24:55 am Hey guys, this is more just a rule that i'm asking, but I remember that when there are parallel lines and two transversal intersect them, the ration between the two transversals are equal or something? Could someone just rephrase that properly or enlighten me on whether that is the correct rule? Cheers, ALSO GOOD LUCK PEEPS and a MASSIVE shoutout to all the moderators and members that make revising for 2U that much easier. :P You can call it "Ratio of intercepts on transversals of parallel lines" when you write out the ratio relevant. Or just "intercepts on transversals of parallel lines" TBH - I've never needed that theorem yet even though it's in the course Glad we've helped ya Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 11:25:35 am is using the @ symbol ok? In what sense? But probably not a good idea Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 11:26:09 am is using the @ symbol ok? That just stands for "at". It's not too much of a hassle to write two letters lol When do you need it in 2/4U though? I don't really recall any scenario I came across it Title: Re: Mathematics Question Thread Post by: g98 on October 21, 2016, 11:31:15 am In what sense? But probably not a good idea I tend to use when i sub points into a line... find the tangent at (1,0) and the gradient equation was m=x-2 then I would write: @ x=1 m=-1 ...I also use it in stationary point questions..I guess i'll try not to in the exam though ;) Title: Re: Mathematics Question Thread Post by: RuiAce on October 21, 2016, 11:33:08 am Yeah might want to write "At" for those. The symbol doesn't really leave a good impression. Of course if you really forget to then don't stress over it when you come out of the exam either. Title: Re: Mathematics Question Thread Post by: dtinaa on October 21, 2016, 11:40:02 am Please help!! Title: Re: Mathematics Question Thread Post by: lha on October 21, 2016, 11:43:02 am Well if you need to describe the motion based off what you have. Suppose you have the graph of the displacement. Velocity is the first derivative of the displacement Therefore if the graph is increasing, velocity is positive If graph is decreasing, velocity is negative If graph is at a stationary point, particle is at rest Acceleration is the second derivative of the displacement Therefore if the graph is concave up, acceleration is positive If the graph is concave down, acceleration is positive If we are at a point of inflexion, acceleration is constant (possibly 0 but we don't know). And say we were given the graph of the velocity. Velocity is the first derivative Hence v > 0 means particle is travelling away from the origin (+'ve direction, to the right, however you memorised it) v < 0 means particle is travelling in the other direction from the origin (-'ve direction, to the left etc.) And those are just some examples. If you know what the derivatives actually MEAN then this is more intuitive than trying to rote learn a few dot points. If velocity is positive, does that mean its going in thd positive direction? Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down? also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph Title: Re: Mathematics Question Thread Post by: imtrying on October 21, 2016, 11:43:15 am are both these formulas right for surface area of a cone? Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 11:51:11 am are both these formulas right for surface area of a cone? Yep they are both correct! Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 11:51:38 am If velocity is positive, does that mean its going in thd positive direction? Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down? also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph Imagine a ball thrown in the air, the exponential either moves to the right or the left. The velocity goes up but then comes to a stop. Then acceleration is applied. Apples. Gyet Yit? Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 11:52:24 am If velocity is positive, does that mean its going in thd positive direction? Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down? also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph Yep, that's pretty much right. However, if the velocity is negative, and the acceleration is negative, the particle is actually speeding up in the negative direction! It changes direction if the displacement starts to decrease (ie. go back towards the origin) or the velocity changes sign (ie. goes from positive to negative). Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 11:52:59 am Yep they are both correct! are we expected to know that? Title: Re: Mathematics Question Thread Post by: BPunjabi on October 21, 2016, 11:54:26 am Good Luck Guys!! We will need it. See you in an hour. Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 11:54:56 am Please help!! Hey! I think it's easier here to take cases where he DOESN'T win in the first three throws. This would only occur if the coins read HTH, or THT. The probability of HTH is (0.5)(0.5)(0.5), as is the probability of THT. Therefore, 2(0.5)(0.5)(0.5)=0.25. The probability of WINNING is 1-P(not winning), therefore the answer to i) is 1-0.25=0.75. Fuck Joe. Why make such a complicated game? Now, we don't want him to win on the second go (even), or on any even goes. The probability of NOT winning on an even go must be something like P(Not winning)=HTHTHTHT...+THTHTHTHTHTHT=2(0.5)^n for n throws, where n is even. Him WINNING on an odd turn is the same as NOT WINNING on an even term (sort of? In terms of infinity, where winning is assured? I'm not actually 100% about that, but let's go with it). Therefore, we need to sum up 2(0.5)+2(0.5)^2+...+2(0.5)^n for n=infinity. Use your sum to infinity formula to get a solution, then your answer should be 1-ans! Let me know if this works out, Jake Title: Re: Mathematics Question Thread Post by: jakesilove on October 21, 2016, 11:55:43 am Good Luck Guys!! We will need it. See you in an hour. Know whatever formulas you think you'll need. You won't need to know multiple variations of volume/surface area for the same shape Title: Re: Mathematics Question Thread Post by: lha on October 21, 2016, 12:01:19 pm Yep, that's pretty much right. However, if the velocity is negative, and the acceleration is negative, the particle is actually speeding up in the negative direction! It changes direction if the displacement starts to decrease (ie. go back towards the origin) or the velocity changes sign (ie. goes from positive to negative). So if velocity is negative but acceleration is positive, the particle is going in the negative direction but slowing down? Title: Re: Mathematics Question Thread Post by: Hua Fei on October 21, 2016, 12:10:06 pm So if velocity is negative but acceleration is positive, the particle is going in the negative direction but slowing down? Yes. If velocity and acceleration are opposite in direction, the particle would be decelerating in the direction of the velocity (although, I'd wait for someone else to second that) Also, I don't get how to do this Q Title: Re: Mathematics Question Thread Post by: ml125 on October 21, 2016, 12:35:52 pm Yes. If velocity and acceleration are opposite in direction, the particle would be decelerating in the direction of the velocity (although, I'd wait for someone else to second that) Also, I don't get how to do this Q The displacement is equivalent to the area under the velocity curve: (8×4)/2=16. In the question, it states that at t=0, x=2. Therefore, the maximum displacement at t=4 would be x=2+16=18 :) Title: Re: Mathematics Question Thread Post by: yogiburay on October 21, 2016, 02:43:56 pm Hi. Sorry to ask silly questions, but I'm having trouble doing these questions. We just started these topic and I'm so confused. I just can't keep up with my teacher, he's just too fast. I hope you can help me. :'( * Find the stationary points on the curve y=(3x-1)(x-2)^4 *Differentiate y=x√x+1. Hence find the stationary point on the curve, giving the exact value. *The curve f(x)=ax^4-2x^3+7x^2-x+5 has a stationary point at x=1. Find the value of a. * Show that f(x) =√x has no stationary points *Show that f(x) = 1/x^3 has no stationary points. I hope you can explain these questions to me. I really want to improve my maths. Thankyou. :) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 21, 2016, 03:15:04 pm Hi. Sorry to ask silly questions, but I'm having trouble doing these questions. We just started these topic and I'm so confused. I just can't keep up with my teacher, he's just too fast. I hope you can help me. :'( ... I hope you can explain these questions to me. I really want to improve my maths. Thankyou. :) Hey Yogi! Welcome to the forums! Definitely don't be sorry for wanting to improve, good on you for working hard! ;D Before we give you a hand with these, I'll get you to read this guide which covers some of the theory that you are confused with. Have a read, and see if it helps you understand the questions you've posted (it's brief, but it might help a tad). Otherwise, pick the question there that is confusing you the most and post it back, and we'll give you a hand! Best to start with the most confusing and then see if it helps you answer the others ;D Title: Re: Mathematics Question Thread Post by: yogiburay on October 21, 2016, 03:20:58 pm Hey Yogi! Welcome to the forums! Definitely don't be sorry for wanting to improve, good on you for working hard! ;D Before we give you a hand with these, I'll get you to read this guide which covers some of the theory that you are confused with. Have a read, and see if it helps you understand the questions you've posted (it's brief, but it might help a tad). Otherwise, pick the question there that is confusing you the most and post it back, and we'll give you a hand! Best to start with the most confusing and then see if it helps you answer the others ;D Thankyou :) Title: Re: Mathematics Question Thread Post by: rinagee12 on October 21, 2016, 05:42:04 pm Well that was a fair exam... I think Title: Re: Mathematics Question Thread Post by: samuels1999 on October 21, 2016, 08:33:16 pm Hi Jake or anyone who wants to answer, I am a 2017 Year 12 Student. This year I did accelerated 2U Mathematics and just 3 and a bit hours ago I did my HSC mathematics. It was ok- but I am certain I was way off 100%, which would probably mean I'll redo it next year. With that in mind, along with my experience of doing past HSC papers, I realised that It was pretty easy for me to get 70% to 80% in a Mathematics Paper since so much of that style of question I had seen before. I found it much trickier getting that last 15-20% since it involves far more application of skills and technique than the earlier questions. I know this is a problem partially due to the fact I kinda rushed the course in one year, but with respect to the next year, I just wanted to know if you have any advice on how I could fine tune maths skills to the level I could actually do those harder questions...or even just have the confident mindset. Whether it's more past papers, or maybe going back to basic text book work, I just want to know how I should approach the next year keeping in mind the fact that I have bascially completed "course work" already. Thanks, Samuel Title: Re: Mathematics Question Thread Post by: anotherworld2b on October 21, 2016, 08:51:54 pm Hi could i get help with these questions? Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 21, 2016, 08:56:21 pm Hi could i get help with these questions? You have been given the displacement equation and are being asked to find the velocity which means you have to differentiate your equations before you substitute Title: Re: Mathematics Question Thread Post by: anotherworld2b on October 21, 2016, 11:06:55 pm Thank you for the help I was also wondering if i could get help with q15 and 16c-e You have been given the displacement equation and are being asked to find the velocity which means you have to differentiate your equations before you substitute Title: Re: Mathematics Question Thread Post by: Blissfulmelodii on October 22, 2016, 09:15:35 am Thank you for the help I was also wondering if i could get help with q15 and 16c-e No worries. For Q15 you would differentiate the displacement equation to get the velocity equation and then let it equal to 0 to find t Once you know what time the velocity is 0 you can go back and substitute the t into the displacement equation to find out the displacement at that time. I hope that makes sense. Let me know if you need me to write up the solution I'll look at 16 now Title: Re: Mathematics Question Thread Post by: jamonwindeyer on October 22, 2016, 10:39:31 am Hi Jake or anyone who wants to answer, I am a 2017 Year 12 Student. This year I did accelerated 2U Mathematics and just 3 and a bit hours ago I did my HSC mathematics. It was ok- but I am certain I was way off 100%, which would probably mean I'll redo it next year. With that in mind, along with my experience of doing past HSC papers, I realised that It was pretty easy for me to get 70% to 80% in a Mathematics Paper since so much of that style of question I had seen before. I found it much trickier getting that last 15-20% since it involves far more application of skills and technique than the earlier questions. I know this is a problem partially due to the fact I kinda rushed the course in one year, but with respect to the next year, I just wanted to know if you have any advice on how I could fine tune maths skills to the level I could actually do those harder questions...or even just have the confident mindset. Whether it's more past papers, or maybe going back to basic text book work, I just want to know how I should approach the next year keeping in mind the fact that I have bascially completed "course work" already. Thanks, Samuel Hey Samuels!! It definitely sounds like the troubles you are having are completely down to just doing the course in one year. You don't quite have the experience to be able to tackle the weird stuff (and that is absolutely not an insult to your abilities or anything of the sort, it's purely just, you haven't been doing the course for two years). So for starters, you have nothing to worry about at all, and you might be happy with how you went this year! You never know! ;D Practice papers of all types are the best way to improve here; and doing higher levels of Math (3U and 4U presumably?) will help there too. They will expose you to problem types that can be emulated in 2U exams to challenge the high achievers; the questions you are struggling with. Practice makes perfect here Sam, but it sounds like you know that already. Keep doing past papers, and watch yourself improve! ;D Title: Re: Mathematics Question Thread Post by: asd987 on October 22, 2016, 11:03:07 pm Hello, Can someone please help me with Q16c ii from the 2015 paper. Thankyou Title: Re: Mathematics Question Thread Post by: RuiAce on October 22, 2016, 11:05:38 pm Hello, Can someone please help me with Q16c ii from the 2015 paper. Thankyou Addressed in post #505 Title: Re: Mathematics Question Thread Post by: katnisschung on October 23, 2016, 02:08:20 pm there are no answers >:( >:( for the practice test i did q6. Novy leaves her base camp, O and walks in the direction 150°T until she reaches point A on the bank of the river. Unable to cross the river, she follows it on a bearing of 030°T and stops at point C. The bearing and distance of C from O are 120°T and 4km respectively ii)show that angle OAC=60° I proved it by drawing the diagram which is the first part of the diagram figured out a few angles proof (co-interior angle sum is 180 on parallel lines) Q5. A person on top of a building 49m in height wishes to find the height of an adjacent building. If the angle of depression of the adjacent building is 47° 21' from where the person is standing and the buildings are a distance of 12 m apart. ii) Calculate the height of the adjacent building (2 dp) my answer=35.97m simplify i) sin^2x - sin^2x cos^x Thank you!! :) Title: Re: Mathematics Question Thread Post by: RuiAce on October 23, 2016, 02:26:23 pm there are no answers >:( >:( for the practice test i did q6. Novy leaves her base camp, O and walks in the direction 150°T until she reaches point A on the bank of the river. Unable to cross the river, she follows it on a bearing of 030°T and stops at point C. The bearing and distance of C from O are 120°T and 4km respectively ii)show that angle OAC=60° I proved it by drawing the diagram which is the first part of the diagram figured out a few angles proof (co-interior angle sum is 180 on parallel lines) Q5. A person on top of a building 49m in height wishes to find the height of an adjacent building. If the angle of depression of the adjacent building is 47° 21' from where the person is standing and the buildings are a distance of 12 m apart. ii) Calculate the height of the adjacent building (2 dp) my answer=35.97m simplify i) sin^2x - sin^2x cos^x Thank you!! :) Not sure what cos^x is meant to mean there. If it was a typo for cos(x) then there are many ways of simplifying that and we cannot say for certain which is "the correct answer". Presumably sin^2x-sin^2x cos^2x which might make some sense. Please put up diagrams that you used in your attempt to complete the other questions (and preferably also your working out for checking). ATARnotes has an app to upload photos or you can attach screenshots if the test was a pdf _____________________ \text{Assuming my suggestion for the typo with the simplify}\\ \begin{align*}\sin^2x -\sin^2x\cos^2x&=\sin^2x(1-\cos^2x)\\ &=\sin^2x\sin^2x\\ &=\sin^4x\end{align*} Title: Re: Mathematics Question Thread Post by: katnisschung on October 23, 2016, 03:40:02 pm thanks ruiAce I've attached it here Title: Re: Mathematics Question Thread Post by: RuiAce on October 23, 2016, 04:05:14 pm thanks ruiAce I've attached it here (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps0bvz0oxt.png) $\text{The information in Q6 has been annotated over this diagram.}\\ \text{It remains to show that }\angle OAN_1=30^\circ\\ \text{to show that }\angle OAC = 60^\circ$ $\text{As you have stated, }\angle OAN_1=30^\circ\text{ by co-interior angles}$ And then part iii) can be done by using the angle sum of a triangle, and the sine rule. Title: Re: Mathematics Question Thread Post by: asd987 on October 24, 2016, 06:40:12 pm How would I integrate x^-3 / 3 Title: Re: Mathematics Question Thread Post by: jakesilove on October 24, 2016, 07:11:10 pm How would I integrate x^-3 / 3 Hey! To integrate something like that, you just use the normal anti-derivative formula $\int{x^n}dx=\frac{x^{n+1}}{{n+1}}+C$ Apply this directly to the equation you've given us! You can bring the 1/3 outside the integral :) Title: Re: Mathematics Question Thread Post by: asd987 on November 01, 2016, 09:57:36 pm hi i've got a question. the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed Title: Re: Mathematics Question Thread Post by: RuiAce on November 01, 2016, 10:05:12 pm hi i've got a question. the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed $\text{From sketching the curves, clearly the line }y=x+2\\ \text{is above the curve }y=x^2$ $\text{We may identify the points of intersection by solving }x^2=x+2\\ \text{The answers will be }x=-1, x=2$ $\text{So just like how the area between two curves is }\int_a^b (y_1-y_2)dx\\ \text{Using the volume between two curves }V=\pi\int_a^b(y_1^2-y_2^2)dx$ \begin{align*}V&=\int_{-1}^2((x^2)^2 - (x+2)^2)dx\\ &= \int_{-1}^2(x^4-x^2-4x-4)dx\\ &=\left[\frac{x^5}{5}-\frac{x^3}{3}-2x^2-4x\right]_{-1}^2\end{align*} Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on November 03, 2016, 12:08:05 pm Hey, I have a series of question I need help with; please and thank you. 1. Find the discriminant of the quadratic equation: A) -x2 - 3x = 0 B) x2 + 4 = 0 2. Find any values of b for which 2x2 + x + b + 1 = 0 Again, thanks!!! :D Title: Re: Mathematics Question Thread Post by: RuiAce on November 03, 2016, 12:12:42 pm Hey, I have a series of question I need help with; please and thank you. 1. Find the discriminant of the quadratic equation: A) -x2 - 3x = 0 B) x2 + 4 = 0 2. Find any values of b for which 2x2 + x + b + 1 = 0 Again, thanks!!! :D $\text{That's just using the formula }\Delta = b^2-4ac\\ \text{For A) you have }a=-1, \, b=-3, \, c=0\\ \text{So }\Delta =9$ Your second question is inconclusive. Because the way you put it, it depends on your value for x. Suspecting something in the question wasn't typed because I can sort of tell what the question should be. Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on November 03, 2016, 12:42:06 pm Oh yep thanks; I forgot the last bit of the question. Find any values of b for which 2x2 + x + b + 1 = 0 has real roots. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on November 03, 2016, 12:44:04 pm Oh yep thanks; I forgot the last bit of the question. Find any values of b for which 2x2 + x + b + 1 = 0 has real roots. Cool! For real roots we require our discriminant to be greater than or equal to zero! $\Delta\ge0\\b^2-4ac\ge0\\1^2-4(2)(b+1)\ge0\\-8b-7\ge0\\b\le\frac{-7}{8}$ Title: Re: Mathematics Question Thread Post by: RuiAce on November 03, 2016, 12:45:45 pm $\text{Recall that the condition for real roots is that }\Delta \ge 0\\ \text{If }\Delta < 0\text{, then we have no real roots.}\\ \text{Whereas otherwise, 1 or 2 roots are real.}$ $\Delta = (1)^2 - 4(2)(b+1) = 1-8(b+1) = -8b-7$ $\text{So }\Delta \ge 0\implies -8b-7 \ge 0 \implies b \le- \frac78$ Cool! For real roots we require our discriminant to be greater than or equal to zero! $\Delta\ge0\\b^2-4ac\ge0\\1^2-4(2)(b+1)\ge0\\-8b-7\ge0\\b\le\frac{-7}{8}$ Well damn I didn't type fast enough Title: Re: Mathematics Question Thread Post by: Hplovers on November 03, 2016, 03:36:23 pm Would be great if you could help me out with 4 (ii) please :) Thankyou! Title: Re: Mathematics Question Thread Post by: RuiAce on November 03, 2016, 03:44:48 pm Would be great if you could help me out with 4 (ii) please :) Thankyou! $\text{Given }\triangle EBF ||| \triangle AED\text{ from the proportional sides we have}\\\frac{EF}{FB}=\frac{AD}{DE}\\ \text{But }CFED\text{ is a rhombus, so }CF=FE=ED=DC=x$ \text{So subbing it in we have}\\ \begin{align*}\frac{x}{a-x}&=\frac{b-x}{x}\\ \implies x^2&=(b-x)(a-x)\\ x^2&=ab-(a+b)x+x^2\\ \implies (a+b)x&=ab\\ \therefore x&=\frac{ab}{a+b}\end{align*} Title: Re: Mathematics Question Thread Post by: Hplovers on November 03, 2016, 03:55:45 pm $\text{Given }\triangle EBF ||| \triangle AED\text{ from the proportional sides we have}\\\frac{EF}{FB}=\frac{AD}{DE}\\ \text{But }CFED\text{ is a rhombus, so }CF=FE=ED=DC=x$ \text{So subbing it in we have}\\ \begin{align*}\frac{x}{a-x}&=\frac{b-x}{x}\\ \implies x^2&=(b-x)(a-x)\\ x^2&=ab-(a+b)x+x^2\\ \implies (a+b)x&=ab\\ \therefore x&=\frac{ab}{a+b}\end{align*} Thankyou so so much! Title: Re: Mathematics Question Thread Post by: teapancakes08 on November 04, 2016, 09:43:08 am Probably should be able to answer this, but the second term is throwing me off: 4. Simplify: (h) 1 - sin^2 (180-x) If it was just 1-sin^2x it would be cos^2x, but... If someone could help clarify that would be great, thanks ^^ Title: Re: Mathematics Question Thread Post by: RuiAce on November 04, 2016, 09:55:49 am Probably should be able to answer this, but the second term is throwing me off: 4. Simplify: (h) 1 - sin^2 (180-x) If it was just 1-sin^2x it would be cos^2x, but... If someone could help clarify that would be great, thanks ^^ $\text{Note that }\sin(180^\circ-x)=\sin x\quad \text{(second quadrant angle)}$ Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on November 04, 2016, 08:52:26 pm Hey!!! I have two new questions that I need help with. :-\ I think I know how to do them but I just need some reassurance; please and thank you!!! 1. Express x2 - 4x + 5 in the form Ax (x+2) + B (x +1) + C + 4 2. Show that x2 + 2x + 9 can be written in the form a (x-2)(x+3) + b (x-2) + c where a=1, b=1, and c=17 Title: Re: Mathematics Question Thread Post by: jakesilove on November 04, 2016, 08:55:23 pm Hey!!! I have two new questions that I need help with. :-\ I think I know how to do them but I just need some reassurance; please and thank you!!! 1. Express x2 - 4x + 5 in the form Ax (x+2) + B (x +1) + C + 4 2. Show that x2 + 2x + 9 can be written in the form a (x-2)(x+3) + b (x-2) + c where a=1, b=1, and c=17 Hey! In both methods, you just need to expand the 'form' equation (ie. Ax (x+2) + B (x +1) + C + 4). Then, equate the x^2 terms, the x terms and the constant terms with the original question. For instance, if you expand, the only x^2 term is Ax^2. Therefore, for the first one, A=1! Let me know if there's something specific you're struggling with, or if you need the whole answer; sounds like you're on the right track! Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on November 04, 2016, 09:04:40 pm Can you please try explaining that another way? It didn't make sense at all to me. Thank you!!! :D Title: Re: Mathematics Question Thread Post by: jakesilove on November 04, 2016, 09:11:25 pm Can you please try explaining that another way? It didn't make sense at all to me. Thank you!!! :D $Ax(x+2)+B(x+1)+C+4=Ax^2+2Ax+Bx+B+C+4$ We equate x squared terms in each equation $Ax^2=x^2$ $A=1$ $2Ax+Bx=-4x$ $2(1)+B=-4$ $B=-6$ $B+C+4=5$ $-6+C+4=5$ $C=7$ Therefore, the expression can be written as $x(x+2)-6(x+1)+11$ Title: Re: Mathematics Question Thread Post by: RuiAce on November 04, 2016, 09:24:48 pm Can you please try explaining that another way? It didn't make sense at all to me. Thank you!!! :D \text{Note that the point is to use the logic of two things being IDENTICAL}\\ \text{i.e. }Ax^2+Bx+C = Dx^2+Ex+F\\ \text{IF and ONLY IF:}\\ \begin{align*}A&=D\\ B&=E\\ C&=F\end{align*}\\ \text{So we can EQUATE the coefficients.} $\text{i.e. }x^2+4x+2 = Ax^2+Bx+C\\ \text{Only if we can guarantee that }A=1, B=4, C=2$ Title: Re: Mathematics Question Thread Post by: Aussie1Italia2 on November 05, 2016, 02:52:16 pm Hey again! Can we tell I suck at maths yet? Thanks for the help! Find values of A, B, and C if x2 + x - 2 = A (x-2)2 + Bx + C Title: Re: Mathematics Question Thread Post by: RuiAce on November 05, 2016, 02:55:57 pm Hey again! Can we tell I suck at maths yet? Thanks for the help! Find values of A, B, and C if x2 + x - 2 = A (x-2)2 + Bx + C $\text{Functionally the same to the above.}\\ \text{Expand the RHS out to have something useful,}$ \begin{align*}x^2+x-2&\equiv A(x-2)^2+Bx+C\\ &\equiv A(x^2-4x+4)+Bx+C\\ &\equiv Ax^2+(-4A+B)x+(4A+C)\end{align*} \text{So by equating the components, because this is an identity}\\ \begin{align*}1&=A\\ 1&=-4A+B\\ -2&=4A+C\end{align*} Title: Re: Mathematics Question Thread Post by: katnisschung on November 05, 2016, 09:37:20 pm need help solving this trig identity Title: Re: Mathematics Question Thread Post by: jakesilove on November 05, 2016, 09:52:46 pm need help solving this trig identity Hey! The trick to a question like this is to let sin(x) be equal to a variable; I like using u. $\sin(x)=u$ $2u^2-7u+3=0$ Now, we can just solve this like any old quadratic! You can either use the quadratic equation, or any other method you can implement. It ends up simplifying to $(2u-1)(u-3)=0$ Great! We can now let each brackets be equal to zero $2u-1=0$ $u=\frac{1}{2}$ $u-3=0$ $u=3$ Now, let's change our u's back into sin(x)s $\sin(x)=\frac{1}{2}$ $\sin(x)=3$ Can you see any issues? You definitely will as you study this area more :) sin(x) oscilates between -1 and 1, so will never equal 3! If you don't believe me, try typing shift sin of 3 into your calculator :) That leaves us with the equation sin(x)=1/2. We can solve this in the normal way, which gives us $x=30, 150$ So those will be our answers! Title: Re: Mathematics Question Thread Post by: RuiAce on November 06, 2016, 08:58:23 am need help solving this trig identity Regarding the website you used, that was a trigonometric equation; not proving an identity. So if you chose the next option on that list on the left you would've gotten the computerised solution(http://i.imgur.com/OrdUL13.png) Title: Re: Mathematics Question Thread Post by: katnisschung on November 06, 2016, 12:31:40 pm i cant find it in exact value form :( Title: Re: Mathematics Question Thread Post by: RuiAce on November 06, 2016, 12:59:53 pm i cant find it in exact value form :( $\text{Denote the triangle as }\triangle ABC\text{ from left to right}\\ \text{and let the altitude }h=BD$ $\text{Observe that }AD + CD = AC = 40$ \text{From trigonometry:}\\ \begin{align*}\tan 30^\circ = \frac{h}{AD} &\implies AD=h\sqrt{3}\\ \tan 45^\circ = \frac{h}{CD} &\implies CD =h\end{align*} $\therefore h\sqrt{3}+h=40 \implies h=\frac{40}{1+\sqrt3}$ Title: Re: Mathematics Question Thread Post by: katnisschung on November 07, 2016, 05:57:54 pm thanks in advance :) question on series The sum of the first 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160 find the first term and the common difference Title: Re: Mathematics Question Thread Post by: RuiAce on November 07, 2016, 06:08:03 pm thanks in advance :) question on series The sum of the first 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160 find the first term and the common difference $\text{Using the formula }S_n=\frac{n}{2}[2A+(n-1)D]$ \text{Let }a\text{ be the first term and }d\text{ be the common difference.}\\\text{The sum of the first five terms gives us}\\ \begin{align*}S_5&=\frac{5}{2}[2a+4d]\\ 35&=5a+10d \tag{1}\end{align*} $\text{Then there's many things we can do. One way (and I'd say the more clever way)}\\ \text{Is to consider the sum from a slightly different angle.}\\ \text{We are still summing 5 terms, so }n=5\\ \text{The common difference is still }D=d\\ \textit{But the starting point changes.}$ \text{The first term is no longer }T_1\text{, it is }T_6=a+5d\\ \text{So instead, subbing THIS into our formula:}\\ \begin{align*}S&=\frac{5}{2}[2(a+5d)+4d]\\ 160&=5a+35d\tag{2}\end{align*} $\text{Solving (1) and (2) simultaneously, we arrive at}\\ \boxed{a=-3, d=5}\\ \text{You can also check that this is the answer}$ Title: Re: Mathematics Question Thread Post by: RuiAce on November 08, 2016, 02:13:30 pm I know that nobody really cares, but I just felt the need to share this exponential growth question which could've gotten 4-6 marks in the HSC... as opposed to in ACTL ;D (http://i.imgur.com/jS1KfIa.png) Any thoughts Jamon? :P Title: Re: Mathematics Question Thread Post by: jamonwindeyer on November 08, 2016, 02:28:28 pm I know that nobody really cares, but I just felt the need to share this exponential growth question which could've gotten 4-6 marks in the HSC... as opposed to in ACTL ;D (http://i.imgur.com/jS1KfIa.png) Any thoughts Jamon? :P I think they'd have to step you through that one for it to be included in the HSC ;) Title: Re: Mathematics Question Thread Post by: katnisschung on November 08, 2016, 08:30:46 pm $\text{Using the formula }S_n=\frac{n}{2}[2A+(n-1)D]$ \text{Let }a\text{ be the first term and }d\text{ be the common difference.}\\\text{The sum of the first five terms gives us}\\ \begin{align*}S_5&=\frac{5}{2}[2a+4d]\\ 35&=5a+10d \tag{1}\end{align*} $\text{Then there's many things we can do. One way (and I'd say the more clever way)}\\ \text{Is to consider the sum from a slightly different angle.}\\ \text{We are still summing 5 terms, so }n=5\\ \text{The common difference is still }D=d\\ \textit{But the starting point changes.}$ \text{The first term is no longer }T_1\text{, it is }T_6=a+5d\\ \text{So instead, subbing THIS into our formula:}\\ \begin{align*}S&=\frac{5}{2}[2(a+5d)+4d]\\ 160&=5a+35d\tag{2}\end{align*} $\text{Solving (1) and (2) simultaneously, we arrive at}\\ \boxed{a=-3, d=5}\\ \text{You can also check that this is the answer}$ Ruiace thanks for your reply but im having trouble grasping the second part of your answer. would u mind elaborating on s=5/2 [2(a+5d) +4d] why is it +4d when the formula is sn=n/2 [2a+(n-1)d] i dont understand :P Title: Re: Mathematics Question Thread Post by: RuiAce on November 08, 2016, 08:33:51 pm Ruiace thanks for your reply but im having trouble grasping the second part of your answer. would u mind elaborating on s=5/2 [2(a+5d) +4d] why is it +4d when the formula is sn=n/2 [2a+(n-1)d] i dont understand :P T6, T7, T8, T9, T10 There are 5 terms So n=5 So n-1=4 So (n-1)d=4d Title: Re: Mathematics Question Thread Post by: katnisschung on November 08, 2016, 08:41:10 pm thanks Ruiace got it ! Title: Re: Mathematics Question Thread Post by: RuiAce on November 08, 2016, 08:46:20 pm thanks Ruiace got it ! Excellent :) Title: Re: Mathematics Question Thread Post by: teapancakes08 on November 08, 2016, 11:50:11 pm When proving that: (sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to: (sin^2x + cos^2x)(tanx +cotx) Or do you have to show the whole working out from A to B? Title: Re: Mathematics Question Thread Post by: Dolphax on November 08, 2016, 11:59:28 pm When proving that: (sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to: (sin^2x + cos^2x)(tanx +cotx) Or do you have to show the whole working out from A to B? I was taught that (and pretty sure that for the HSC): "Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary. "Show that" questions: you can only start off with LHS. ... however this seems to go against RuiAce's guide Verbs and Maths. Title: Re: Mathematics Question Thread Post by: jamonwindeyer on November 09, 2016, 12:35:20 am I was taught that (and pretty sure that for the HSC): "Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary. "Show that" questions: you can only start off with LHS. ... however this seems to go against RuiAce's guide Verbs and Maths. Yeah, they don't make that distinction at the HSC level, prove and show are for all purposes identical ;D When proving that: (sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to: (sin^2x + cos^2x)(tanx +cotx) Or do you have to show the whole working out from A to B? Hey pancakes! In this case, I would say that's a tad few steps to skip. If you are doing it anyway, you might as well write it down! Especially in proof questions, where the marker needs to verify that you didn't just fudge steps based on a guess/intuition, it's best to show all but the most trivial steps :) in this case, try and show how you got to that next result; you've skipped virtually all of the hard work :) Title: Re: Mathematics Question Thread Post by: RuiAce on November 09, 2016, 04:30:36 am \begin{align*}\sin^2x\tan x + \cos^2x\cot x+2\sin x\cos x&=\frac{\sin^3 x}{\cos x}+2\sin x \cos x + \frac{\cos^3x}{\sin x}\\ &= \sin x \cos x\left(\frac{\sin^2x}{\cos^2x}+2+\frac{\cos^2x}{\sin^2x}\right)\\ &= \sin x \cos x (\tan x + \cot x)^2\\ &= \sin x \cos x (\tan x +\cot x)(\tan x +\cot x)\\ &= (\sin^2 x + \cos^2 x)(\tan x + \cot x)\end{align*}\\ \text{It's 4:30 in the morning so my brain isn't fully switched on}\\ \text{But the fact it took me about FIVE steps to reach this means that jumping straight here}\\ \text{is just far too rapid. There's an invisible boundary}\\ \text{between what's ok to skip and what isn't.} $\text{If anything, in proofs the only thing safe to skip is trivial algebra.}\\ \text{When something not immediately obvious is used, that's calling for some attention.}$ I was taught that (and pretty sure that for the HSC): "Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary. "Show that" questions: you can only start off with LHS. ... however this seems to go against RuiAce's guide Verbs and Maths. $\text{As far as the 3U course goes}\\ \text{When examiners (for I don't know why) use SHOW that }N=P+Ae^{kt}\\ \text{is a solution to }\frac{dN}{dt}=k(N-P)\\ \text{Most of the time they don't expect you to integrate the differential equation}\\ \text{They just want you to assume the result and differentiate and rearrange.}$ $\text{They will never say that you are required to PROVE such a result.}$ $\text{In reality, prove is a much more powerful word than 'show'.}\\ \text{The motivation behind }\textit{show}\text{ is just to give a demonstration}\\ \text{Whereas }\textit{prove}\text{ is more for describing a series of logically correct steps to achieve a result.}$ $\text{The problem is, in the HSC trying to force you to work both ways will hurt}\\ \text{because not everyone is trained to think at THAT level.}\\ \text{So they relax the strictness on these definitions a bit.}$ Title: Re: Mathematics Question Thread Post by: katnisschung on November 09, 2016, 07:10:33 am The sum of 12 terms of an arithmetic series is 186 and the 20th term is 83 find the sum of 40 term....I tried Title: Re: Mathematics Question Thread Post by: RuiAce on November 09, 2016, 07:57:39 am The sum of 12 terms of an arithmetic series is 186 and the 20th term is 83 find the sum of 40 term....I tried Is that supposed to mean the sum of the first 12 and the first 40 terms? Title: Re: Mathematics Question Thread Post by: katnisschung on November 09, 2016, 09:48:43 am Is that supposed to mean the sum of the first 12 and the first 40 terms? thats the issue i had RuiAce so i just assumed it was the first 12 Title: Re: Mathematics Question Thread Post by: RuiAce on November 09, 2016, 09:59:24 am thats the issue i had RuiAce so i just assumed it was the first 12 Well doing a check with WolframAlpha this gives you a=-12 d=5 Which gives S40 = 3420 Is that meant to be the answer? (Where confusion arises supplying the answer helps.) Title: Re: Mathematics Question Thread Post by: katnisschung on November 09, 2016, 12:13:35 pm Well doing a check with WolframAlpha this gives you a=-12 d=5 Which gives S40 = 3420 Is that meant to be the answer? (Where confusion arises supplying the answer helps.) yes that is the correct answer Title: Re: Mathematics Question Thread Post by: RuiAce on November 09, 2016, 12:35:24 pm \text{Considering both the sum of first 12th and the 20th term}\\ \begin{align*}a+19d&=83\\ \frac{12}{2}(2a+11d)&=186\end{align*} $\text{So }a=-12, d=5\text{ solving the equation}$ $\text{So }S_{40}=\frac{40}{2}[2(-12)+39(5)]$ Title: Re: Mathematics Question Thread Post by: katnisschung on November 10, 2016, 06:16:56 pm Lucia currently earns25,000
her wage increases by 5% each year.
Find
a) her wage after 6 years
b) her total earnings (before tax)
in 6 years
Post by: RuiAce on November 10, 2016, 06:21:14 pm
Lucia currently earns $25,000 her wage increases by 5% each year. Find a) her wage after 6 years b) her total earnings (before tax) in 6 years $\text{The current wage increases by 0.05, so her wage in one year}\\ \text{will be }1.05\text{ times her wage the year before.}\\ \text{This is just understanding the question. It's the basis of compound interest.}$ $\text{Wage right now: }25000\\ \text{Wage 1 year later: }25000(1.05)\\ \text{2 years later: }25000(1.05)(1.05)=25000(1.05)^2$ $\text{Repeating the process, her wage in 6 years will be }25000(1.05)^6\\ \text{Note that this is a geometric sequence with }a=25000,\, r=1.05$ Part b) is just weird. Although it looks to me like they meant the sum, their wording isn't clear. I'm not providing a solution to that one without knowing what's going on (so provide the answer if you want a solution) Title: Re: Mathematics Question Thread Post by: jamonwindeyer on November 10, 2016, 06:26:25 pm Lucia currently earns$25,000
her wage increases by 5% each year.
Find
a) her wage after 6 years
b) her total earnings (before tax)
in 6 years

I'll take a stab at what it means! This is a geometric series question, where $$a=25000$$ and ](r=1.05\) (the 5% interest rate).

The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.

$T_6=ar^5=25000(1.05)^5=31907.04$

Then the total earned is just the sum of the first six terms:

$S_6=\frac{a(1-r^n)}{1-r}=151887.55$

This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!
Post by: katnisschung on November 10, 2016, 06:28:25 pm
thanks ruiace and jamon
also i think the answers are wrong
the common ratio of a geometric series is 4 and the sum of
the first 5 terms is 3069 find the first term
Post by: RuiAce on November 10, 2016, 06:31:08 pm
I'll take a stab at what it means! This is a geometric series question, where $$a=25000$$ and ](r=1.05\) (the 5% interest rate).

The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.

$T_6=ar^5=25000(1.05)^5=31907.04$

Then the total earned is just the sum of the first six terms:

$S_6=\frac{n}{2}(a+l)=3(25000+31907.04)=170721.12$

This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!
Yeah the question was just irritating with its wording. Upsetting when you're studying business at uni haha

thanks Ruiace! it popped up in a series exercise?

also i think the answers are wrong
the common ratio of a geometric series is 4 and the sum of
the first 5 terms is 3069 find the first term
\begin{align*}S_5=\frac{a(r^5-1)}{r-1}&=3069\\ \frac{a(4^5-1)}{3}&=3069\\ a&=9\end{align*}
Post by: katnisschung on November 10, 2016, 06:41:03 pm
I'll take a stab at what it means! This is a geometric series question, where $$a=25000$$ and ](r=1.05\) (the 5% interest rate).

The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.

$T_6=ar^5=25000(1.05)^5=31907.04$

Then the total earned is just the sum of the first six terms:

$S_6=\frac{n}{2}(a+l)=3(25000+31907.04)=170721.12$

This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!

question why did you use the arithmetic series formula?

a
Post by: RuiAce on November 10, 2016, 06:41:43 pm
question why did you use the arithmetic series formula?

a
Haha lol I thought something was weird but I didn't pick up on it. I think he made a mistake there.
Post by: katnisschung on November 10, 2016, 06:47:36 pm
Haha lol I thought something was weird but I didn't pick up on it. I think he made a mistake there.

phew thats why there was a discrepancy between both your answers  ;D
Post by: jamonwindeyer on November 10, 2016, 07:40:46 pm
phew thats why there was a discrepancy between both your answers  ;D

Sorry just fixed it up! Note also that Rui and I interpreted Part (a) differently, which could lead to different answers for (b) (it's a bad question) :P
Post by: katnisschung on November 10, 2016, 08:32:15 pm
should i write notes for my maths test
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.

tips?
note i am not a maths genius i take 2u and get around 80s
(my teacher keeps on saying my results have fluctuated and
i have 'potential'  ;D let's hope so)

oh and when i mean writing notes i mean like transcribing
everytime after i have a maths lesson
Post by: RuiAce on November 10, 2016, 08:35:23 pm
should i write notes for my maths test
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.

tips?
Back when there was no "reference sheet" I found it acceptable for some people to write a formula sheet or formula book. Things in there were meant to be really basic and serve no more purpose than be a refresher, and their only advantage was that they're going to be easier to use than textbooks.

The fact that the reference sheet has all the formulas you actually need really downs on this. I don't see any more that you can do than just printing it off and dropping in a few annotations here and there (A3 paper is fine for this though). I am with them in that in general doing "notes" for maths is not advisable, because it really doesn't help you remember as much as you think it would.

The furthest you should ever go with notes is about 5 dot points per topic, just highlighting some key areas of concern. If you ask me, that is.
Post by: jamonwindeyer on November 10, 2016, 09:01:00 pm
should i write notes for my maths test
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.

tips?
note i am not a maths genius i take 2u and get around 80s
(my teacher keeps on saying my results have fluctuated and
i have 'potential'  ;D let's hope so)

oh and when i mean writing notes i mean like transcribing
everytime after i have a maths lesson

You should write as much as you need to to remember your concepts! Like, if you need to have a set of notes specifically proving differentiation by first principles, then have that. It's all about whatever works for you! :)

That said, whatever your choice, you need to be doing practice as well. No one ever got a strong result in a HSC Math subject without doing any practice at all. Do what you need to do, but make sure practice is mixed in :)
Post by: jamesq on November 11, 2016, 11:17:11 pm
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments. (a) What is the amount of each instalment? Answer is$1835.68.
I tried the question and got $1712.87, any help is appreciated. Thanks in advance! Title: Re: Mathematics Question Thread Post by: jamonwindeyer on November 11, 2016, 11:20:27 pm A loan of$6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
(a) What is the amount of each instalment?

Answer is $1835.68. I tried the question and got$1712.87, any help is appreciated. Thanks in advance!

Hey James! So two steps here; find the total amount owed, and then divide that into the 5 repayments :) I'd be happy to lend a hand, but we need to know whether this is simple interest or compound interest? Indeed, perhaps you used the wrong one in your working? Or did the question not say? :)

If you let me know whether its simple or compound I'd be happy to show you the working for it! ;D

Edit: Yeah actually, sorry, these are always compound. That said, reckon you could snap a pic of your working and pop it as an attachment?Since you are close I reckon you might have just made a small mistake, I'll try and spot it for you! :)
Post by: RuiAce on November 11, 2016, 11:33:02 pm
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments. (a) What is the amount of each instalment? Answer is$1835.68.
I tried the question and got \$1712.87, any help is appreciated. Thanks in advance!
$\text{Note that the effective interest rate per period is }1.25\text{ percent per month}\\ \text{Hence every month, the loan gains interest of }1.0125\times\text{ every month}$
$\text{Observe that the payments are made in years.}\\ \text{Hence, despite monthly interest, we must think in years as that's how we make payments}\\ \text{Let }A_n\text{ be the amount owing after }n\textbf{ years}.$
\text{Let }M\text{ be the annual payment}\\ \text{Note that interest gets compounded }\textbf{12 times}\text{ (once per month) every year} \\ \begin{align*}A_0&=6000\\ A_1&=6000(1.0125)^{12} - M\\ A_2&=A_1(1.0125)^{12}-M\\ A_2&= 6000(1.0125)^{24}-M(1+1.0125^{12})\end{align*}
\text{Continuing the pattern, we eventually get}\\\begin{align*} A_n&=6000(1.0125)^{12n}-M(1+1.0125^{12}+1.0125^{24}+\dots+1.0125^{12(n-1)})\\&=6000(1.0125)^{12n}-M\left(\frac{1.0125^{12n}-1}{1.0125^{12}-1}\right)\end{align*}
$\text{Since }A_5=0\text{ we have}\\ 6000(1.0125)^{60}-M\left(\frac{1.0125^{60}-1}{1.0125^{12}-1}\right)=0\\ \text{giving us }M=1835.68$
Hey James! So two steps here; find the total amount owed, and then divide that into the 5 repayments :) I'd be happy to lend a hand, but we need to know whether this is simple interest or compound interest? Indeed, perhaps you used the wrong one in your working? Or did the question not say? :)

If you let me know whether its simple or compound I'd be happy to show you the working for it! ;D

Edit: Yeah actually, sorry, these are always compound. That said, reckon you could snap a pic of your working and pop it as an attachment?Since you are close I reckon you might have just made a small mistake, I'll try and spot it for you! :)
When you're an actuary and just chuck it into the present value formula before doing the question 8)
Post by: teapancakes08 on November 12, 2016, 09:06:36 am
Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d]  and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^
Post by: RuiAce on November 12, 2016, 09:21:48 am
Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d]  and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^
\text{With correct calculations, the following system of equations:}\\ \begin{align*}S_6=\frac62(2a+5d)&=-12\\ S_{14}=\frac{14}{2}(2a+13d)&=196\end{align*}\\ \text{yields }a=-12, d=4
\text{To get an expression for a general }S_n\text{, upon substituting back in:}\\ \begin{align*}S_n&=\frac{n}{2}(-24+(n-1)4)\\ &= \frac{n}{2}(4n-28)\\ &=2n^2-14\end{align*}\\ \textbf{Check that you didn't make an algebra mistake somewhere}
$\text{In b), note that what we }\textbf{actually}\text{ want to solve is }\\ S_n > 250\\ \text{We just replace > with = because it looks neater that way.}$
\text{And we now consider the quadratic.}\\ \begin{align*}S_n&=250\\ 2n^2-14n-250&=0\\ n^2-7n-125&=0\\ n&=\frac{7\pm \sqrt{49+500}}{2}\\ &= -8.215..., 15.215...\end{align*}\\ \text{Indeed, the need for the quadratic formula is NOT surprising.}
$\text{Clearly }n\text{ can't be a negative number, so we have }n=15.215...\\ \text{So because we want it to }\textbf{exceed}\text{, clearly we must go up by 1.}\\ \text{So the answer is }n=16$
$\textbf{Note that solving the inequality }S_n > 250\textbf{ makes it 'mathematically' tidier and more fluent}\\ \text{But it becomes visually less appealing, because it's a quadratic inequality nonetheless}$
Post by: jakesilove on November 12, 2016, 09:23:12 am
Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d]  and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^

Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!

As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!

Let me know if that all makes sense.

Jake
Post by: teapancakes08 on November 12, 2016, 09:45:28 am
Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!

As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!

Let me know if that all makes sense.

Jake

I think I got it. I checked back and saw that I made an algebra mistake when subbing in the values for a and d. Ruiace also helped me out on that one. As for the part b, I should be good. The explanations helped clear things up a lot. Thanks so much for the help ^^
Post by: teapancakes08 on November 12, 2016, 09:59:38 am
\text{With correct calculations, the following system of equations:}\\ \begin{align*}S_6=\frac62(2a+5d)&=-12\\ S_{14}=\frac{14}{2}(2a+13d)&=196\end{align*}\\ \text{yields }a=-12, d=4
\text{To get an expression for a general }S_n\text{, upon substituting back in:}\\ \begin{align*}S_n&=\frac{n}{2}(-24+(n-1)4)\\ &= \frac{n}{2}(4n-28)\\ &=2n^2-14\end{align*}\\ \textbf{Check that you didn't make an algebra mistake somewhere}
$\text{In b), note that what we }\textbf{actually}\text{ want to solve is }\\ S_n > 250\\ \text{We just replace > with = because it looks neater that way.}$
\text{And we now consider the quadratic.}\\ \begin{align*}S_n&=250\\ 2n^2-14n-250&=0\\ n^2-7n-125&=0\\ n&=\frac{7\pm \sqrt{49+500}}{2}\\ &= -8.215..., 15.215...\end{align*}\\ \text{Indeed, the need for the quadratic formula is NOT surprising.}
$\text{Clearly }n\text{ can't be a negative number, so we have }n=15.215...\\ \text{So because we want it to }\textbf{exceed}\text{, clearly we must go up by 1.}\\ \text{So the answer is }n=16$
$\textbf{Note that solving the inequality }S_n > 250\textbf{ makes it 'mathematically' tidier and more fluent}\\ \text{But it becomes visually less appealing, because it's a quadratic inequality nonetheless}$

Guess I did make an algebraic error ^^; but I suppose better now than during the test.

Thanks for the explanation, it cleared my understanding of it by a large margin. Thank you so much for the help ^^
Post by: julzzz on November 12, 2016, 10:45:12 am
(1/27)^(3n+1)=sqrt{3}/81

This is a prelim question which i couldnt manage to do
Using indices, not log, thanks
Post by: jakesilove on November 12, 2016, 11:01:55 am
(1/27)^(3n+1)=sqrt{3}/81

This is a prelim question which i couldnt manage to do
Using indices, not log, thanks

So, you're question is

$(\frac{1}{27})^{3n+1}=\frac{\sqrt{3}}{81}$

We need to convert everything into powers of three;

$(\frac{1}{3^3})^{3n+1}=\frac{3^{0.5}}{3^4}$

$(3^{-3})^{3n+1}=3^{0.5}*3^{-4}$

$3^{-9n-3}=3^{-3.5}$

Therefore, we can let

$-9n-3=-3.5$

$n=\frac{0.5}{9}$

So our answer is n=1/18! Hope that all makes sense :)
Post by: teapancakes08 on November 12, 2016, 09:51:00 pm

20. Show that the sum of the first n odd natural numbers is a perfect square.

The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?
Post by: RuiAce on November 12, 2016, 09:54:22 pm

20. Show that the sum of the first n odd natural numbers is a perfect square.

The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?
$\text{Just consider }S_n=1+3+5+\dots+(2n-1)\\ \text{The idea is to consider an A.P. with }a=1, \,d=2$
\text{So by summing the A.P. using }S_n=\frac{n}{2}(a+\ell)\\ \begin{align*}S_n&=\frac{n}{2}(1+(2n-1))\\ &=n^2\end{align*}
Post by: RuiAce on November 12, 2016, 09:57:53 pm
$\text{If you wanted to do a claim, followed by a proof by induction}\\ \text{You can't do it like that, because that's subject to experimental error}\\ \text{You have to properly induce it then. (Which is obviously not 2U)}$
$\text{i.e. You can trial some values of }n\text{ to get a proposition}\\ \text{But you can't just keep testing to prove it holds for ALL }n$
Post by: samuels1999 on November 12, 2016, 11:15:54 pm
Hi everyone

I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.

Thanks,
Samuel
Post by: RuiAce on November 13, 2016, 06:03:59 am
Hi everyone

I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.

Thanks,
Samuel
$\text{For a), if the score happened to be 1}\\ \text{Analysing the scenario, this can ONLY occur}\\ \text{if both dice ended up with a 1.}$
$\text{The probability of (1,1) is clearly }\frac{1}{36}$
________________________
$\text{Essentially if we want three consecutive scores of 1}\\ \text{Not only do we have to get }(1,1)\text{ in the first throw}\\ \text{But we must get it in the subsequent throws as well.}\\ \text{So we have }\frac{1}{36}\times\frac{1}{36}\times\frac{1}{36}$
________________________
$\text{Analyse the scenario carefully.}\\ \text{For c), if the score were to be a 6, then:}$
$\textit{Whilst both dice CAN show a 6}\\ \textit{We only care that at least ONE of the dice shows a 6}$
$\text{There are many ways of tackling this.}\\ \text{The easiest way is to consider the }\textit{complement}\\ \text{(This is due to the presence of 'at least')}$
\begin{align*}Pr(\text{At least 1 six})&=1-Pr(\text{No sixes})\\ &= 1-\frac{25}{36}\\ &= \frac{11}{36}\end{align*}
$\text{Note: }\frac{25}{36}\text{ comes from the fact that}\\ \text{Probability of an event}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{Think about why favourable outcomes = 25}$
Post by: Aussie1Italia2 on November 13, 2016, 09:42:21 am

Alpha & beta are roots of the quadratic equation x2 - 6x + 3 = 0.

Find alpha x beta2 + alpha2 x beta
Post by: jamonwindeyer on November 13, 2016, 09:59:30 am

Alpha & beta are roots of the quadratic equation x2 - 6x + 3 = 0.

Find alpha x beta2 + alpha2 x beta

Hey Aussie!

$\alpha+\beta=-\frac{b}{a}=6\\\alpha\beta=\frac{c}{a}=3\\\alpha\beta^2+\alpha^2\beta=\alpha\beta(\alpha+\beta)$

And then you just sub and go to get 18 as your answer ;D
Post by: katnisschung on November 13, 2016, 11:55:58 am
I probably got it completely wrong but don't know becos they're arent any answers

Q) ap: 52,46,40
Find the smallest value of n so Sn<0
so i got n=19

i couldn't get it any bigger becos the limit for file size

Moderator Edit: Merged question w/ answer
Post by: RuiAce on November 13, 2016, 12:02:41 pm
so i got n=19

i couldn't get it any bigger becos the limit for file size
Snipping tool (automatically in Windows 7 and above) works miracles for captures; saves you having to use PrtScn to get a full screenshot

A quick check on WolframAlpha confirms that answer.
Post by: katnisschung on November 13, 2016, 12:04:12 pm
Thanks Ruiace
Yeah I should start using wolfram alpha more
Post by: katnisschung on November 13, 2016, 12:25:50 pm
A gp has 1st term 12
the terms of the sequence are all different and the sum of the 1st 2 two terms is equal to twice the third term

Find the 1st 3 terms of the sequence

So I interpreted this as
a=12
S2=2T3
So how do I know which formula to use for Sn becos I don't know what r is ? Or do I test them into both formulas
Post by: RuiAce on November 13, 2016, 01:08:30 pm
A gp has 1st term 12
the terms of the sequence are all different and the sum of the 1st 2 two terms is equal to twice the third term

Find the 1st 3 terms of the sequence

So I interpreted this as
a=12
S2=2T3
So how do I know which formula to use for Sn becos I don't know what r is ? Or do I test them into both formulas
$r\text{ is something you need to }\textit{find}$
\text{Using }2T_3=S_2\text{ we have:}\\ \begin{align*}2ar^{3-1}&= \frac{a(r^2-1)}{r-1}\\ 2r^2(r-1)&=(r^2-1)\\ 2r^2(r-1)&=(r-1)(r+1)\\ 2r^2&=r+1\tag{*} \\ (2r+1)(r-1)&=0\\ r&=-\frac{1}{2}\tag{**}\end{align*}\\ *-\text{ Note that }r\neq 1\text{ as all the terms are different, so }r-1\text{ is cancellable}\\ **-\text{ For similar reasons, }r=1\text{ was discarded.}
Post by: teapancakes08 on November 13, 2016, 09:22:44 pm
Having trouble visualising this:

31. The track of a gramophone record is in the shape of a spiral curve and may be considered as a number of concentric circles of inner and outer radius 5.25cm and 10.5cm respectively. The record rotates at 33 1/3 rev/min and takes 18 minutes to play. Find the length of the track.

I'm presuming you have to use Sn = (n/2)(a+l) as it's possible to find both values by using C = 2πr (if I'm doing this correctly...), but the other values throw me off. Do I use d = st to help solve the question, or is there a entirely different way of solving it? The question also suggests to take π = 22/7, although I'm not sure what to do with it...any suggestions? Thanks to anyone who can help ^^
Post by: jakesilove on November 13, 2016, 10:12:52 pm
Having trouble visualising this:

31. The track of a gramophone record is in the shape of a spiral curve and may be considered as a number of concentric circles of inner and outer radius 5.25cm and 10.5cm respectively. The record rotates at 33 1/3 rev/min and takes 18 minutes to play. Find the length of the track.

I'm presuming you have to use Sn = (n/2)(a+l) as it's possible to find both values by using C = 2πr (if I'm doing this correctly...), but the other values throw me off. Do I use d = st to help solve the question, or is there a entirely different way of solving it? The question also suggests to take π = 22/7, although I'm not sure what to do with it...any suggestions? Thanks to anyone who can help ^^

Weird question! Here's how I interpret it; if it takes 18 minutes to play, and the speed is 33 1/3 rev/min, then the number of revolutions will be (33.3333)*(18)=600 revolutions. So, we have 600 circles with radius' between 5.25cm and 10.5cm. That means that the radius will increase by (10.5-5.25)/600= 0.0525cm with each subsequent circle. This is a seriously weird question.

Our first circumference is 2π(5.25), then 2π(5.325) etc. etc. Clearly, this is increasing by 2*0.0525π with each circle, so that is going to be our value for d! We have n, a, l and d, so we can sum up the equation :)

Let me know if that all made sense!
Post by: teapancakes08 on November 13, 2016, 10:55:17 pm
Weird question! Here's how I interpret it; if it takes 18 minutes to play, and the speed is 33 1/3 rev/min, then the number of revolutions will be (33.3333)*(18)=600 revolutions. So, we have 600 circles with radius' between 5.25cm and 10.5cm. That means that the radius will increase by (10.5-5.25)/600= 0.0525cm with each subsequent circle. This is a seriously weird question.

Our first circumference is 2π(5.25), then 2π(5.325) etc. etc. Clearly, this is increasing by 2*0.0525π with each circle, so that is going to be our value for d! We have n, a, l and d, so we can sum up the equation :)

Let me know if that all made sense!

It seriously does...

My working out kind of looked like this:

full rev = speed*time
C = 2πr
a = 2π*5.25
l = 2π10.5

d (total revolutions) = speed*time
hence n = vt
n = 100/3 x 18
n = 600

...and then I pretty much plug in everything into Sn = (n/2) (a+l) and then BOOM. Answer. (Which is 296.88m for 2.d.p or 297m if you round it up to the nearest whole.)

To be honest the first thing that came to mind was physics, haha...

But yeah, it makes a lot more sense now - thanks so much for the clarifications ;D
Post by: jakesilove on November 14, 2016, 11:19:26 am
It seriously does...

My working out kind of looked like this:

full rev = speed*time
C = 2πr
a = 2π*5.25
l = 2π10.5

d (total revolutions) = speed*time
hence n = vt
n = 100/3 x 18
n = 600

...and then I pretty much plug in everything into Sn = (n/2) (a+l) and then BOOM. Answer. (Which is 296.88m for 2.d.p or 297m if you round it up to the nearest whole.)

To be honest the first thing that came to mind was physics, haha...

But yeah, it makes a lot more sense now - thanks so much for the clarifications ;D

No problem! Glad I could help :)