September 19, 2020, 09:51:10 pm

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#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18690 on: August 11, 2020, 07:39:57 pm »
0
I'm just copying my past message because I don't want it to get lost in the past messages lol

Hello! I hope you all are doing well

I have a question on sinusoidal graphs, it's a small part of a larger question.
What is h if d=0 when h=0 for the equation d= 2 +/- 3sin(2π/15(t-h))

I wrote the 3 as either negative or positive as we only know that the amplitude is 3. I'm unsure whether this is enough information to find h, and I am confused since wouldn't there be an unlimited values of h? The answer states that h = 1.74202.

I would really appreciate some help!  (•◡•) /
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#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18691 on: August 11, 2020, 07:47:20 pm »
+1
I'm just copying my past message because I don't want it to get lost in the past messages lol

Hello! I hope you all are doing well

I have a question on sinusoidal graphs, it's a small part of a larger question.
What is h if d=0 when h=0 for the equation d= 2 +/- 3sin(2π/15(t-h))

I wrote the 3 as either negative or positive as we only know that the amplitude is 3. I'm unsure whether this is enough information to find h, and I am confused since wouldn't there be an unlimited values of h? The answer states that h = 1.74202.

I would really appreciate some help!  (•◡•) /

I mean, if when d=0, h=0, then h=0? I think you've mislabeled something here.

As for whether the 3 is positive or negative, I think we're going to need to know more context. Can you give us the information from earlier parts? Eg, you've said it's +/-3 because you only know the amplitude is 3, but that's information that us on AN don't even know, so we can only trust your word which may have misinterpreted the actual question
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#### humanbeing

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##### Re: VCE Methods Question Thread!
« Reply #18692 on: August 11, 2020, 08:04:54 pm »
0
Hi,

Methods 1/2 student here. For the image that I've attached and the specific part I've highlighted, when I do that on my CAS, I get {0,0} instead of the highlighted answer. As far as I know, I've done all of the above steps the same, so if anyone could pls help me that would be much appreciated.

#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18693 on: August 11, 2020, 08:21:48 pm »
+1
Hi,

Methods 1/2 student here. For the image that I've attached and the specific part I've highlighted, when I do that on my CAS, I get {0,0} instead of the highlighted answer. As far as I know, I've done all of the above steps the same, so if anyone could pls help me that would be much appreciated.

Assuming you've done everything exactly the same as the image, there's no reason it shouldn't work. Maybe try subbing the values in individually and see if it works?

Hope this helps.
VCE Class of 2021
2020: Maths Methods [?]
2021: English [?] Biology [?] Chemistry [?] Physics [?] Specialist Maths [?]

#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18694 on: August 12, 2020, 12:18:12 pm »
0
I mean, if when d=0, h=0, then h=0? I think you've mislabeled something here.

As for whether the 3 is positive or negative, I think we're going to need to know more context. Can you give us the information from earlier parts? Eg, you've said it's +/-3 because you only know the amplitude is 3, but that's information that us on AN don't even know, so we can only trust your word which may have misinterpreted the actual question

Ughhhh, why do I always make mistakes when writing my questions on AN??
Yep! I meant 'find h if d=0 when t=0', whoops! The actual question is much longer, but I'll include the answers from the previous questions for extra information. All I really know is that d is the distance from the water surface to point P on a water wheel with radius 3, and t is the time is seconds. Also, the wheel rotates at 4 revolutions per minute.

What is h if d=0 when t=0 for the equation d= 2 +bsin(2π/15(t-h)). For this equation, the period is 15 and the amplitude is 3. I put 'b' because we don't know the exact value, right? (It's either 3 or -3. Correct me if I'm wrong I'm just confused on whether the value of h is indefinite, so any explanation on why h = 1.74 would be great  Thank you for your help
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#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #18695 on: August 12, 2020, 12:52:11 pm »
0
Ughhhh, why do I always make mistakes when writing my questions on AN??
Yep! I meant 'find h if d=0 when t=0', whoops! The actual question is much longer, but I'll include the answers from the previous questions for extra information. All I really know is that d is the distance from the water surface to point P on a water wheel with radius 3, and t is the time is seconds. Also, the wheel rotates at 4 revolutions per minute.

What is h if d=0 when t=0 for the equation d= 2 +bsin(2π/15(t-h)). For this equation, the period is 15 and the amplitude is 3. I put 'b' because we don't know the exact value, right? (It's either 3 or -3. Correct me if I'm wrong I'm just confused on whether the value of h is indefinite, so any explanation on why h = 1.74 would be great  Thank you for your help
h is that value since if you shift it, it each minute it begins and ends at y = 0 after 4 rotations.

If you graph the funciton you'll notice.

That at 0 seconds and 60 seconds it is at y = 0, I'm assuming that's the reason for the shift is there any part of the quesetion that states that d starts at 0?
If so, t = 0 and d = 0 that is (0, 0). There are technically infinte solutions yes. That's why people write it as a function like cos (45 + 360k) where k is an integer. But they probably chose the closest solution or something. It's basically asking 'for what value of h does the function intersect (0, 0)'

The example cos (45 + 360k) was made up btw.

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18696 on: August 12, 2020, 01:02:17 pm »
+5
Ughhhh, why do I always make mistakes when writing my questions on AN??
Yep! I meant 'find h if d=0 when t=0', whoops! The actual question is much longer, but I'll include the answers from the previous questions for extra information. All I really know is that d is the distance from the water surface to point P on a water wheel with radius 3, and t is the time is seconds. Also, the wheel rotates at 4 revolutions per minute.

What is h if d=0 when t=0 for the equation d= 2 +bsin(2π/15(t-h)). For this equation, the period is 15 and the amplitude is 3. I put 'b' because we don't know the exact value, right? (It's either 3 or -3. Correct me if I'm wrong I'm just confused on whether the value of h is indefinite, so any explanation on why h = 1.74 would be great  Thank you for your help

All good! At least you're honest about it - I've had some students (not from AN) come to me and go, "no!! Of course I copied the question right, maybe you're just a bad tutor?!", and then when I gave them proof that the question they asked couldn't be answered, they'd come back with, "oh, turns out I copied it wrong. It's actually this" without even an apology.

Okay, so now that I've got a better understanding of what's going on, let's work with what we know. d=0 when t=0, this gives us:

$0=2+b\sin\left(\frac{2\pi}{15}(0-h)\right)$

Since h is a translation factor, we know that it can have an INFINITE possible number of values - but all of them should be spaced 15 seconds apart. The reason it's 15 seconds apart is because the period is 15 seconds (feel free to confirm this for yourself). So, let's find the first value of h for 0<t<15, so first we solve for h:

$\frac{-2}{b}=\sin\left(\frac{2\pi}{15}(-h)\right)$

Okay, so this is immediately frustrating because we can't use exact values, and don't know if b is negative or positive!!. But, that's not as difficult as you think. Here, consider the two sin waves below:

https://www.desmos.com/calculator/9cvmthhh0u (note that you can turn off graphs by clicking the red or blue circle)

So, there is logic I could use to solve this. Notice that the red curve immediately goes up, while the right curve immediately goes down. That would mean that the sign of b is going to depend on whether that point P initially goes up or down. But that logic will rely on information you don't necessarily know - so instead, let's just be simple, and assume b=+3. If we do that, we get:

$\frac{-2}{3}=\sin\left(\frac{2\pi}{15}(-h)\right)\\
\sin^{-1}\left(\frac{-2}{3}\right)=-h\frac{2\pi}{15}\\
-0.729=-h\frac{2\pi}{15}\\
-1.742=-h\\
h=1.742$

But if that doesn't sit well with you, it shouldn't - it doesn't with me, either. So, let's see what happens if we let b=-3:

$\frac{-2}{-3}=\sin\left(\frac{2\pi}{15}(-h)\right)\\
\sin^{-1}\left(\frac{2}{3}\right)=-h\frac{2\pi}{15}\\
0.729=-h\frac{2\pi}{15}\\
1.742=-h\\
h=-1.742$

How weird - we get the exact same number, just a different sign! If you're curious, that's because sin(x) is an odd function (f(-x)=-f(x), I think this definition unfortunately got removed from the methods curriculum, but it's still kinda cool!), and doesn't really have bearing on this question. So, how do our graphs look with the h value added in?

https://www.desmos.com/calculator/sbrggivf0k

Well, the truth is, they both have d=0 when t=0, so as far as I'm concerned - both are valid answers. See your teacher for any reason as to why one of these answers isn't correct - it might be due to some information that you didn't even realise was information!

h is that value since if you shift it, it each minute it begins and ends at y = 0 after 4 rotations.

If you graph the funciton you'll notice.
(Image removed from quote.)
That at 0 seconds and 60 seconds it is at y = 0, I'm assuming that's the reason for the shift is there any part of the quesetion that states that d starts at 0?
If so, t = 0 and d = 0 that is (0, 0). There are technically infinte solutions yes. That's why people write it as a function like cos (45 + 360k) where k is an integer. But they probably chose the closest solution or something. It's basically asking 'for what value of h does the function intersect (0, 0)'

The example cos (45 + 360k) was made up btw.

You're not wrong, the problem is your answer misses the question that was being asked. In your graph, you've chosen there are two points that intersect with the x-axis - one in which the curve then starts going down, and one in which the curve starts going up. We don't know which of these points is making the intersection in Azila's answer, hence why I've explored what happens for both cases.
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

Previous Study:
Bachelor of Science Advanced (Research) - Monash University, majoring in Mathematical Statistics and Chemistry
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#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18697 on: August 12, 2020, 09:59:17 pm »
0
All good! At least you're honest about it - I've had some students (not from AN) come to me and go, "no!! Of course I copied the question right, maybe you're just a bad tutor?!", and then when I gave them proof that the question they asked couldn't be answered, they'd come back with, "oh, turns out I copied it wrong. It's actually this" without even an apology.

Okay, so now that I've got a better understanding of what's going on, let's work with what we know. d=0 when t=0, this gives us:

$0=2+b\sin\left(\frac{2\pi}{15}(0-h)\right)$

Since h is a translation factor, we know that it can have an INFINITE possible number of values - but all of them should be spaced 15 seconds apart. The reason it's 15 seconds apart is because the period is 15 seconds (feel free to confirm this for yourself). So, let's find the first value of h for 0<t<15, so first we solve for h:

$\frac{-2}{b}=\sin\left(\frac{2\pi}{15}(-h)\right)$

Okay, so this is immediately frustrating because we can't use exact values, and don't know if b is negative or positive!!. But, that's not as difficult as you think. Here, consider the two sin waves below:

https://www.desmos.com/calculator/9cvmthhh0u (note that you can turn off graphs by clicking the red or blue circle)

So, there is logic I could use to solve this. Notice that the red curve immediately goes up, while the right curve immediately goes down. That would mean that the sign of b is going to depend on whether that point P initially goes up or down. But that logic will rely on information you don't necessarily know - so instead, let's just be simple, and assume b=+3. If we do that, we get:

$\frac{-2}{3}=\sin\left(\frac{2\pi}{15}(-h)\right)\\
\sin^{-1}\left(\frac{-2}{3}\right)=-h\frac{2\pi}{15}\\
-0.729=-h\frac{2\pi}{15}\\
-1.742=-h\\
h=1.742$

But if that doesn't sit well with you, it shouldn't - it doesn't with me, either. So, let's see what happens if we let b=-3:

$\frac{-2}{-3}=\sin\left(\frac{2\pi}{15}(-h)\right)\\
\sin^{-1}\left(\frac{2}{3}\right)=-h\frac{2\pi}{15}\\
0.729=-h\frac{2\pi}{15}\\
1.742=-h\\
h=-1.742$

How weird - we get the exact same number, just a different sign! If you're curious, that's because sin(x) is an odd function (f(-x)=-f(x), I think this definition unfortunately got removed from the methods curriculum, but it's still kinda cool!), and doesn't really have bearing on this question. So, how do our graphs look with the h value added in?

https://www.desmos.com/calculator/sbrggivf0k

Well, the truth is, they both have d=0 when t=0, so as far as I'm concerned - both are valid answers. See your teacher for any reason as to why one of these answers isn't correct - it might be due to some information that you didn't even realise was information!

h is that value since if you shift it, it each minute it begins and ends at y = 0 after 4 rotations.

If you graph the funciton you'll notice.
(Image removed from quote.)
That at 0 seconds and 60 seconds it is at y = 0, I'm assuming that's the reason for the shift is there any part of the quesetion that states that d starts at 0?
If so, t = 0 and d = 0 that is (0, 0). There are technically infinte solutions yes. That's why people write it as a function like cos (45 + 360k) where k is an integer. But they probably chose the closest solution or something. It's basically asking 'for what value of h does the function intersect (0, 0)'

The example cos (45 + 360k) was made up btw.

Thank you guys so much!  I really appreciate the effort you put in to help, it's great!
Just someone who likes to learn a lot of questions.

Aspiring medical practitioner! ʕ•́ᴥ•̀ʔっ

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18698 on: August 13, 2020, 11:28:15 am »
+1
First off, I'd like to thank James for his last response to my question
It helped.

Second, I have a new question regarding simultaneous equations.
I've gone wrong somewhere in the process of answering this question on the Elimination method. I can't see where.
If anyone here is willing to help I would much appreciate it
Many thanks,
Corey

#### The Cat In The Hat

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##### Re: VCE Methods Question Thread!
« Reply #18699 on: August 13, 2020, 11:32:55 am »
0
I tried looking at it, but I'm afraid I got completely lost in the first step. Where do you get x - 2y + 4 = 10 from?
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#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18700 on: August 13, 2020, 11:39:52 am »
+1
First off, I'd like to thank James for his last response to my question
It helped.

Second, I have a new question regarding simultaneous equations.
I've gone wrong somewhere in the process of answering this question on the Elimination method. I can't see where.
If anyone here is willing to help I would much appreciate it
Many thanks,
Corey

Making a2 was unnecessary, but not wrong, I guess. Your issue is you said 8y=4 means that y=2

I tried looking at it, but I'm afraid I got completely lost in the first step. Where do you get x - 2y + 4 = 10 from?

They added 4 to both sides of the first equation
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#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18701 on: August 13, 2020, 11:53:34 am »
0
Making a2 was unnecessary, but not wrong, I guess. Your issue is you said 8y=4 means that y=2

They added 4 to both sides of the first equation

Ah I see, that was redundant. Thank you
Another silly operations error. I seem to make a lot of those. Anything you would recommend that could reduce the amount I make?
Gratefully,
Corey

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18702 on: August 13, 2020, 12:27:22 pm »
+1
Ah I see, that was redundant. Thank you
Another silly operations error. I seem to make a lot of those. Anything you would recommend that could reduce the amount I make?
Gratefully,
Corey

So like, this might sound a little odd, but my suggestion is to just whenever you make a mistake - don't just figure out what you did wrong, do the question again from scratch. Hell, if you know you're making mistakes like these, don't check your working to see what you did wrong (because hell, even for me it took me a while to see what you did wrong, and I'm not biased by your own thinking of, "but I did everything right?" that everyone has when analysing their own questions or proof-reading their own essays), just do the question again blindly and see if you get the same answer.

I had a teacher who used to say, "practice doesn't make perfect - perfect practice makes perfect". And so, if you're constantly making silly little calculation errors, that's not perfect practice. But if you make sure that every question you do, you've done it at least once without making this little mistakes, you might be able to slowly ease them out of your own thinking so that you don't make them in an exam. Hell, if you have time in an exam, find the questions worth the most marks and do them a second time to see if you get the same answer. Can't hurt, as long as you're not doing it on your actual exam booklet (you should be able to get spare paper in your exams)

I have no proof that this will work, but tbh the only other advice I have is, "make sure you double check your working, always be careful, read every line twice, etc.", which we all know isn't that effective.
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

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#### Bri MT

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##### Re: VCE Methods Question Thread!
« Reply #18703 on: August 13, 2020, 02:35:25 pm »
0
Ah I see, that was redundant. Thank you
Another silly operations error. I seem to make a lot of those. Anything you would recommend that could reduce the amount I make?
Gratefully,
Corey

I've found using a log of mistakes helps me & as meithy said making sure to redo questions correctly.  This might be obvious but I also find that slowing down and writing the steps out properly rather than trying to rush through it in my head makes a big difference.

Also going to add in that when checking an answer try seeing if you can use a different method so you don't fall into the same thinking traps.

Hope this helps

#### Coolgalbornin03Lo

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##### Re: VCE Methods Question Thread!
« Reply #18704 on: August 15, 2020, 01:29:54 pm »
0
Why is loge(4)/2 the same as loge(2)??
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