Post by: Joseph41 on January 30, 2019, 03:17:59 pm
QCE MATHS METHODS Q&A THREAD
What is this thread for?
If you have general questions about the QCE Maths Methods course (both Units 1&2 and 3&4) or how to improve in certain areas, this is the place to ask! 👌
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Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.
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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.
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Post by: Declan.B on November 03, 2019, 07:05:41 pm
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule
product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln(b)×b^t
b^' (x)=e^t×ln(e)=e^t
product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln(b)+1)
let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt
p^' (x)=(da^(d-1)-da^(d-1)×ln(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln(b)×d(be)^(d-1×t) )]
Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B
Post by: DrDusk on November 03, 2019, 07:34:27 pm
Hey guys,Sorry but I can't make sense of it written like this =( If you could perhaps write it down physically and upload it as an attachment or even better write it in LateX(any of which would be fine), then I could help you =)
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule
product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln(b)×b^t
b^' (x)=e^t×ln(e)=e^t
product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln(b)+1)
let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt
p^' (x)=(da^(d-1)-da^(d-1)×ln(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln(b)×d(be)^(d-1×t) )]
Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B
Post by: RuiAce on November 03, 2019, 08:11:50 pm
Hey guys,Hey,
I'm just trying to differentiate the following function:
y=1000×b^t×e^t (c(a(1-b^t×e^t ))^d ).
The "t" values are the x values (I put them as t to prevent confusion with multiplication symbols)
I tried to break up sections of it and solve using product rule
product rule:let f(x)= 1000c(be)^t,let g(x)=a^d (1-(be)^dt)
y' =f^' (x)g(x)+f(x) g^' (x)
let z(x)=f(x)
let 1000cb^t=a(x),e^t=b(x)
a^' (x)=1000×c×ln(b)×b^t
b^' (x)=e^t×ln(e)=e^t
product rule: z^' (x)=a^' (x)×b(x)+a(x) b^' (x)
z^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
z(x)=f(x)∴z^' (x)=f^' (x)
f^' (x)=(1000×c×ln(b)×b^t ) e^t+(1000×c×b^t ) e^t
f^' (x)=e^t×1000×c×b^t (ln(b)+1)
let p(x)=g(x)
p(x)=a^d (1-(be)^dt )
p(x)=(a^d-a^d (be)^dt
p^' (x)=(da^(d-1)-da^(d-1)×ln(b)×d(be)^(d-1×t) )
p^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
p(x)=g(x)∴p^' (x)=g^' (x)
g^' (x)=da^(d-1) (1- ln(b)×d(be)^(d-1×t) )
y^'=f^' (x)g(x)+f(x) g^' (x)
y^'=[e^t×1000×c×b^t (ln(b)+1)][a^d (1-(be)^dt )]+[1000c(be)^t ][da^(d-1) (1- ln(b)×d(be)^(d-1×t) )]
Sorry if it's a bit messy (I'm new to differentiating exponentials) but could someone let me know if my working is correct. I didn't want to simplify my answer if it wasn't correct. Also, the additional coefficients are just variables that I wanted to do a general case for before substituting.
~Declan.B
Before we jump into this, I just want to check if this was the function you were after?
\[ y = 100 b^t e^t \left( c\left(a (1-b^t e^t) \right)^d\right) \]
That, and:
a) you're after a derivative in terms of \(t\)?
b) do I assume that \(a\), \(b\), \(c\) and \(d\) are constants?
(It may be really hard to type it out without LaTeX ngl. If you feel it's too hard, please feel welcome to just post pictures/screenshots instead. :))
Post by: Declan.B on November 03, 2019, 08:51:50 pm
~Declan Bolster
Post by: RuiAce on November 04, 2019, 01:01:30 pm
I've attached a write-out of the question and my answer.I'm taking a more closer look now and there's a few things I should comment on from my first glance:
~Declan Bolster
- You're already using \(a\) and \(b\) for constants already. It's not recommended to now define stuff like \(a(x)\) \(b(x)\), where you use the same letter.
- Your derivative is in terms of \(t\). Use functions in terms of \(t\), not \(x\).
- \( 1 - (be)^{dt}\) isn't the same thing as \( (1-be^t)^d\).
So I feel the problem got made a bit more convoluted than what it had to be! This is how I would've done it. First rewrite:
\[ y = 1000 a^d c (be)^t \left( 1- (be)^t\right)^d. \]
Since \(a\), \(c\) and \(d\) are constants here, we can just leave the \(1000 a^d c\) bit in front, and focus on everything else for the product rule.
Set up for the product rule:
\begin{align*}
f(t) &= (be)^t \\
g(t) &= \left( 1 - (be)^t\right)^d.
\end{align*}
Judging by your working out, I'm assuming you're allowed to use the rule \( \boxed{\frac{d}{dx}a^x = a^x \ln (a)} \). Applying it here, we have
\begin{align*}
f^\prime (t) &= (be)^t \ln (be)\\
g^\prime(t) &= (be)^t\ln (be) \times d\left(1 - (be)^t \right)^{d-1}
\end{align*}
having used the chain rule on the second computation.
Note that \(\ln(be) = \ln(b)+\ln(e) = \ln(b)+1\), which does not conflict with some of the stuff I see you did there.
Hence, putting everything together, the derivative is
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c \left[g(t)f^\prime(t) + f(t)g^\prime(t) \right]\\
&= 1000 a^d c\left[ \left(1 - (be)^t\right)^d (be)^t \ln (be) + (be)^t (be)^t \ln (be) d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
This certainly can be simplified further, but it's still going to be ugly for sure. If you want me to simplify it, feel free to ask :)
Post by: Declan.B on November 04, 2019, 07:56:33 pm
I'm taking a more closer look now and there's a few things I should comment on from my first glance:Thanks for the help. If you're not too busy simplifying it would be great!
- You're already using \(a\) and \(b\) for constants already. It's not recommended to now define stuff like \(a(x)\) \(b(x)\), where you use the same letter.
- Your derivative is in terms of \(t\). Use functions in terms of \(t\), not \(x\).
- \( 1 - (be)^{dt}\) isn't the same thing as \( (1-be^t)^d\).
So I feel the problem got made a bit more convoluted than what it had to be! This is how I would've done it. First rewrite:
\[ y = 1000 a^d c (be)^t \left( 1- (be)^t\right)^d. \]
Since \(a\), \(c\) and \(d\) are constants here, we can just leave the \(1000 a^d c\) bit in front, and focus on everything else for the product rule.
Set up for the product rule:
\begin{align*}
f(t) &= (be)^t \\
g(t) &= \left( 1 - (be)^t\right)^d.
\end{align*}
Judging by your working out, I'm assuming you're allowed to use the rule \( \boxed{\frac{d}{dx}a^x = a^x \ln (a)} \). Applying it here, we have
\begin{align*}
f^\prime (t) &= (be)^t \ln (be)\\
g^\prime(t) &= (be)^t\ln (be) \times d\left(1 - (be)^t \right)^{d-1}
\end{align*}
having used the chain rule on the second computation.
Note that \(\ln(be) = \ln(b)+\ln(e) = \ln(b)+1\), which does not conflict with some of the stuff I see you did there.
Hence, putting everything together, the derivative is
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c \left[g(t)f^\prime(t) + f(t)g^\prime(t) \right]\\
&= 1000 a^d c\left[ \left(1 - (be)^t\right)^d (be)^t \ln (be) + (be)^t (be)^t \ln (be) d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
This certainly can be simplified further, but it's still going to be ugly for sure. If you want me to simplify it, feel free to ask :)
~Declan.B
Post by: RuiAce on November 05, 2019, 02:19:12 pm
Thanks for the help. If you're not too busy simplifying it would be great!Sure. :)
~Declan.B
Basically the simplifying relies on some careful factorisation. Firstly it should be clear that \( (be)^t\) and \(\ln (be)\) are common factors, so we first pull them out.
\begin{align*}
\frac{dy}{dt} &= 1000a^d c (be)^t \ln (be) \left[ \left( 1 - (be)^t\right)^d + (be)^t d \left(1-(be)^t \right)^{d-1} \right]\\
&= 1000a^d c(be)^t \left( \ln(b) + 1 \right)\left[ \left( 1 - (be)^t\right)^d + (be)^t d \left(1-(be)^t \right)^{d-1} \right]
\end{align*}
A common thing with product rule computations involving powers is that a power can also be factored out. Here, observe that \( \left(1-(be)^t\right)^{d-1}\) can be factored out.
For the second term, this should be clear. For the first term, the reason this works is because of index laws: \( \left( 1-(be)^t\right)^d = \left( 1-(be)^t\right) \times \left( 1-(be)^t\right)^{d-1} \).
\begin{align*}
\frac{dy}{dt} &= 1000 a^d c (be)^t (\ln(b) +1) \left( 1-(be)^t\right)^{d-1} \left[1 + (be)^t d \right].
\end{align*}
Once here, I can't see any more simplification possible :)
Post by: A.Rose on March 12, 2020, 07:33:06 pm
Would anyone be able to help me with this Unit 4 Topic 2 (Trig with sine and cosine rule) Question:
"A surveyor measures the angle of elevation of the top of a mountain from a point at sea level as 20∘. She then travels 1000 m along a road that slopes uniformly uphill towards the mountain. From this point, which is 100 m above sea level, she measures the angle of elevation as 23∘. Find the height of the mountain above sea level, correct to the nearest metre."
I mainly need to make sure I am drawing it correctly and then I can see how I go with finding the height.
Thank you!
Post by: fun_jirachi on March 13, 2020, 12:24:00 am
This should be a rough diagram :)
Spoiler
(https://i.imgur.com/Ek4j7Qa.png)
Hope this helps :)
Post by: A.Rose on March 13, 2020, 10:30:46 pm
Although I got h = 1839 m plus 100 for height above sea level would be 1939 m and I don't know if that is right? Would you be able to check, please?
Also if possible there is another question I am stuck on:
From a clifftop 11 m above sea level, two boats are observed. One has an angle of depression of 45∘ and is due east, the other an angle of depression of 30∘ on a bearing of 120∘. Calculate the distance between the boats.
I'm a bit confused about where the 2nd boat is positioned - can you help me with the drawing?
Thank you!
Post by: fun_jirachi on March 13, 2020, 11:47:12 pm
This one is a little tougher to draw - it's a 3D diagram on a 2D plane! I don't trust my hand to draw things that other people can understand, so hopefully my MS Paint work is good enough :)
Spoiler
(https://i.imgur.com/UaiZtCy.png)
Hope this helps :)
Post by: Specialist_maths on March 14, 2020, 09:01:46 am
Thank you yes that did help!
Although I got h = 1839 m plus 100 for height above sea level would be 1939 m and I don't know if that is right? Would you be able to check, please?
Hi A.Rose,
I also got 1939 m for the height of the mountain, by solving the following equations simultaneously.
\[ \begin{align*} \tan(23^{\circ})&=\frac{a}{b} \tag{1} \\ \tan(20^{\circ})&=\frac{a+100}{b+c} \tag{2}\\
c^2+100^2&=1000^2 \tag{3} \end{align*} \]
Post by: fun_jirachi on March 14, 2020, 09:22:32 am
Hi A.Rose,
I also got 1939 m for the height of the mountain, by solving the following equations simultaneously.
\[ \begin{align*} \tan(23^{\circ})&=\frac{a}{b} \tag{1} \\ \tan(20^{\circ})&=\frac{a+100}{b+c} \tag{2}\\
c^2+100^2&=1000^2 \tag{3} \end{align*} \]
I've made an error - you would both be correct! Just realised I misread my working from one line to the next :)
Post by: fun_jirachi on March 19, 2020, 10:54:30 pm
I think the reason that you're a bit stuck on this question is that it's impossible to have a triangle with those specified dimensions, then have the third side equal to 325m! The question should really have angle ACB (which essentially means you have a single application of the cosine rule to do to get the answer), instead of angle CAB, which results in the third side being 677m.
Hope this helps :)
Post by: A.Rose on April 27, 2020, 05:56:57 pm
I'm stuck on a question about Bernoulli sequences and the binomial distribution from Unit 4 Topic 3 Methods.
If it says 'at least' does this mean P(X>9)? And could you calculate that by doing P(X>9)=1-P(0≤X≤9)? I tried that and I didn't achieve the answer in the textbook.
Thank you!
Post by: Bri MT on April 27, 2020, 08:35:51 pm
Hi!
I'm stuck on a question about Bernoulli sequences and the binomial distribution from Unit 4 Topic 3 Methods.
If it says 'at least' does this mean P(X>9)? And could you calculate that by doing P(X>9)=1-P(0≤X≤9)? I tried that and I didn't achieve the answer in the textbook.
Thank you!
Hi!
Your thinking is good here except that "at least" means greater than or equal to - not just greater than :)
So in your subtraction you shouldn't include the case where x=9
Post by: RuiAce on May 27, 2020, 06:20:44 pm
Hi!At first glance, I actually think that it is really simple. Technically speaking, "the distance from the foot of the cliff to a boat", and "the height of the cliff above sea level", are exactly the ingredients of a right-angled triangle formulation!
I have a question I'm stuck on from Unit 4 Topic 2 Methods. It's either really simple and I'm missing something or its generally a bit tricky. First time I read the question I thought of a right triangle but you can't assume that can you? Plus this Unit 4 methods, not Unit 1. ;D
So if you can't assume it is a right triangle then how do you find the height? It reminded me of the ambiguous case but I'm not sure how/if to use that property to answer the question. I would greatly appreciate the help as I am doing a revision of Unit 4 Topic 1&2 in class this week.
Thank you!
I am surprised that this is considered a Unit 4 question instead of Unit 1. For all I know I could be missing something too, but my logic came from the wording of the question entirely.
Perhaps there's follow-up questions to this problem that are more Unit 4 based? Or perhaps there's an answer to aim for?
Post by: orla007 on July 09, 2020, 05:31:19 pm
I am a little confused as to what I need to do to complete this question- as it is a proof the textbook doesn't have a solution.
Thanks in advance :)
Post by: Bri MT on July 09, 2020, 05:49:59 pm
Hi!
I am a little confused as to what I need to do to complete this question- as it is a proof the textbook doesn't have a solution.
Thanks in advance :)
Hi!
Welcome to the forums!
It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.
Once you have part a part b should be pretty straightforward.
Hope this helps and please let me know if this clarifies things enough :)
Post by: orla007 on July 09, 2020, 05:59:17 pm
Hi!
Welcome to the forums!
It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.
Once you have part a part b should be pretty straightforward.
Hope this helps and please let me know if this clarifies things enough :)
Ahhhh thank you!!!
Post by: Bri MT on July 09, 2020, 06:04:41 pm
Ahhhh thank you!!!
No worries! I was actually teaching the sine rule on Monday so it was very fresh in my memory :)
Post by: 1729 on July 09, 2020, 06:15:24 pm
Hi!
Welcome to the forums!
It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.
Once you have part a part b should be pretty straightforward.
Hope this helps and please let me know if this clarifies things enough :)
a+b/c can be separated into a/c +b/c
sinA/a=sinC/c
So sinA/sinC= a/c
sinB/b=sinC/c
So sinB/sinC=b/c
So sinB/sinC+ sinA/sinC=( sinA + sinB)/sinC
sinA/a=sinC/c
So sinA/sinC= a/c
sinB/b=sinC/c
So sinB/sinC=b/c
So sinB/sinC+ sinA/sinC=( sinA + sinB)/sinC
From then you should be able to get the second one.
Post by: matthew hay on August 26, 2020, 06:42:21 pm
Do you or anyone know if there is an annotated syllabus? or even annotated Units 4 syllabus?? Please reply no if not (it saves me the extravagant search!
Thanks and kind regards, Matt
Post by: Bri MT on August 26, 2020, 07:48:03 pm
I'm pretty sure there aren't any annotated QCE methods syllabuses. There are other methods resources listed here but an annotated syllabus isn't one of them.
Post by: snr.mmorris4.19 on September 15, 2020, 06:07:31 pm
P is a movable point on the line y=7-x
Find the coordinates of P when it is closest to the origin (0,0).
Post by: keltingmeith on September 15, 2020, 06:12:36 pm
whats the process behind solving these? Never seen them before i'm not sure where to start
P is a movable point on the line y=7-x
Find the coordinates of P when it is closest to the origin (0,0).
This is a really interesting question! What I'm going to do is give you some starting points, then why don't you let us know how far you get?
Firstly, you may remember that the straight-line distance between any two points is given by the equation:
which is secretly just Pythagoras' theorem in disguise. Because you don't know the exact points of P, you could actually pretend that it lies on the point x_2=x and y_2=y.
Another thing to think about is that closest and furthest have some synonyms that would normally raise alarm bells to any student studying differentiation - maybe you can use differentiation in some way, here?
EDIT: See below for some pointers on using a non-calculus approach - I tried to generalise for other curves, not just for straight lines, but their approach is WAY simpler and still worth knowing. Particularly if you're not up to differentiation, yet
Post by: fun_jirachi on September 15, 2020, 06:17:54 pm
There are multiple ways of solving this - the two I'd recommend you have a look at are a) using the perpendicular distance formula or b) finding a perpendicular line passing through the given point, then finding the point of intersection to determine the coordinates of P. Expressing the distance as a polynomial function to exploit it as a max/min problem seems excessive for a question that involves a point and a line, but is worth exploring as suggested by keltingmeith as it will provide practice for tougher locus problems :)
EDIT: beaten by keltingmeith
Post by: orla007 on October 02, 2020, 04:15:09 pm
I would love some help with this question.
Thanks in advance
Post by: dzach0 on December 20, 2020, 09:30:21 am
Does anyone know what specific topics and subtopics I have to revise for unit 1 at least? (I have attached the unit outlines given by my school)
Happy Holidays!
Post by: Bri MT on December 20, 2020, 10:05:12 am
I have just finished year 10 and got a very unhappy mark for my methods exam (D+). I feel like it is due to me switching from general midway through the year and not being able to keep up with the pace. I want to get ahead on the topics for year 11 so that I am very confident in what I do. However I do not understand what I need to specifically revise on as the unit outlines only consist of hard to understand learning goals.
Does anyone know what specific topics and subtopics I have to revise for unit 1 at least? (I have attached the unit outlines given by my school)
Happy Holidays!
Hey,
Welcome to the forums!
Great to see that you're taking a proactive approach to tackling this.
This is actually a pretty detailed lesson outline, I think it might just be unfamiliarity with the technical maths language that's making it a bit confusing.
For example, when it says "recognise and determine features of the graphs of 𝑦=𝑥2, 𝑦=𝑎𝑥2+𝑏𝑥+𝑐, 𝑦=𝑎(𝑥−𝑏)2+𝑐, and 𝑦=𝑎(𝑥−𝑏)(𝑥−𝑐), including their parabolic nature, turning points, axes of symmetry and intercepts"
You should be able to: take something that looks like \( ax^{2} + bx + c \) where a, b and c can be any numbers (but a won't be 0 otherwise it's linear rather than a quadratic) and know that this has a parabolic shape (go here and play around with different values of a, b & c to get a feel for the shape); it can have 0,1, or 2 x-intercepts and 1 y-intercept; and be able to find the turning point. As with other graph forms, if you want to find the y intercept you set x equal to zero & if you want to find the x intercept you set y equal to zero. Finding x-intercepts can be a bit trickier with quadratics than it is for linear equations so you need to learn how to factorise the equation in different ways and use that to help you + the quadratic equation .
I recommend that you do the quadratic section first (you're given textbook chapter numbers and there are heaps of online resources that teach people about quadratics) and make sure you have solid understanding before moving on because it's going to be very hard to understand other polynomials well if you don't get quadratics.
To go super-specific (skip the ones you already know well):
- look at binomial expansion e.g. (3 + a) (2+ b) or ( x - 5) (x+4)
- look at the reverse, doing basic factorising and rules for this (e.g. difference of two squares, perfect squares)
- be able to factorise things like \( x^{2} + 5x + 6 \)
- use the null factor law to see what the x intercepts are
- be able to deal with having a coefficient of \(x^{2}\) that isn't 1. e.g. multiply the whole above example by 2 or 3 or 5
- be able to use completing the square for factorisation
- be able to use completing the square for factorisation with a coefficient of \(x^{2}\) that isn't 1
- be able to use the null factor law on the above
- be able to read from, and make quadratics into turning point form
- be able to use the quadratic equation
- be able to plot quadratics using the above techniques (could have this dot point earlier) & find the equation if given a graph
- be able to use and find the discriminant
^^ All of the above are covered in year 10 maths lectures I gave earlier this year so they might be a good place to look, the slides are available in the free notes section
I hope this helps!
Post by: jasmine24 on March 22, 2021, 07:08:44 am
Post by: fun_jirachi on March 22, 2021, 07:33:28 am
Once you get this answer, all you really have to do is make sure you obtain a value for the mass and the end of the second stage that seems accurate (ie. by adding the result of the integral to the trunk radius then calculating the new mass.) Note that the density is irrelevant as the volume is almost analogous to the mass. Just remember to have your ratio the right way around as well.
Answer should be 4:1.
Post by: jasmine24 on March 22, 2021, 09:01:04 am
If you're given the rate of change for the trunk's growth, you can find out the amount it has grown by choosing an appropriate upper and lower bound then integrating the rate of change between those bounds.
Once you get this answer, all you really have to do is make sure you obtain a value for the mass and the end of the second stage that seems accurate (ie. by adding the result of the integral to the trunk radius then calculating the new mass.) Note that the density is irrelevant as the volume is almost analogous to the mass. Just remember to have your ratio the right way around as well.
Answer should be 4:1.
Thank you so much!
I used 10 and 0 as the bounds which I’m assuming is wrong considering I got 15 as the answer. Also, the method I originally tried was finding the indefinite interval then substituting t=0, r=15 to find c but since it didn’t work, I was wondering if u knew why this wouldn’t work.
Post by: fun_jirachi on March 22, 2021, 09:37:26 am
Might've just been an algebraic error, unfortunately - it seems to work here :D
EDIT: having thought about it again, you probably haven't read the question properly? This is for radial growth after 10 years, not for the actual radius. While using the method shown above does get you the actual radius, using an upper and lower bound will get you the radial change because we are taking R(10)-R(0) rather than the actual radius using our predefined radius function R(t) at t=10.
Post by: justsomerandom21 on May 12, 2021, 08:05:57 pm
Post by: fun_jirachi on May 12, 2021, 11:41:52 pm
I can't really give specific tips since I don't have any details, but some general tips:
- Do your best to calm down; it may be difficult to do so (and it's especially easy for me to say) but maintaining a level head is going to do you so much more good. Also, this is your first exam, so don't feel the need to put so much pressure on yourself; base your future performance and tactics on what happens this time
- Remember that whoever is issuing the exam cannot (at least, in theory they shouldn't) test you on something you haven't learned. If there's something that doesn't seem quite right, break it down into smaller bits so you can use the knowledge you do have to solve the question
- Use any reading time you do get effectively
- Any other previous exam tips like if you get stuck move on, etc. also apply here.
Hope this helps!
Post by: justsomerandom21 on June 06, 2021, 08:55:48 pm
Welcome to the forums! ;DSorry for the long reply, but thanks so much for the advice!!
I can't really give specific tips since I don't have any details, but some general tips:
- Do your best to calm down; it may be difficult to do so (and it's especially easy for me to say) but maintaining a level head is going to do you so much more good. Also, this is your first exam, so don't feel the need to put so much pressure on yourself; base your future performance and tactics on what happens this time
- Remember that whoever is issuing the exam cannot (at least, in theory they shouldn't) test you on something you haven't learned. If there's something that doesn't seem quite right, break it down into smaller bits so you can use the knowledge you do have to solve the question
- Use any reading time you do get effectively
- Any other previous exam tips like if you get stuck move on, etc. also apply here.
Hope this helps!
Post by: jinx_58 on July 14, 2021, 07:09:13 pm
Could someone please give me some tips on how to do well in a PSMT?
Many thanks. :)
Post by: K.Smithy on July 18, 2021, 06:00:25 pm
Howdy.
Could someone please give me some tips on how to do well in a PSMT?
Many thanks. :)
Hey Jinx_58,
There are a few things you need to make sure that you do:
- You must state any assumptions and observations you make
- You have to make evident the mathematical and technical procedures used
- Your report must address why your designs are reasonable
- You must discuss recommendations for improvement in your evaluation of design and touch on any strengths and limitations
I hope this helps, if you have any questions please don't hesitate to give us a shout :)
Katelyn
[/list]
Post by: jasmine24 on July 20, 2021, 10:48:44 pm
TIA :)
Post by: fun_jirachi on July 20, 2021, 11:40:38 pm
Note that we are given that \(b^2>3c\), so we can see that the case where there is/are one/none stationary points doesn't happen ie. two distinct roots to the derivative at \(\frac{-b\pm \sqrt{b^2-3c}}{3}.\)
\(f''(x)=6x+2b \implies \text{one solution to } f''(x)=0 \implies \text{one inflection point at } x=-\frac{b}{3}.\)
Clearly, the inflection point is the midpoint of the two stationary points, a small of computation will let you see that each stationary point is \(\frac{\sqrt{b^2-3c}}{3}\) units away from the point of inflection.
Let me know if I've missed anything / screwed up; should be correct! :)
Post by: jasmine24 on August 12, 2021, 09:03:35 pm
Post by: Bri MT on August 13, 2021, 10:02:46 am
Hi, my math methods exam is tomorrow and I was wondering if anyone knew how to solve this :)
Hey,
Are we told anything that student race times follow a normal distribution?
In that case, what you would want to do is:
- find the distribution using the information from each end ( number in time range / 900 = probability of the time being in that time range)
- then find the place on the distribution for 100/900 (make sure to pick the right end of the distribution :) )
Hope this helps and let me know if you have any questions :)
Post by: Gracey1415 on October 20, 2021, 11:38:18 am
Q: The Apache Orchard grows a very juicy apple called the Fuji apple. Fuji apples are picked and then sorted by diameter into three categories:
• small — diameter less than 60 mm
• jumbo — the largest 15% of the apples
• standard — all other apples.
f) Some apples are selected before sorting and are packed into bags of 6 to be sold at the front gate of the orchard. Determine the probability, correct to 4 decimal places, that one of these bags contains at least 2 jumbo apples.
The answer given is
Post by: Phytoplankton on October 20, 2021, 01:34:16 pm
Hi, I'm struggling with this question I'm not sure how the textbook gets to the answer. Would anyone be able to help?
Q: The Apache Orchard grows a very juicy apple called the Fuji apple. Fuji apples are picked and then sorted by diameter into three categories:
• small — diameter less than 60 mm
• jumbo — the largest 15% of the apples
• standard — all other apples.
f) Some apples are selected before sorting and are packed into bags of 6 to be sold at the front gate of the orchard. Determine the probability, correct to 4 decimal places, that one of these bags contains at least 2 jumbo apples.
The answer given is
Hey Gracey1415,
I believe the distribution they give i.e. Y~Bi(6,0.15) is correct as the number of apples (n) is 6 and the probability of having a jumbo apple (p) is 15%=0.15. However, the probability the textbook states is incorrect. As they ask for the probability that one of these bags contains at least 2 jumbo apples, this means that the probability is actually P(Y≥2) and not P(Y≤2). From here, you can either use the binomial probability formula (on the QCAA formula sheet) or use the Bcd function on your calculator to figure the answer out. I entered it into my calculator (as I'm too lazy to algebraically work it out) and the answer was indeed 0.2235.
Hope this helps! If you have any other questions or want me to elaborate, feel free to reply back! Good luck!
Post by: Amity H on October 26, 2021, 11:09:37 am
Sorry I tried to attach stuff but it was either too big or was the wrong file type.
Question was x = t^2 - 6t + 8, t greater than or equal to zero, t is in mins, x is in cm
A find its initial velocity
B when does its velocity equal zero, and what is its position at this time?
C what is its average velocity for the first 4 seconds?
D determine its average speed for the first 4 seconds.
I was fine with a-c but not sure about d.
External is on Thursday so any help would be greatly appreciated. Thanks heaps!
Post by: Gracey1415 on October 27, 2021, 07:20:33 pm
It is a binomial probability
with p = 0.18
and sample of 10
calculate Pr(P' < 0.2)
I have
P' = X/n = X/10
Pr(P' < 0.2) = Pr(X/10 < 0.2) = Pr(X<2)
The answers tell me in the calculator use binomial distribution function with n = 10, p=0.18
and the answer given is 0.4392.
But I can't get that I have no idea what to put in the x value part of the calculator, and should I be using binompdf or binom cdf?
Post by: Commercekid2050 on October 27, 2021, 08:52:54 pm
I have a question with binomial probabilities and inputting into the calculator.
It is a binomial probability
with p = 0.18
and sample of 10
calculate Pr(P' < 0.2)
I have
P' = X/n = X/10
Pr(P' < 0.2) = Pr(X/10 < 0.2) = Pr(X<2)
The answers tell me in the calculator use binomial distribution function with n = 10, p=0.18
and the answer given is 0.4392.
But I can't get that I have no idea what to put in the x value part of the calculator, and should I be using binompdf or binom cdf?
Hi,
It says Pr(x<2)
This does not include 2 and as this is discrete the x would become 1.
Also as it says x<2 you would be using Binominal CDF. It would be lower being 0 and higher being 1. N as you said would be 10 and the probability would be 0.18
Hope this helps.
Post by: bia234 on November 21, 2021, 03:07:00 pm
I've attached the function below.
Cheers! :)
Post by: fun_jirachi on November 21, 2021, 03:27:09 pm
If you need extra help ask again! :D
Post by: GreenNinja on December 27, 2021, 07:06:50 pm
Could someone please show how to answer this question. I would show some working out if I am able to, however I am thoroughly confused on even beginning to answer this.
Thanks.
Post by: fun_jirachi on December 27, 2021, 07:49:17 pm
\(\frac{d}{dx} e^{-3x}\sin (2x) = -3e^{-3x}\sin (2x) + 2e^{-3x}\cos (2x) = e^{-3x} (2\cos (2x) -3\sin (2x))\)
\(\frac{d}{dx} e^{-3x}\cos (2x) = -3e^{-3x}\cos(2x) - 2e^{-3x}\sin(2x) = e^{-3x} (-2\sin (2x) -3\cos(2x))\)
There's not much else to do for part b) if you get that differentiation is pretty much an inverse operation to integration.
For part c), try and manipulate the equations so that you can eliminate \(\int e^{-3x} \cos (2x) \ dx\), and thus get an integral of \(\int e^{-3x} \sin(2x) \ dx\) only, in terms of \(e^{-3x} \sin(2x) \ dx\) and \(e^{-3x} \cos (2x) \ dx\).
Post by: jinx_58 on April 20, 2022, 07:47:12 pm
I am having trouble with a trig integration question as I have no idea how to get to the answer. Could someone please show me working?
The question is attached.
Thanks,
- jinx_58
Post by: 1729 on April 20, 2022, 10:04:40 pm
Hello.Hi,
I am having trouble with a trig integration question as I have no idea how to get to the answer. Could someone please show me working?
The question is attached.
Thanks,
- jinx_58
Consider how \(\int f'\left(x\right)\text{d}x=f\left(x\right)+c\) and \(\int f\left(x\right)+g\left(x\right)\text{d}x=\int \:f\left(x\right)\text{d}x+\int g\left(x\right)\text{d}x\)
This means that \(\int \sin \left(x\right)+x\cos \left(x\right)\text{d}x=x\sin \left(x\right)\)
\(\Longrightarrow \int \sin \left(x\right)\text{d}x+\int x\cos \left(x\right)\text{d}x=x\sin \left(x\right)\)
\(\Longrightarrow \int x\cos \left(x\right)\text{d}x=x\sin \left(x\right)-\int \:\sin \left(x\right)\text{d}x\)
\(\Longrightarrow \int x\cos \left(x\right)\text{d}x=x\sin \left(x\right)+\cos \left(x\right)\)
Note: I've excluded the constant of integration because it is asking for an antiderivative.
Hope this helps