Author Topic: [Specialist] - Maths questions of the week thread! (Read 9014 times) Tweet Share

zsteve · « **on:** July 16, 2016, 10:18:33 pm »

VCE Specialist Mathematics - Question of the week thread

About
Every week, three challenging mathematics questions will be posted here, for your nerdish enjoyment

Questions posted here are all accessible using Specialist Maths knowledge that is, all accessible using specialist mathematics knowledge.... jk if you got what i mean!!.

Feel free to discuss solutions and post your answers here... solutions will be added at the end of each week, and new answers posted!

Think questions are too easy? too hard? Want a specific topic covered? Feedback is welcome in this thread as well!

Massive thanks to SlyDev for enabling inline $\mathrm\LaTeX$ by the way!

Posting rules
To keep it fair, answer posting is restricted to current Year 12 (or year 11s or really bright year 10s

, AND graduates who are not pursuing higher-level mathematics. All you university maths nerds, feel free to offer 'hints' at your discretion, but spoiler tags MUST be used to avoid giving it away!

Week 1 | Week 1 Solutions

Spoiler

Q1. Calculate the indefinite integral $\int \sqrt{1-x^2} dx$ by making an appropriate substitution for $x$. Full mathematical justification is required

Q2. Polar coordinates are defined as follows: $x=r\cos(\theta), y=r\sin(\theta)$. For the polar graph $r=10\sin(5\theta))$, find the gradient $\dfrac{dy}{dx}$ when $\theta = \dfrac{\pi}{10}$.

Q3. Using calculus, find an expression for a the volume of the truncated cone with minimal and maximal radii respectively $q, r$, and height $h$. Give your answer in the form $\dfrac{a\pi}{b}$ where a and b are real expressions (that is, $a, b \in \mathbf{R}$)

Week 2

Spoiler

Q1. [Complex exponential] Euler's formula states that
$e^{ix} = \operatorname{cis}(x) = \cos(x) + i\sin(x)$

Show that $\sin(x) = \dfrac{e^{ix} - e^{-ix}}{2i} \text{ and } \cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$

Using these results, write $\cos^4(x)$ as a sum of cosines of integer multiples of x

Q2. A family of curves is called an orthogonal trajectory of another family of curves if each member of one family intersects at right angles with every member of the other family.

Find the orthogonal trajectories of the family of parabolas $y^2=4ax$, i.e. the family of curves that have a gradient perpendicular to the gradient of the ellipses at every point in the plane.

Q3. [Implicit differentiation]

Let us introduce the concept of partial derivatives. Take $z = x^2+y^2+2xy$ for instance.
Then the partial derivative with respect to x is denoted
$\dfrac{\partial z}{\partial x} = 2x+2y$
(i.e. differentiate with respect to x, treating y as a constant)
And the partial derivative with respect to y is denoted
$\dfrac{\partial z}{\partial y} = 2y + 2x$
(i.e. differentiate with respect to y, treating x as a constant)

Take the implicitly defined relation $x^2-2xy+3y^2=2$. Find $\dfrac{dy}{dx}$ using the method you learned in Specialist Maths.

The relation can be rearranged into the form $F(x, y) = x^2-2xy+3y^2-2=0$. Find $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$.

Hence show that $\dfrac{dy}{dx} =- \dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}}$.

Try some other relations to convince yourself that this is generally true.

Week 3

I thought I'd try a slightly easier question this week, as the response to previous questions hasn't been that hot (possibly because they're hard or advanced lol)

A particle has position: $\vec{r}(t) = r\cos(\theta(t))\vec{i}+r\sin(\theta(t))\vec{j}$, where $\theta(t)$ is some function. (This models a particle in accelerating circular motion, with angular acceleration $\alpha$, not necessary for solving this question)

(a) Find $\dot{\vec{r}}, \text{ and } \ddot{\vec{r}}$, the velocity and acceleration vectors.

(b) A result from VCE Physics (LOL) states that for uniform circular motion, the velocity vector and the acceleration vector are perpendicular. Show that if $\theta(t) = \omega t$, (i.e. the motion is uniform), that $\vec{v}\cdot\vec{a} = 0$.

(c) Given that $\theta(t)$ is any function (in which case the particle may be accelerating), find the scalar component of $\vec{a}$ in the radial direction

(d) Hence, given that $\theta(t) = \omega t$, show that the component of acceleration in the radial direction is $a_r = \omega^2 r$.

(e) Now, using $\omega = \dot{\theta}\(t)$, show that the radial acceleration (i.e. towards the centre of the circle) is still $a_r = r[\omega(t)]^2$.

Have fun... and let me know if this is still too hard

Hopefully this is a bit of a more applied taste to the thread...

Not going to force people to post their working here, but let us know how you go. I'll stick up solutions for Week 2 and this question once I get around to it

This isn't a super hard question btw

§§§

qazser · « **Reply #1 on:** July 18, 2016, 07:47:03 am »

Quote from: zsteve on July 16, 2016, 10:18:33 pm

VCE Specialist Mathematics - Question of the week thread

About
Every week, three challenging mathematics questions will be posted here, for your nerdish enjoyment Questions posted here are all accessible using Specialist Maths knowledge that is, all accessible using specialist mathematics knowledge.... jk if you got what i mean!!.

Feel free to discuss solutions and post your answers here... solutions will be added at the end of each week, and new answers posted!

Think questions are too easy? too hard? Want a specific topic covered? Feedback is welcome in this thread as well!

Massive thanks to SlyDev for enabling inline $\mathrm\LaTeX$ by the way!

Posting rules
To keep it fair, answer posting is restricted to current Year 12 (or year 11s or really bright year 10s , AND graduates who are not pursuing higher-level mathematics. All you university maths nerds, feel free to offer 'hints' at your discretion, but spoiler tags MUST be used to avoid giving it away!

Week 1
Q1. Calculate the indefinite integral $\int \sqrt{1-x^2} dx$ by making an appropriate substitution for $x$. Full mathematical justification is required

Q2. Polar coordinates are defined as follows: $x=r\cos(\theta), y=r\sin(\theta)$. For the polar graph $r=10\sin(5\theta))$, find the gradient $\dfrac{dy}{dx}$ when $\theta = \dfrac{\pi}{10}$.

Q3. Using calculus, find an expression for a the volume of the truncated cone with minimal and maximal radii respectively $q, r$, and height $h$. Give your answer in the form $\dfrac{a\pi}{b}$ where a and b are real expressions (that is, $a, b \in \mathbf{R}$)

§§§

Well, i guess i'll start this one off. Might be wrong, just learnt this on Khan Academy

LaTeX looks scrappy, don't know how to format well

Question 1

Using U-Substitution

Let $x=sin(u)=arcsin(x)$

Derivative of $sin(x)=cos(u)$

$\int cos(u)\sqrt{(-sin^2(u)+1} du$

Using "fundamental' Pythagorean Identity

$sin^2(x)+cos^2(x)=1$

$\int cos(u)\sqrt{cos^2(u)} du$

$\int cos^2(u) du$

$\int \frac{cos(u)sin(u)}{2} + \frac{1}{2}u$

Substitute $arcsin(x)$ back into equation

$\dfrac{\arcsin\left(x\right)+x\sqrt{1-x^2}}{2}$

Alternatively, the CAS calculator states: $\frac{sin^{-1}(x)}{2}+\frac{x\sqrt(1-x^2)}{2}$
Let me know if working is wrong, i'm in Year 11, don't know how to set it out correctly

RuiAce · « **Reply #2 on:** July 18, 2016, 09:42:27 am »

Quote from: qazser on July 18, 2016, 07:47:03 am

Well, i guess i'll start this one off. Might be wrong, just learnt this on Khan Academy

Correct!

With your LaTeX, in front of sin, cos and arcsin you should always use a backslash (\) so that the typeface looks neater. Here's an appropriate way of setting it out
$\text{Let }x=\sin \theta \implies dx = \cos \theta\, d\theta$
$\begin{align*}\int \sqrt{1-x^2}dx&= \int \sqrt{1-\sin^2\theta} \, \cos \theta \, d\theta\\ &= \int \cos^2\theta \, d\theta\\ &= \frac{1}{2}\int (1+\cos 2\theta)d\theta\\ &= \frac{1}{2}\left(\theta+\frac{1}{2}\sin 2\theta\right)+C\\ &= \frac{1}{2}\left(\arcsin x +{\sin \theta \cos \theta}\right)\\ &=\frac{1}{2}\left(\arcsin x + {x\sqrt{1-x^2}}\right) \tag{QED} \end{align*}$

zsteve · « **Reply #3 on:** July 18, 2016, 09:58:16 am »

qazser - excellent work! Note that, in the question, I stated that full justification is required. Hint: you're using a lot of trig functions here (which is of course correct) as well as their inverses!

Spoiler

So... in tossing around trig functions and their inverses, as well as going from a positive square root to the function itself... what justifications might be necessary?

I've had a PM or two about these questions - please post discussion in the thread, the questions here are all more or less above VCAA level so discussion is helpful.

Oh yeah, and it might be a good idea to avoid quoting massive posts/solutions

RuiAce · « **Reply #4 on:** July 18, 2016, 09:59:57 am »

That might be a bit demanding since √(1-sin²(x)) = +cos(x) was done without loss of generality here even though it should actually be |cos(x)|
____________________________
<Merged Posts>

Quote from: zsteve on July 18, 2016, 10:02:31 am

Hey RuiAce!
Spoiler
The key here is to specify $x=\sin(u)$ where $u\in[-\pi/2, \pi/2]$, the principal domain of sine. This allows x to be in the domain [-1, 1] which is correct.

Oh right. Forgot about that restriction nullifying the problem it seems

zsteve · « **Reply #5 on:** July 18, 2016, 10:02:31 am »

Quote from: RuiAce on July 18, 2016, 09:59:57 am

That might be a bit demanding since √(1-sin²(x)) = +cos(x) was done without loss of generality here even though it should actually be |cos(x)|

Hey RuiAce!

Spoiler

The key here is to specify $x=\sin(u)$ where $u\in[-\pi/2, \pi/2]$, the principal domain of sine. This allows x to be in the domain [-1, 1] which is correct.

Jayz2398 · « **Reply #6 on:** July 18, 2016, 06:37:16 pm »

I tried typing this in Latex, but I might have gotten fed up and just tried to upload my handwriting instead.

Not too sure if I started even mildly on the right track, but I'm crossing my fingers at the least. Answer is under spoiler below.

Question 2:

Spoiler

PS In year 12, currently doing 3/4 spesh.

RuiAce · « **Reply #7 on:** July 18, 2016, 06:46:52 pm »

Q2
$\begin{align*}x&=r\cos \theta\\ &= 10\sin 5\theta \cos \theta\\ \frac{dx}{d\theta}&=50\cos 5\theta\cos\theta-10\sin5\theta\sin\theta\\ \\ y&=r\sin\theta\\&=10\sin 5\theta \sin \theta \\ \frac{dy}{d\theta}&=50\cos 5\theta \sin \theta+10\sin 5\theta \cos \theta\end{align*}$
$\text{Using parametric differentiation}\\\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{50\cos 5\theta \sin \theta+10\sin 5\theta \cos \theta}{50\cos 5\theta\cos\theta-10\sin5\theta\sin\theta}$
$\begin{align*}\frac{dy}{dx}|_{\theta=\frac{\pi}{10}}&=\frac{50\cos \frac{\pi}{2}\sin\frac{\pi}{10}+10\sin \frac{\pi}{2} \cos\frac{\pi}{10}}{50\cos \frac{\pi}{2}\cos\frac{\pi}{10}-10\sin\frac{\pi}{2}\sin\frac{\pi}{10}}\\ &= \frac{10\cos \frac{\pi}{10}}{-10\sin \frac{\pi}{10}}\\ &= -\cot \frac{\pi}{10}\end{align*}$
Haven't checked your working in detail but your final answer is in agreement with what I got.
$\text{Note that parametric differentiation falls out of the chain rule }\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\\ \text{and the inverse function theorem }\frac{dy}{dx}\times\frac{dx}{dy}=1 \iff \frac{dy}{dx}=\frac{1}{dx/dy}$

Jayz2398 · « **Reply #8 on:** July 18, 2016, 07:08:55 pm »

Awesome!

I don't think I've done parametric differentiation, but glad to know the logic was somewhat right.

RuiAce · « **Reply #9 on:** July 18, 2016, 07:47:05 pm »

Quote from: Jayz2398 on July 18, 2016, 07:08:55 pm

Awesome! I don't think I've done parametric differentiation, but glad to know the logic was somewhat right.

Yeah it might be foreign, so I included where it comes from as a reference.
_____________
$\text{For anyone interested, the first part of the question could be done by multivariable calculus (albeit more annoyingly).}\\ \begin{align*}x&=r\cos \theta\\ \frac{dx}{d\theta}&=\frac{\partial x}{\partial r}\, \frac{dr}{d\theta}+\frac{\partial x}{\partial <br />\theta}\, \frac{d\theta}{d\theta}\\ &= (\cos\theta)(50\cos 5\theta)+(-r\sin\theta)(1)\\ &= 50\cos 5\theta\cos\theta - 10\sin 5\theta \sin \theta\end{align*}$

RuiAce · « **Reply #10 on:** July 20, 2016, 07:59:17 pm »

Q3 hint

There are (at least) three ways of approaching this question to begin with.
1. Compare relevant areas
2. Use similar triangles
3. Fit a line (y=mx+b) through two points

zsteve · « **Reply #11 on:** July 23, 2016, 01:01:25 am »

Nice work this week qazser and Jayz2398, you both got the right answer for Q1 and Q2 respectively!

Solutions for Week 1 attached

And thanks to Rui for all that advice

zsteve · « **Reply #12 on:** July 23, 2016, 06:46:07 pm »

Bump: Week 2 questions are up now!

Spoiler

Q3 might be a big help in the Spesh exam!

Edit: thanks Rui for the heads up, fixed now

RuiAce · « **Reply #13 on:** July 23, 2016, 07:42:18 pm »

Quote from: zsteve on July 23, 2016, 01:01:25 am

Nice work this week qazser and Jayz2398, you both got the right answer for Q1 and Q2 respectively!

Solutions for Week 1 attached

And thanks to Rui for all that advice

____________________________________
Week 2:

Q1

For the proof, work on the RHS

Hey btw Steve, I think you meant
$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$

sweetiepi · « **Reply #14 on:** July 24, 2016, 10:20:35 am »

Q3.
Am I on the right track here?

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Author Topic: [Specialist] - Maths questions of the week thread! (Read 9014 times) Tweet Share

zsteve

[Specialist] - Maths questions of the week thread!

qazser

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RuiAce

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zsteve

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RuiAce

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zsteve

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Jayz2398

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RuiAce

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RuiAce

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zsteve

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zsteve

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