Author Topic: 4U Maths Question Thread (Read 664237 times) Tweet Share

massive · « **Reply #675 on:** October 20, 2016, 06:46:15 pm »

guys whats the difference between uniform and non-uniform circular motion??

RuiAce · « **Reply #676 on:** October 20, 2016, 06:46:50 pm »

Quote from: massive on October 20, 2016, 06:46:15 pm

guys whats the difference between uniform and non-uniform circular motion??

Non-uniform circular motion is not in the HSC 4U course.

It means that the angular velocity is not constant; it varies.

Brenda0708 · « **Reply #677 on:** October 20, 2016, 06:55:44 pm »

Quote from: RuiAce on October 20, 2016, 06:30:07 pm

$\text{I'll just use part (i) as a demo. Try doing etc. by yourself and post up whatever you have.}$
$\text{Note that we are basically proving }x-\sin x > 0\\ \text{We move everything to one side to make it easier on ourselves.}\\ \text{That way, we only have to compute 1 derivative, not 2.}$
$\text{Consider }g(x) = x-\sin x\text{ defined for all }x \ge 0\\ \text{Then }g^\prime(x) = 1-\cos x$
$\text{Now, }1-\cos x \ge 0\text{ always.}\\ \text{This is simply due to the fact that }\cos x \le 1$
$\text{So }g^\prime(x) \ge 0\text{ for all }x\\ \text{This means, by definition, }g(x)\text{ is monotonic increasing.}$
$\text{If it's monotonic increasing, then }g(x)\text{ never goes down. It either sits or always goes up.}\\ \text{So where does it start?}\\ \text{Find }g(0) = 0 - \sin 0 = 0$
$\text{So }g(x)\text{ starts increasing, from }x=0, y=0\\ \text{Hence }g(x) \ge 0 \text{ for all }x \ge 0\\ \text{i.e. }x - \sin x \ge 0 \text{ for all }x\ge 0$
$\text{Then the }\ge \text{ becomes }> \text{ cause we really just care about }x> 0$

Ohhhhh so is it like this: if we didn't show that g(0)=0, then the curve could be below the x axis somewhere between 0<x<infinity. But seeing that is is =0 at 0 then after x=0 the curve must be >0. Which is what you pretty much said-omg took me a while. LMAO wow thank you!!!

RuiAce · « **Reply #678 on:** October 20, 2016, 06:57:27 pm »

Quote from: Brenda0708 on October 20, 2016, 06:55:44 pm

Ohhhhh so is it like this: if we didn't show that g(0)=0, then the curve could be below the x axis somewhere between 0<x<infinity. But seeing that is is =0 at 0 then after x=0 the curve must be >0. Which is what you pretty much said-omg took me a while. LMAO wow thank you!!!

Lel glad that we figured out the issue

jakesilove · « **Reply #679 on:** October 20, 2016, 07:09:19 pm »

Quote from: katherine123 on October 20, 2016, 05:38:55 pm

image 1: how to do part v)
image 2: how to do part ii) and iii)

Other than the equations given, we also know from basic trig that

$r=Rsin\theta$

Thus, subbing into the first equation,

$Nsin\theta=m(Rsin\theta)w^2$
$sin\theta(N-mw^2R)=0$
$sin\theta=0, N-mw^2R=0$

This directly results in

$\theta=0, N=mw^2R$

Halfway there! Let's sub this new relationship into the second proven equation.

$mg=Ncos\theta=(mw^2R)cos\theta$
$cos\theta=\frac{g}{Rw^2}$

As required

Okay, onto the last part. We know that, if theta does not equal zero,

$cos\theta=\frac{g}{Rw^2}$

This last part is tricky to see, but obvious once you've seen it; we need to introduce an inequality, so I'll just recall that

$cos\theta<1$
$\frac{g}{Rw^2}<1$
$w>(\frac{g}{R})^{0.5}$

As required! Lot's of subbing in, lot's of thinking, but not too difficult a question considering how annoying it first seems

Brenda0708 · « **Reply #680 on:** October 20, 2016, 07:18:55 pm »

Would this be the correct layout to part iii) of the previous question?

RuiAce · « **Reply #681 on:** October 20, 2016, 07:35:03 pm »

Quote from: Brenda0708 on October 20, 2016, 07:18:55 pm

(Image removed from quote.)
Would this be the correct layout to part iii) of the previous question?

Where you said since f"(x) > 0 for all x > 0 you might want to put in brackets (by part (i))

I think, for that one it's like a step by step process. Because we consider two derivatives here, not one.
Since f"(x) > 0 for all x > 0 we know that f'(x) is increasing.
Since f'(0) = 0 we know that f'(x) > 0 for all x > 0 as well now

So since f'(x) > 0, going further back up f(x) is increasing
And then finish it off with whatever you wrote at the end

amandali · « **Reply #682 on:** October 20, 2016, 08:00:50 pm »

how to do part i) im confused with the angles for normal force

RuiAce · « **Reply #683 on:** October 20, 2016, 08:32:07 pm »

Quote from: amandali on October 20, 2016, 08:00:50 pm

(Image removed from quote.)

how to do part i) im confused with the angles for normal force

Note that F is NOT perpendicular to N here. Here's some things done to the diagram to make it clearer. Note: Corresponding angles on parallel lines used.

The diagram is misleading because it goes against the norm.

Neutron · « **Reply #684 on:** October 20, 2016, 08:41:03 pm »

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you

jakesilove · « **Reply #685 on:** October 20, 2016, 09:25:20 pm »

Quote from: Neutron on October 20, 2016, 08:41:03 pm

(Image removed from quote.)

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you

Let's start with i)

$(\cos\alpha+i\sin\alpha)^{2n}$

Using the standard binomial expansion, we get

$\binom{n}{0}\cos^n\alpha+\binom{n}{1}\cos^{n-1}\alpha i\sin{\alpha}+....+\binom{n}{n}(i)^{2n}\sin^{2n}\alpha$

It might be a good idea to write out some more terms, I just can't be bothered using LaTeX.

Now, using standard DeMoivre's theorum,

$(\cos\alpha+i\sin\alpha)^{2n}=(cis\alpha)^{2n}=cis2n\alpha$

If we equate just REAL parts, we get

$\cos2n\alpha=\binom{n}{0}\cos^n\alpha+....+\binom{n}{n}(i)^{2n}\sin^{2n}\alpha=\binom{n}{0}\cos^n\alpha+....+\binom{n}{n}(-1)^{n}\sin^{2n}\alpha$

As required (obvs add some more middle terms, just couldn't be bothered here)

Okay, clearly here we need to let

$\cos\alpha=x$
$\cos^{-1}x=\alpha$

Therefore, from part i)

$\cos(2n\cos^{-1}x=x^{2n}+....(-1)^n(1-\cos^2\alpha)^n=x^{2n}+....(-1)^n(1-x^2)^n$

As required. Rui is coming along with the last part of the question

Let me know if I can expand (haha) on anything

(Rui also fixed some of the $ \LaTeX $ up

)

RuiAce · « **Reply #686 on:** October 20, 2016, 09:35:26 pm »

Quote from: Neutron on October 20, 2016, 08:41:03 pm

(Image removed from quote.)

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you

$\text{Starting from the bottom because part (iv) is surprisingly neat.}\\ \text{Ignore part (iii) and look closely at part (i).}$
$\text{Somehow, BOTH }x^2\text{ and }(1-x^2)\\ \text{Also note how weirdly enough, a }2^n\text{ appears on the RHS.}$
$\text{The RHS has a }\cos\text{, and the LHS has a ton of binomial coefficients}\\ \text{This implies that we need to use part (ii).}\\ \text{In fact, we are evaluating }T(something)$
$\text{When does }x^2=1-x^2\text{ I wonder?}\\ \text{The answer is when }\boxed{x=\frac{1}{\sqrt{2}}}\\ \text{We need to evaluate }T\left(\frac{1}{\sqrt2}\right)$
$\text{Let }x=\frac{1}{\sqrt2}, \text{ then }x^2 = 1-x^2 = \frac{1}{2^n}$
$\begin{align*}\cos (2n\cos^{-1}x)&= x^{2n}-\binom{2n}{2}x^{2n-2}(1-x^2)+\binom{2n}{4}x^{2n-4}(1-x^2)^2+\dots+(-1)^n(1-x^2)^n\\ \therefore \cos \left(2n \cos^{-1}\frac{1}{\sqrt{2}}\right)&=2^{-n} -\binom{2n}{2}2^{-n+1}2^{-1}+\binom{2n}{4}x^{-n+2}x^{-2}+\dots+(-1)^n2^{-n}\\ implies 2^n\cos \left(\frac{n\pi}{2}\right)&=1-\binom{2n}{2}+\binom{2n}{4}-\binom{2n}{6}+\dots+(-1)^n\binom{2n}{2n} \end{align*}\\ \text{noting }\cos^{-1}\frac{1}{\sqrt2}=\frac{\pi}{4}$
I was so happy when I got this question out in the exam. I was fanboying for 10 seconds...

RuiAce · « **Reply #687 on:** October 20, 2016, 10:18:25 pm »

$\text{Part (iii) is the big bad and probably the hardest question of the paper.}\\ \text{We will firstly check that }\cos\frac{\pi}{4n}\text{ and etc. ARE the product of roots of }T_{2n}(x)\text{, as }\textit{implied.}$
$T_{2n}(x)=0\text{ solves for }\\ \cos (2n\cos^{-1}x) = 0\\ \text{which has solutions}\\ 2n\cos^{-1}x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\dots, \frac{(4n-1)\pi}{2}$
$\text{Note that we go up to }\frac{(4n-1)\pi}{2}\\ \text{as the polynomial in (ii) implies that we have }2n\text{ distinct roots.}$
$\text{So rearranging:}\\ \cos^{-1}x=\frac{\pi}{4n},\frac{3\pi}{4n},\dots, \frac{(4n-1)\pi}{4n}\\ x=\cos\frac{\pi}{4n},\cos\frac{3\pi}{4n},\dots ,\cos\frac{(4n-1)\pi}{4n}\\ \text{So we HAVE the correct roots.}$
______________________________
$\text{But the question now comes to what the product of roots actually IS.}\\ \text{This is HARD.}$
$\text{Firstly, to present the pattern, consider a degree 6 polynomial}\\ ax^6+bx^5+cx^4+dx^3+ex^2+fx+g\\ \text{Then we have}\\ \Sigma \alpha = -b/a\\ \Sigma \alpha\beta = c/a\\ \Sigma \alpha\beta\gamma = -d/a \\ \Sigma \alpha\beta\gamma\delta=e/a \\ \text{i.e. for every even power, the sign is positive.}\\ \text{Whereas for every odd power, the sign is negative.}$
$\text{So since this polynomial degree is even, it is positive.}$
$\text{Note also that in the above case, product of roots }\Pi \alpha = \frac{g}{a}\\ \text{i.e. }\frac{\text{constant term}}{\text{coefficient on }x^6}$
$\text{With all that background information out of the way}\\ \text{The product of roots here is equal to }\frac{\text{constant term}}{\text{coefficient on }x^{2n}}$
_____________________________
$\text{The constant term is }(-1)^n\\ \text{That's dangling off the last term in part (ii)}$
$\text{The coefficient on }x^{2n}\text{ is trickier. It is}\\ 1+\binom{2n}{2}+\binom{2n}{4}+\binom{2n}{6}+\dots+\binom{2n}{2n}$
$\text{To visualise THIS, try expanding out some terms.}\\ \text{The }1\text{ comes straight from }x^{2n}\\ \text{The }\binom{2n}{2}\text{ comes from expanding term 2 to give}\\ -\binom{2n}{2}x^{2n-2}(1-x^2) = \binom{2n}{2}x^{2n} - \binom{2n}{2}x^{2n-2}$
$\text{The }\binom{2n}{4}\text{ comes from expanding term 3 to give}\\ \binom{2n}{4}x^{2n-4}(1-x^2)^2=\binom{2n}{4}x^{2n}-2\binom{2n}{4}x^{2n-2}+\binom{2n}{4}x^{2n-4}\\ \text{And so on.}$
$\text{So our answer is}\\ \frac{(-1)^n}{1+\binom{2n}{2}+\binom{2n}{4}+\dots+\binom{2n}{2n}}$
_____________________________
$\text{Except we STILL aren't done.}\\ \text{Recall this:}$
$(1+x)^{2n}-(1-x)^{2n}= 2\binom{2n}{0}+2\binom{2n}{2}x^2+2\binom{2n}{4}x^4+2\binom{2n}{6}x^6+\dots+2\binom{2n}{2n}x^{2n}$
$\text{If we let }x=1\text{ then }$
$2^n = 2\binom{2n}{0}+2\binom{2n}{2}x^2+2\binom{2n}{4}x^4+2\binom{2n}{6}x^6+\dots+2\binom{2n}{2n}x^{2n}$
$\text{i.e. }1+\binom{2n}{2}+\binom{2n}{4}+\dots+\binom{2n}{2n}=2^{n-1}$
$\text{So our final answer is }\frac{(-1)^n}{2^{n-1}}$

katherine123 · « **Reply #688 on:** October 21, 2016, 03:47:04 am »

how to do part ii)

Ali_Abbas · « **Reply #689 on:** October 21, 2016, 05:00:35 am »

Quote from: katherine123 on October 21, 2016, 03:47:04 am

how to do part ii)

Expand the LHS using the Binomial Theorem and observe that the imaginary terms (which occur when the index is of odd parity) cancel out leaving twice the sum of the real terms. This 2 out the front cancels with the 2 on the RHS.

Next, simply note that if n is divisible by 4, then n/4 = k, k integer. That means that cos(n*pi/4) = cos(k*pi) = (-1)^k = (-1)^(n/4). Combining all these and you will arrive at the required identity.

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massive

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