Author Topic: 4U Maths Question Thread (Read 665250 times) Tweet Share

massive · « **Reply #421 on:** August 15, 2016, 09:38:44 pm »

far out thanks!!

Also how do you do part ii for this q?

RuiAce · « **Reply #422 on:** August 15, 2016, 09:55:40 pm »

Quote from: massive on August 15, 2016, 09:38:44 pm

far out thanks!!

Also how do you do part ii for this q?

$\text{By the product rule }\frac{d}{dx}xe^x=(x+1)e^x\text{ so }\int (x+1)e^xdx=xe^x+C$
$\text{The point is that }\\ \begin{align*}I+J&=\int_0^\pi xe^x\cos x \, dx + \int_0^\pi e^x\cos x\, dx\\ &= \int_0^\pi (x+1)e^x\cos x \, dx\end{align*}$
$\text{So when we use integration by parts, we can now integrate the }(x+1)e^x\text{ and differentiate the }\cos x\\ \begin{align*}I+J&=\left[xe^x\cos x\right]_0^\pi-\int_0^\pi xe^x(-\sin x)dx\\ &= -\pi e^\pi + \int_0^\pi xe^x\sin x\, dx\end{align*}$

massive · « **Reply #423 on:** August 15, 2016, 10:02:14 pm »

Thank you Rui!!!

I have another question. Part ii - i don't get how you do it, i tried using part i but there's also a B, what do you with that?

RuiAce · « **Reply #424 on:** August 15, 2016, 10:07:48 pm »

Quote from: massive on August 15, 2016, 10:02:14 pm

Thank you Rui!!!

I have another question. Part ii - i don't get how you do it, i tried using part i but there's also a B, what do you with that?

$\text{This is a telescoping sum. Watch how adjacent terms cancel out when you sub it all in.}\\ \text{With the }B\text{, if you look at the result you're trying to prove and }\textbf{compare it with the expression in part a)}\\ B=1x=x$
$\begin{align*}&\quad \cos x + \cos 3x + \cos 5x + \dots + \cos [(2n-3)x] +\cos [(2n-1)x]\\ &= \frac{\sin 2x - \sin 0x}{2\sin x}+\frac{\sin 4x - \sin 2x}{2\sin x}+\frac{\sin 6x - \sin 4x}{2\sin x}+\dots+\frac{\sin [(2n-2)x]-\sin[(2n-4)x]}{2\sin x}+\frac{\sin (2nx) - \sin [(2n-2)x]}{2\sin x}\\ &= \frac{\sin 2nx}{2\sin x}\end{align*}$

massive · « **Reply #425 on:** August 15, 2016, 10:19:50 pm »

Quote from: RuiAce on August 15, 2016, 10:07:48 pm

$\text{This is a telescoping sum. Watch how adjacent terms cancel out when you sub it all in.}\\ \text{With the }B\text{, if you look at the result you're trying to prove and }\textbf{compare it with the expression in part a)}\\ B=1x=x$
$\begin{align*}&\quad \cos x + \cos 3x + \cos 5x + \dots + \cos [(2n-3)x] +\cos [(2n-1)x]\\ &= \frac{\sin 2x - \sin 0x}{2\sin x}+\frac{\sin 4x - \sin 2x}{2\sin x}+\frac{\sin 6x - \sin 4x}{2\sin x}+\dots+\frac{\sin [(2n-2)x]-\sin[(2n-4)x]}{2\sin x}+\frac{\sin (2nx) - \sin [(2n-2)x]}{2\sin x}\\ &= \frac{\sin 2nx}{2\sin x}\end{align*}$

Woaah thats so cool! This is probably gonna be a dumb q, but how do you know you're left with sin2nx, like how do you know that that term doesn't get cancelled out?

RuiAce · « **Reply #426 on:** August 15, 2016, 10:40:13 pm »

Quote from: massive on August 15, 2016, 10:19:50 pm

Woaah thats so cool! This is probably gonna be a dumb q, but how do you know you're left with sin2nx, like how do you know that that term doesn't get cancelled out?

Because the last term has nothing to pair up and vanish with.

Neither does the first term. It just so happens that the first term was sin(0) though.

massive · « **Reply #427 on:** August 15, 2016, 10:44:41 pm »

Oh yeaah i get it!

Hey i just have one more question, last question for the night (hopefully). How do you do it?

RuiAce · « **Reply #428 on:** August 15, 2016, 10:59:28 pm »

Quote from: massive on August 15, 2016, 10:44:41 pm

Oh yeaah i get it!

Hey i just have one more question, last question for the night (hopefully). How do you do it?

$\text{When it comes to integration by reduction I would actually bother to write out the negative index for IBP.}$
$\begin{align*}I_n&=\int_0^t (1+x^2)^{-n}dx\\ &= \left[x(1+x^2)^{-n}\right]_0^t-\int_0^tx(-2nx)(1+x^2)^{-n-1}dx\\ &= t(1+t^2)^{-n}+2n \int_0^t\frac{x^2}{(1+x^2)^{n+1}}dx\\ &= \frac{t}{(1+t^2)^n}+2n\int_0^t\frac{1+x^2}{(1+x^2)^{n+1}}dx-2n\int_0^t\frac{1}{(1+x^2)^n}dx\\ &= \frac{t}{(1+t^2)^n}+2nI_n-2nI_{n+1}\\ \therefore 2nI_{n+1}&=(2n-1)I_n+\frac{t}{(1+t^2)^n}\end{align*}\\$
Also, pretty sure that 1 in the boundary is supposed to be a t.
$\text{Intuition to start on }I_n\text{ and not }I_{n+1}:\\ \text{Upon integrating, }-(n+1)\text{ appears, not }-(n-1)$

massive · « **Reply #429 on:** August 16, 2016, 02:04:52 pm »

Guys when doing volumes, when do you split the area into two sections and then find the volume of each and add the two separate volumes to get the total volume?

massive · « **Reply #430 on:** August 16, 2016, 02:54:20 pm »

how do you do this q guys?

RuiAce · « **Reply #431 on:** August 16, 2016, 06:02:51 pm »

Quote from: massive on August 16, 2016, 02:04:52 pm

Guys when doing volumes, when do you split the area into two sections and then find the volume of each and add the two separate volumes to get the total volume?

Don't quite understand this question. I analyse what areas are relevant before I analyse the relevant volume. And the methods for area analysis should be familiar from 2U.

Quote from: massive on August 16, 2016, 02:54:20 pm

how do you do this q guys?

Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?

massive · « **Reply #432 on:** August 16, 2016, 06:43:07 pm »

Quote from: RuiAce on August 16, 2016, 06:02:51 pm

.Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?

Yeah, but idk where the pi came from

RuiAce · « **Reply #433 on:** August 16, 2016, 07:23:40 pm »

$\text{Rearrange the equation to make }y\text{ the subject due to the orientation of the cross sections.}\\ y^2=16x-x^2 \implies y=\pm\sqrt{16x-x^2}$
$\text{Clearly, the positive case corresponds to the upper semicircle and the bottom case to the lower.}\\ \text{Hence the base of our rectangle will have length }l=2\sqrt{16x-x^2}$
$\text{Since the height is just double the distance from the origin, it is literally }b=2x$
$\text{So we have the volume of a typical cross section being }\\ \begin{align*}\delta V &= lb \, \delta x\\ &= 4x\sqrt{16x-x^2}\delta x\end{align*}\\ \text{for some thickness }\delta x$
$\text{Note that our region goes from }x=0\text{ to }x=16.\\ \text{So our required volume is:}\\ \begin{align*} V&=\lim_{\delta x \to 0}\sum_{x=0}^{16}4x\sqrt{16x-x^2}\delta x\\ &= \int_0^{16}4x\sqrt{16x-x^2}dx\end{align*}$
$\text{Then just apply the substitution }u=x-8\text{, complete the square under the square root}\\ \text{and use properties of odd function and }\int_{-a}^{a}\sqrt{a^2-x^2}dx\text{ equals area of a semicircle.}$
If your diagram happened to be wrong and consequently you couldn't obtain the correct integral, you might want to read what I said at the start a few times. Because the π comes out of direct integration.

massive · « **Reply #434 on:** August 16, 2016, 08:00:19 pm »

OHHH mate thanks i get it!!

Also need help in the one attached. Wth is it even ?? :S

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