Author Topic: 4U Maths Question Thread (Read 665219 times) Tweet Share

RuiAce · « **Reply #435 on:** August 16, 2016, 08:34:34 pm »

Quote from: massive 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

This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.

$\text{The second part is just your classic max/min question. So you have}\\ \begin{align*}36V^2&=A^2x^2-2Ax^4\\ 72V\frac{dV}{dx}&=2A^2x-8Ax^3\\ &=0\text{ for max/min}\\ 2Ax(A-4x^2)&=0\end{align*}\\ \text{etc.}$

massive · « **Reply #436 on:** August 16, 2016, 08:48:21 pm »

Quote from: RuiAce on August 16, 2016, 07:23:40 pm

$\text.\\ \text{So our required volume is:}\\ \begin{align*} V&=\lim_{\delta x \to 0}\sum_{x=0}^{16}4x\sqrt{16x-x^2}\delta x\\ &= \int_0^{16}4x\sqrt{16x-x^2}dx\end{align*}$

do we have to put this line for hsc (the line with the sigma notation), like do we lose marks if we don't, because in our school our teacher never told us to put this line in :S

massive · « **Reply #437 on:** August 16, 2016, 08:49:17 pm »

Quote from: RuiAce on August 16, 2016, 08:34:34 pm

This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.(Image removed from quote.)
$\text{The second part is just your classic max/min question. So you have}\\ \begin{align*}36V^2&=A^2x^2-2Ax^4\\ 72V\frac{dV}{dx}&=2A^2x-8Ax^3\\ &=0\text{ for max/min}\\ 2Ax(A-4x^2)&=0\end{align*}\\ \text{etc.}$

Also thanks heaps for this!!

jakesilove · « **Reply #438 on:** August 16, 2016, 08:52:14 pm »

Quote from: massive on August 16, 2016, 08:48:21 pm

do we have to put this line for hsc (the line with the sigma notation), like do we lose marks if we don't, because in our school our teacher never told us to put this line in :S

Nup it's fine if you go straight to the integral

massive · « **Reply #439 on:** August 16, 2016, 09:00:35 pm »

guys how do you do this :S

RuiAce · « **Reply #440 on:** August 16, 2016, 11:42:42 pm »

Quote from: massive on August 16, 2016, 09:00:35 pm

guys how do you do this :S

$\text{The chord }PB\text{ is the line through }P\text{ with gradient }-1\text{ so it has equation }\\ y-\frac{1}{x_0}=-x+x_0\\ \text{using the point-gradient formula.}$
$\text{Where does the chord meet the hyperbola }xy=1\text{ again?}\\ \text{Sub }xy=1\text{ into the equation of the chord.}\\ \frac{1}{x}-\frac{1}{x_0}=x_0-x\\ \text{By inspection the other point is }x=\frac{1}{x_0}\\ \textit{Two ways to prove this: One is to just multiply through by }x\textit{ and solve the quadratic.}\\ \textit{Two is to multiply through as well but realise that the product of roots is 1, so one root is the reciprocal of the other.}$
$\text{Notice how }2r\text{ is just the distance between where the chord intersects the hyperbola.} \\ \begin{align*}(2r)^2&=\left(x_0-\frac{1}{x_0}\right)^2+\left(\frac{1}{x_0}-x_0\right)^2\\ 4r^2&=2\left(x_0-\frac{1}{x_0}\right)^2\\2r^2&=\left(x_0-\frac{1}{x_0}\right)^2 \end{align*}$
$\textit{Important observation: }xy=1\text{ is symmetric about }y=x$
$\text{Note that }OP^2=\left(x_0-0\right)^2+\left(\frac{1}{x_0}-0\right)^2=x_0^2+\frac{1}{x_0^2} \\ \text{Expanding out the expression in (i)} \\ \begin{align*}2r^2&= x_0^2-2+\frac{1}{x_0^2}\\ 2(1+r^2)&=x_0^2+\frac{1}{x_0^2}\end{align*}\\ \text{Equate with above to get the result for part ii)}$
$A\text{ is just the point }(1,1)\text{. This is easy to prove.}\\ \text{So we can use Pythagoras' Theorem to introduce the }t\text{ in }\triangle AOB\\ \begin{align*}OP^2&=OB^2+BP^2\\ 2(1+r^2)&=(\sqrt{2}+t)^2+r^2\\ r^2&=(t+\sqrt{2})^2-2\end{align*}$
$\text{Note: }OA=\sqrt{2}\text{ since }A\text{ is the point }(1,1)$

massive · « **Reply #441 on:** August 17, 2016, 01:09:42 am »

Thanks Rui!!

Guys the question attached is so simple but I can't seem to get it :/
The answer to part i is: 4pi.a.f(x).dx How?? its so weird

RuiAce · « **Reply #442 on:** August 17, 2016, 08:38:39 am »

Quote from: massive on August 17, 2016, 01:09:42 am

Thanks Rui!!

Guys the question attached is so simple but I can't seem to get it :/
The answer to part i is: 4pi.a.f(x).dx How?? its so weird

$\text{Very briefly}\\ \text{Radius of rotation is }r=x+a\\ \text{Height is obviously }h=f(x)\\ \text{So I reckon the answer to part (i) is }\delta V = 2\pi (x-a)f(x)\delta x$
$\text{But the volume is interesting.}\\ \begin{align*}V&=2\pi\int_{-a}^a(x+a)f(x)dx\tag{split integral up}\\ &= \underbrace{2\pi \int_{-a}^a xf(x)dx}_{\text{odd function}} + \underbrace{2\pi \int_{-a}^a f(x)dx}_{\text{even function}}\\ &= 0 + 4\pi\int_{0}^{a}f(x)dx\end{align*}$

Quote from: jakesilove on August 16, 2016, 08:52:14 pm

Nup it's fine if you go straight to the integral

Are you sure?

massive · « **Reply #443 on:** August 17, 2016, 10:53:18 am »

Thanks Rui

You see for the q attached, i keep getting 32 for my ans, but the actual answer is 24 (part ii) :S

RuiAce · « **Reply #444 on:** August 17, 2016, 11:25:25 am »

Quote from: massive on August 17, 2016, 10:53:18 am

Thanks Rui

You see for the q attached, i keep getting 32 for my ans, but the actual answer is 24 (part ii) :S

I also got 32

massive · « **Reply #445 on:** August 17, 2016, 02:36:55 pm »

Guys how do you do this badboy ? $:-\$
(just part ii)

massive · « **Reply #446 on:** August 17, 2016, 02:55:33 pm »

wait part ii is easy, because you just use part a (which is attached)

But i have a new problem, how in the world do you do part a. ??

znaser · « **Reply #447 on:** August 17, 2016, 03:20:52 pm »

I tried following the answers but yh this question is something. Thanks in advance

massive · « **Reply #448 on:** August 17, 2016, 04:13:56 pm »

Guys how do you even do this question?

:S

RuiAce · « **Reply #449 on:** August 17, 2016, 05:00:05 pm »

Quote from: massive on August 17, 2016, 02:55:33 pm

wait part ii is easy, because you just use part a (which is attached)

But i have a new problem, how in the world do you do part a. ??

$\text{For positive }k\text{ we have the immediate domain and range restrictions }x\ge 0, y\ge 0\\ \text{As a matter of fact, }0 \le x \le k, \, 0 \le y \le 1\\ \textbf{Do you see why?}$
$\text{To prove that it is monotonic decreasing, we need only consider the first derivative.}\\ \text{At this point, it should be very obvious that you need implicit differentiation.}\\ \frac{1}{2\sqrt{x}}+\frac{\sqrt{k}}{2\sqrt{y}} \, \frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}\\ \text{Which is obviously always negative or equal to 0, so the curve monotonic decreases.}$
$\text{There are obviously no corners as the curve is smooth, so the only critical points }\\ \text{are the endpoints on the curve (note: one endpoint is simultaneously a stationary point).}$
$\text{At }(0,1)\text{ there is a vertical tangent.}\\ (k,0)\text{ is a stationary point.}\\ \textbf{Make sure you figure out why.}$
$\text{And then you can sketch the curve.}$
Use this GeoGebra simulation to help you.

I will leave the area computation as your exercise

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