Author Topic: 4U Maths Question Thread (Read 663297 times) Tweet Share

birdwing341 · « **Reply #240 on:** June 08, 2016, 06:38:30 pm »

Quote from: RuiAce on June 08, 2016, 06:23:34 pm

You should be decently skilled (if not already mastered) circle geometry at the Extension 1 level before attempting problems that target Extension 2 students.

At the Extension 1 level, I was already familiar with what my theorems "LOOKED LIKE". What does this mean? Examples:
- Alternate angles look like Z angles on parallel lines
- Angles standing on same arc theorem looked like a nice M shape to me
- Alternate segment theorem involves a tangent and a triangle.
However how you visualise it may differ.

When I tackle an Extension 2 question, I don't label 50 thousand things at once if it's way too irrelevant. I work in one direction.

That is, I look at what I am trying to prove (or find).

Then I try to either work forwards, or backwards. That is, I look at the LHS or the RHS, then I start looking for an angle/side that equals to THAT ONLY. THEN, I keep going.

At the Extension 2 level you should also be prepared to manipulate lots of things. Similar triangles and cyclic quadrilaterals are examples of cliches, however even base angles of isosceles triangle are important. Equal radii is something I always keep at the back of my head regardless of if I need to use it.

Progressively, I build up to the final answer. But in doing so, I only consider RELEVANT information, not EXTRANEOUS information that may be useful for another part but not this present part.

A rare trap is the occurrence of trigonometry in circle geometry. When that happens, always look out for any right angled triangles BEFORE you attempt sine/cosine rule.

Thanks for the advice! Prompt as usual

katherine123 · « **Reply #241 on:** June 09, 2016, 01:03:26 pm »

Ques: the region enclosed between the curve y=cos^-1(x) , the lines x=-1 and y=pie/2 is made to revolve about the axis y=-1 . find the volume of solid generated

im not sure what the height is

RuiAce · « **Reply #242 on:** June 10, 2016, 07:01:11 am »

Quote from: katherine123 on June 09, 2016, 01:03:26 pm

Ques: the region enclosed between the curve y=cos^-1(x) , the lines x=-1 and y=pie/2 is made to revolve about the axis y=-1 . find the volume of solid generated

im not sure what the height is

$\text{The region to be rotated is tiny.}$
$\text{Given that you said the word "height", I am assuming you seek to use the shells method. } \\ \text{Doing so requires rearranging the equation: }x=\cos{y}$
$\text{Take a typical shell with respect to the line }y=-1\\ \text{The boundaries of integration are }y=\frac{\pi}{2}, y=\pi\\ \text{The radius is }(1+y)\\ \text{and the }\textbf{height}\text{ is }h=1+x = 1+\cos{y}$
Do you see why?

$\text{I'll leave the rest of the question to you. However, keep in mind computation of }\int_{\frac{\pi}{2}}^\pi{(1+y)(1+\cos{y})dy}\\ \text{is definitely not nice.}$

katherine123 · « **Reply #243 on:** June 13, 2016, 11:37:47 am »

the ans: 480

I did 3 cases 1 M, 2M and 3M : so 5!+ 6P5/2! + 7P5/5!
but it's not working

RuiAce · « **Reply #244 on:** June 13, 2016, 01:15:43 pm »

Quote from: katherine123 on June 13, 2016, 11:37:47 am

the ans: 480

I did 3 cases 1 M, 2M and 3M : so 5!+ 6P5/2! + 7P5/5!
but it's not working

$\text{Case one: 1 M }\\ \text{Your letters are }M,A,O,T,H\\ \text{ which can obviously be arranged in }5!\text{ ways as you said.}$
$\text{Case two: 2 M's} \\ \text{Your letters are }M, M, \_ , \_ , \_ \\ \text{The blanks can be any three letters chosen out of the bundle }A, O, T, H\\ \text{to which there are }\binom{4}{3}\text{ ways of doing so.} \\ \text{So there are }\frac{5! \times \binom{4}{3}}{2!}=240 \text{ such arrangements.}$
$\text{Case three: 3 M's}\\ \text{Your letters are }M, M, M, \_ , \_ \\ \text{Now, the blanks are any two out of the four, i,e, }\binom{4}{2}\text{ choices.} \\ \text{So there are }\frac{5! \times \binom{4}{2}}{3!}=120\text{ ways of doing so.}$
$120+240+120=480$
$\text{Obviously, the division by }2!\text{ and }3!\text{ for case 2 and 3 respectively} \\ \text{arise due to the repeated letter }M$

Disclaimer: I thoroughly dislike combinatorics. My explanations for them are usually suboptimal to those for other topics. If the explanation is insufficient, please allow someone else to supplement my answer.
My explanations for combinatorics will improve after I do discrete maths, which may be a while.

I wasn't able to justify the method you used either.

amandali · « **Reply #245 on:** June 15, 2016, 11:46:27 pm »

need help with this ques
A ring of altitude 2h is generated by revolving about y-axis the area of the segment bounded by the circle x^2+ y^2=a^2 and the chord of length 2h that is parallel to the y-axis. By using cylindrical shell method, show volume is given by v=[4π(h)^3]/3

can you briefly explain the method of pairing cylindrical shell and when is it applicable?
where v=2πsA
s=distance from axis of rotation
A=area

RuiAce · « **Reply #246 on:** June 16, 2016, 09:17:53 am »

Quote from: amandali on June 15, 2016, 11:46:27 pm

need help with this ques
A ring of altitude 2h is generated by revolving about y-axis the area of the segment bounded by the circle x^2+ y^2=a^2 and the chord of length 2h that is parallel to the y-axis. By using cylindrical shell method, show volume is given by v=[4π(h)^3]/3

can you briefly explain the method of pairing cylindrical shell and when is it applicable?
where v=2πsA
s=distance from axis of rotation
A=area

$\text{I don't quite get this terminology of 'paring' cylindrical shells.}\\ \text{Please elaborate.}$
Also, whilst I've seen that formula before, I've never found a use for it in MX2 where they get you to derive everything from scratch. So you'll have to show me an example of a question+answer that your teacher said to use it on.
(This is because normally you can't just assume the area. You can only assume the height.)

$\text{I'm not in favour of using shells here as if you read the question carefully}\\ h\text{ is just the value of the }y\text{-coordinate above the }x\text{-axis.}\\ \text{This implies that the corresponding }x\text{-coordinate is just}\\ x=\sqrt{a^2-h^2}$
$\textit{Do you see why? This is just the equation of the circle rearranged.}$

$\text{Take a typical shell with respect to the }x\text{-axis}\\ \text{Rotation about the }y\text{-axis means the radius is just }r=x\\ \text{The height goes from the top of the circle to the bottom, so it is just }2h=2\sqrt{a^2-x^2}\\ \textit{Once again, do you see why?}$
$\text{So the volume of a typical shell is }\\ \delta V = 2\pi(x)(2\sqrt{a^2-x^2})\delta x = 4\pi x \sqrt{a^2-x^2}\, \delta x \\ \text{for some thickness }\delta x$
$\begin{align*}\therefore V&=\lim_{\delta x \to \infty}{\sum_{x=\sqrt{a^2-h^2}}^{a}{4\pi x \sqrt{a^2-x^2} \,\delta x}} \\ &= \int_{\sqrt{a^2-h^2}}^a{4\pi x \sqrt{a^2-x^2} \, dx}\\ &= \left[-\frac{4\pi {\left( a^2-x^2\right)}^\frac{3}{2}}{3}\right]_{\sqrt{a^2-h^2}}^a \\ &= -\frac{4\pi {\left( a^2-a^2\right)}^\frac{3}{2}}{3} + \frac{4\pi {\left( a^2-\left(a^2-h^2\right)\right)}^\frac{3}{2}}{3} \\ &= \frac{4\pi {\left(h^2\right)}^{\frac{3}{2}}}{3} \\ &= \frac{4\pi h^3}{3} \end{align*}$

amandali · « **Reply #247 on:** June 16, 2016, 03:05:09 pm »

how do prove this thanks

jakesilove · « **Reply #248 on:** June 16, 2016, 04:14:47 pm »

Quote from: amandali on June 16, 2016, 03:05:09 pm

(Image removed from quote.)
how do prove this thanks

Hey!

I'm terrible at inequality questions like this; I've had a bit of a crack, didn't get to an answer but got close. I can see the answer if we are give that
$a+b+c+d=4$

or something similar, but at the moment I don't THINK I have enough information. Just posting this here in case anyone can make sense of it and move forward from there!

Jake

RuiAce · « **Reply #249 on:** June 16, 2016, 07:48:44 pm »

$\text{Admittedly, I genuinely got stuck on this one. Here's an EXTREMELY tidy solution:}$
$\text{AM-GM inequality for two variables: For positive }a,b,c,d\\ \begin{align*}{(a-b)}^2 &\ge 0 \\ \iff a^2+b^2 & \ge 2ab \\ \iff {(a+b)}^2 &\ge 4ab \\ \implies \frac{a+b}{2} &\ge \sqrt{ab}\end{align*}$
$\text{AM-GM inequality for four variables:} \\ \begin{align*} \frac{w+x+y+z}{4} &= \frac{\frac{w+x}{2}+\frac{y+z}{2}}{2} \\ & \ge \frac{\sqrt{wx}+\sqrt{yz}}{2} \\ &\ge \frac{2\times \sqrt{wxyz}}{4} \\ &= \sqrt[4]{wxyz} \end{align*}$
$\text{The point: Let }\\ w=\frac{a}{b}, x=\frac{b}{c}, y=\frac{c}{d}, z=\frac{d}{a}\\ \text{Then }\\ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge 4\sqrt[4]{\frac{a}{b}\times\frac{b}{c}\times\frac{c}{d}\times\frac{d}{a}}=4\sqrt[4]{1}=4$

Can't take credit for this. Whilst I considered the AM-GM inequality several times it went way over my head how it could be used. And I can't spend forever on this question myself cause my exam is tomorrow.

amandali · « **Reply #250 on:** June 16, 2016, 10:14:03 pm »

got stuck on proving this ques

RuiAce · « **Reply #251 on:** June 17, 2016, 09:09:27 am »

Quote from: amandali on June 16, 2016, 10:14:03 pm

got stuck on proving this ques(Image removed from quote.)

$\text{You were nearly there}$
$\begin{align*}a^2\sin{\theta}\left(-2\sin{\left(\frac{\theta+\phi}{2}\right)} \sin{\left(\frac{\theta-\phi}{2}\right)}\right)+b^2\sin{(\phi-\theta)}&=0 \\ 2a^2\sin{\theta}\sin{\left(\frac{\theta+\phi}{2}\right)}\sin{\left(\frac{\phi-\theta}{2}\right)}+b^2\left(2\sin{\left(\frac{\phi-\theta}{2}\right)}\cos{\left(\frac{\phi-\theta}{2}\right)}\right)&=0\\ a^2\sin{\theta}\sin{\left(\frac{\theta+\phi}{2}\right)}+b^2\cos{\left(\frac{\theta-\phi}{2}\right)}&=0\end{align*}$
$\text{The last part requested you identify the double angle}\\ \text{and that sine and cosine are respectively, odd and even functions}$

katherine123 · « **Reply #252 on:** June 18, 2016, 10:36:37 am »

how do i do part (iii)

RuiAce · « **Reply #253 on:** June 18, 2016, 01:26:54 pm »

Quote from: katherine123 on June 18, 2016, 10:36:37 am

how do i do part (iii)

$\text{Join }OQ\text{, then similar triangles fall out:}$
$O\text{ is the midpoint of }S^\prime S\\ \text{and }Q\text{ is the midpoint of }RS\text{ from the congruent triangles you proved in part i)} \\ \textit{(This is the hard part to see)}$
$\text{In }\triangle S^\prime RS\text{ and }\triangle OQS\\ \begin{align*} \angle S^\prime SR &= OSQ & \text{(common }\angle S) \\ \frac{S^\prime S}{OS}&=\frac{RS}{QS}=2 &\text{(midpoints to their respective intervals)}\\ \therefore \triangle S^\prime RS & ||| \triangle OQS & \text{(two sides in proportion, included angles equal)}\end{align*}$
$\text{Hence }\frac{RS^\prime}{OQ}=2 \quad \text{(matching ratio of sides on similar triangles)} \Rightarrow OQ=a \\ \text{which implies }Q\text{ lies on a circle centred at the origin with radius }a$

birdwing341 · « **Reply #254 on:** June 20, 2016, 02:10:39 pm »

Hello!

I think I remember somewhere that one moderator had a small checklist that they went through after completing each question (to check for stuff like units, constants, dxs etc.).

Would you be able to remind me what that is?

Thanks!

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