Author Topic: 3U Maths Question Thread (Read 1236419 times) Tweet Share

RuiAce · « **Reply #2475 on:** July 20, 2017, 09:07:45 pm »

Quote from: junzhang on July 20, 2017, 08:18:20 pm

Hello
Could you help me with this question, particularly the last part?
Ty!

$\text{Substituting }\alpha = \tan 45^\circ\text{ reduces the Cartesian equation to}\\ y=x\left(1-\frac{x}{R}\right)\\ \text{This can be rewritten as}\\ \boxed{Ry=x(R-x)}$
$\text{Suppose }X\text{ is the first }x\text{-coordinate where }y=4\\ \text{Then, }(X,4)\text{ and }(X+6,4)\text{ pass through the equation of motion}\\ \text{So we have:}\\ \begin{align*}4R&=X(R-X)\tag{1}\\ \tag{2}4R&=(X+6)(R-X-6)\end{align*}$
$\text{Upon equating,}\\ \begin{align*}x(R-x)&=(x+6)(R-x-6)\\ Rx-x^2&=Rx-x^2-6x+6R-6x-36\\ 2(x+3)&=R\\ x&=\frac{R}{2}-3\end{align*}$
$\text{Subbing this back into (1) we have}\\ 4R = \left(\frac{R}2-3\right)\left(\frac{R}2+3\right)\\ \text{This is just a quadratic equation, which I leave as your exercise}\\ \text{The final answer is }R=18.$

winstondarmawan · « **Reply #2476 on:** July 20, 2017, 11:03:51 pm »

Hi! Would appreciate help with Qb, TIA.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20226751_1653608764681716_168153538_o.jpg?oh=46eca3b4c0a6fb0d788820cc7df70e99&oe=5972A1BD

RuiAce · « **Reply #2477 on:** July 20, 2017, 11:19:01 pm »

Quote from: winstondarmawan on July 20, 2017, 11:03:51 pm

Hi! Would appreciate help with Qb, TIA.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20226751_1653608764681716_168153538_o.jpg?oh=46eca3b4c0a6fb0d788820cc7df70e99&oe=5972A1BD

$\text{Using the gradient formula this is just}\\ \begin{align*}m&=\frac{ap^4-ap^2}{2ap^2-2ap}\\ &= \frac{p^2(p^2-1)}{2p(p-1)}\\ &= \frac{p(p+1)(p-1)}{2(p-1)}\\ &= \frac{p^2+p}2\end{align*}$
$\text{Then follow the usual approach: }\frac{dm}{dp}=0 \implies \frac{2p+1}{2}=0 \implies p=-\frac12$
$\text{So the minimum gradient is }\frac{1}{2}\left(\left(-\frac12\right)^2-\frac12\right)\\ \text{and the coordinates of }P\text{ are }\left(2a\times -\frac12, a\left(-\frac12\right)^2\right)$

beau77bro · « **Reply #2478 on:** July 21, 2017, 09:50:11 am »

Quote from: Ali_Abbas on July 20, 2017, 07:41:31 pm

Did you assume that TQ is a diameter of circle TPQ?

Quote from: RuiAce on July 20, 2017, 08:35:56 pm

I swear last time I checked it was given. That seriously makes me wonder if this question lacks information for a 2 marker though, because it's damn obvious that we do have a diameter but how are they expecting us to prove it for so little reward?

Yea that is what makes it so strange. is it safe to say we have to assume that, because the question doesnt provide enough information? Because we arent told any information about the secondary circle, the chords arent tangents (not that would help i dont think), tq isnt said to be a diameter, and ps isn't said to meet t when extended.

RuiAce · « **Reply #2479 on:** July 21, 2017, 09:59:44 am »

Quote from: beau77bro on July 21, 2017, 09:50:11 am

Yea that is what makes it so strange. is it safe to say we have to assume that, because the question doesnt provide enough information? Because we arent told any information about the secondary circle, the chords arent tangents (not that would help i dont think), tq isnt said to be a diameter, and ps isn't said to meet t when extended.

I honestly feel like you should. Else it's stretching the bounds for a 2 mark question.

legorgo18 · « **Reply #2480 on:** July 21, 2017, 12:19:53 pm »

Hi! How to integrate dx/[1 + root(1+x)]

beau77bro · « **Reply #2481 on:** July 21, 2017, 12:32:27 pm »

Hope this helps

raymatar · « **Reply #2482 on:** July 21, 2017, 10:07:06 pm »

Hey could someone please help me with ii. The answers arent very thorough with their working out.

beau77bro · « **Reply #2483 on:** July 21, 2017, 10:23:32 pm »

Is that the answer? I think so. It works for both equations. If it's wrong I'll check my answer and get back to u. Hope this helps

raymatar · « **Reply #2484 on:** July 21, 2017, 10:27:29 pm »

Quote from: beau77bro on July 21, 2017, 10:23:32 pm

(Image removed from quote.)

Is that the answer? I think so. It works for both equations. If it's wrong I'll check my answer and get back to u. Hope this helps

These are the answers provided

beau77bro · « **Reply #2485 on:** July 21, 2017, 10:39:27 pm »

OHHH N was supposed to be x1

beau77bro · « **Reply #2486 on:** July 21, 2017, 10:41:12 pm »

Sorry subbed in wrong fixed it though (unless I completely tricked it to the answer)

RuiAce · « **Reply #2487 on:** July 21, 2017, 10:42:33 pm »

Quote from: beau77bro on July 21, 2017, 10:39:27 pm

OHHH N was supposed to be x1

Lol yep. I actually confused myself twice before I got to the same conclusion as you. Once you get to $ap^2 = \frac{3(2a+ap^2)}{5} $ it works.

In the meantime I am going to look at the prescribed answers.

beau77bro · « **Reply #2488 on:** July 21, 2017, 10:45:03 pm »

Quote from: RuiAce on July 21, 2017, 10:42:33 pm

Lol yep. I actually confused myself twice before I got to the same conclusion as you. Once you get to $ap^2 = \frac{3(2a+ap^2)}{5} $ it works.

In the meantime I am going to look at the prescribed answers.

Hahaha twinning

raymatar · « **Reply #2489 on:** July 21, 2017, 10:47:41 pm »

Quote from: beau77bro on July 21, 2017, 10:41:12 pm

(Image removed from quote.)

Sorry subbed in wrong fixed it though (unless I completely tricked it to the answer)

I think you did it correctly the second time round. Thanks for the help I really appreciate it.

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