Author Topic: 3U Maths Question Thread (Read 1500470 times) Tweet Share

raymatar · « **Reply #2535 on:** July 28, 2017, 11:29:37 pm »

Quote from: RuiAce on July 28, 2017, 09:45:46 pm

$\text{Think about the significance of the acceleration}\\ \text{We're told that }\ddot{x}=x-2\text{, yet initially }x=3.$
$\text{This means that initially, }\ddot{x}=1 > 0\\ \text{so the particle starts accelerating towards the right.}$
$\text{But additionally, initially }v=0\\ \text{Hence the particle will be moving towards the right as well.}\\ \text{Because the particle moves to the right, the displacement }\textbf{increases}$
$\text{As the displacement continues to increase, the particle }\textbf{keeps}\text{ accelerating to the right}\\ \text{Because it was already moving to the right, if it keeps accelerating to the right}\\ \text{it will only continue to move to the right}$
$\textbf{i.e. it will never move back to the left.}$

Thanks for the help Rui

seventeenboi · « **Reply #2536 on:** July 29, 2017, 02:29:31 pm »

hello

i need help with question c)
thanks in advanced!!

RuiAce · « **Reply #2537 on:** July 29, 2017, 04:08:29 pm »

Quote from: seventeenboi on July 29, 2017, 02:29:31 pm

hello
i need help with question c)
thanks in advanced!!

$\text{Suppose the particle's acceleration is a fixed constant }a\\ \text{Then }\frac{d^2x}{dt^2} = a \implies \frac{dx}{dt} = at + C_1\implies \boxed{x = \frac{at^2}{2}+C_1t + C_2}$
$\text{We have three conditions, so we can solve for our three constants.}\\ \text{Firstly, when }t=0, x=0\text{ so }C_2=0\\ \text{Hence }\boxed{x=\frac{at^2}{2}+C_1t}$
$\text{When }t=11, x=0.5\\ \text{When }t=18, x=0\\ \text{Substituting this in gives us a system of simultaneous equations:}\\ \begin{align*}0.5&= \frac{121a}{2}+11C_1\\ 0&=\frac{324a}{2}+18C_1\end{align*}\\ \text{I leave it as your exercise to show that }a-\frac1{77},C_1=\frac{9}{77}$
$\text{Hence }x=-\frac{t^2}{154}+\frac{9t}{77}\\ \text{You can now fill in the table of values.}$
________________________________
$\text{For the SHM scenario, commence using intuition.}\\ \text{We know that the period is }12\\ \text{So since when }t=0,x=0\\ \text{we }\textbf{also}\text{ have }t=12, x=0$
$\text{But now we observe that when }t=12\textbf{ and }\text{when }t=18, x=0\\ \text{This means that }x=0\text{is actually obtained exactly once}\\ \textbf{every half-period of motion}\text{, starting at }t=0$
$\text{Normally if a particle moves in SHM we assume }x=a\cos (nt+\alpha)+x_0\text{ or }x=a\sin (nt+\alpha)\\ \textbf{The above information is sufficient to deduce immediately that}\\ \boxed{x=a\sin nt}$
$\text{Furthermore, from }T=\frac{2\pi}{n}\text{ we have }n=\frac{\pi}{6}\\ \text{Hence }\boxed{x=a\sin \frac{\pi t}{6}}$
$\text{It now falls to finding }a\\ \text{When }t=11,a=\frac12\\ \text{Hence }\frac12 = a\sin \frac{11\pi}{6}\implies a=-1\\ \text{So }\boxed{x=-\sin \frac{\pi t}{6}}$
$\text{You can now complete the table of values}$

K9810 · « **Reply #2538 on:** July 29, 2017, 04:37:54 pm »

Hi, how do you do this question: . An upturned cone of semivertical angle 45◦ is being ﬁlled with water at a constant rate of 20cm3/s. Find the rate at which the height, the area of the water surface, and the area of the cone wetted by the water, are increasing when the height is 50cm.

RuiAce · « **Reply #2539 on:** July 29, 2017, 04:46:30 pm »

Quote from: K9810 on July 29, 2017, 04:37:54 pm

Hi, how do you do this question: . An upturned cone of semivertical angle 45◦ is being ﬁlled with water at a constant rate of 20cm3/s. Find the rate at which the height, the area of the water surface, and the area of the cone wetted by the water, are increasing when the height is 50cm.

$\text{Sketch a diagram.}\\ \text{You will observe from trigonometry that }\tan 45^\circ = \frac{r}{h}\implies r=h$
$\text{The volume of the cone is }V=\frac{1}{3}\pi r^2 h\\ \text{Using the above result, }V=\frac{1}{3}\pi h^3\\ \text{and hence }\frac{dV}{dh}=\pi h^2$
$\text{Also, }\frac{dV}{dt}=20\\ \text{So since }\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}$
$\begin{align*}20&=\pi h^2 \frac{dh}{dt}\\ \frac{dh}{dt}&=\frac{20}{\pi h^2}\end{align*}\\ \text{Sub in }h=50.$

K9810 · « **Reply #2540 on:** July 29, 2017, 08:20:08 pm »

Hey, can you please help me with this question

RuiAce · « **Reply #2541 on:** July 29, 2017, 08:28:14 pm »

Quote from: K9810 on July 29, 2017, 08:20:08 pm

Hey, can you please help me with this question

$\begin{align*}\cos^{-1}x-\sin^{-1}x&=k\\ \cos^{-1}x+\sin^{-1}x&=\frac{\pi}2\end{align*}\\ \text{(the second equation being an identity)}$
$\text{Upon adding,}\\ 2\cos^{-1}x = \frac{\pi}{2}+k \implies \boxed{\cos^{-1}x=\frac{\pi}4+\frac{k}2}$
$\text{Taking cosine of both sides and applying the compound angle identity}\\ \begin{align*}x&=\frac{1}{\sqrt2} \cos \frac{k}{2}-\frac1{\sqrt2}\sin \frac{k}2\\ &= \frac{1}{\sqrt2}\left(\cos \frac{k}2-\sin \frac{k}2\right)\end{align*}$

bluecookie · « **Reply #2542 on:** July 30, 2017, 10:40:18 pm »

I don't get the solutions for this :/

RuiAce · « **Reply #2543 on:** July 31, 2017, 10:47:54 pm »

Quote from: bluecookie on July 30, 2017, 10:40:18 pm

I don't get the solutions for this :/

I don't see how -3≤x≤3 allows this question to be doable. What is their proposed answer?

Shadowxo · « **Reply #2544 on:** August 01, 2017, 07:00:55 am »

Quote from: RuiAce on July 31, 2017, 10:47:54 pm

I don't see how -3≤x≤3 allows this question to be doable. What is their proposed answer?

Perhaps they want the students to extrapolate from the graph. It is a weird question though

RuiAce · « **Reply #2545 on:** August 01, 2017, 07:21:40 am »

Quote from: Shadowxo on August 01, 2017, 07:00:55 am

Perhaps they want the students to extrapolate from the graph. It is a weird question though

It doesn't work though. Unless they explicitly specify "over $-3\le x \le 3$" there really is no algebraic method to deduce the domain for which there's exactly three solutions.

E.g. $y=\frac\pi{15}x$ still has three solutions but $y=\frac\pi{16}x$ has about five

Sukakadonkadonk · « **Reply #2546 on:** August 01, 2017, 08:03:26 am »

Hi, how would you do this one?

thanks.

RuiAce · « **Reply #2547 on:** August 01, 2017, 08:12:22 am »

Quote from: Sukakadonkadonk on August 01, 2017, 08:03:26 am

Hi, how would you do this one?

thanks.

$\text{Lemma: }N! = N(N-1)(N-2)!\\ \text{Proof: Exercise}$
$\text{The coefficient of }x^8\text{ is }\binom{n}{8}3^8\\ \text{The coefficient of }x^{10}\text{ is }\binom{n}{10}3^10\\ \text{and we're saying the coefficient of }x^{10}\text{ is twice that of }x^8$
$\text{Hence}\\ \begin{align*}2\binom{n}8 3^8 &=\binom{n}{10}3^{10}\\ 2\frac{n!}{8!(n-8)!}&=\frac{n!}{10!(n-10)!}3^2 \\ \frac{2}{9}&=\frac{8!(n-8)!}{10!(n-10)!}\\ &=\frac{(n-8)(n-9)}{9\times 10} \\ 20&= n^2-17n+72\\ n^2-17n+52&=0\\ (n-13)(n-4)&=0\\ n&=4,13\end{align*}$
$\text{But clearly }n\ge 8\text{ as else the coefficient of }x^8\text{ is }0.\\ \text{Hence }n=13$

junzhang · « **Reply #2548 on:** August 01, 2017, 08:07:22 pm »

Hi,
Could you help me with part ii?
Thanks!

Sorry I have another question.
for part iv, how do I prove it is the shortest beam? I managed to get this value when dy/da=0, but i don't know how to prove minimisation with this question

can't thank you all enough for the help!

Mod edit: Posts merged. You're free to ask a few questions in succession, but please resort to the 'modify' function at the top right corner of a post to refrain from double posting.

RuiAce · « **Reply #2549 on:** August 01, 2017, 09:05:33 pm »

Quote from: junzhang on August 01, 2017, 08:07:22 pm

Hi,
Could you help me with part ii?
Thanks!

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