Author Topic: 3U Maths Question Thread (Read 1236167 times) Tweet Share

Shadowxo · « **Reply #2520 on:** July 25, 2017, 06:53:10 pm »

Quote from: abba9554 on July 25, 2017, 06:04:03 pm

Hi,
How do you do this question?
thanks

I don't think this question can be done, it's just an expression not an equation. It doesn't have an "= number" so they haven't given you enough info

jakesilove · « **Reply #2521 on:** July 25, 2017, 09:19:38 pm »

Quote from: abba9554 on July 25, 2017, 05:52:23 pm

Hello!
Would love it if you could help me with this question. I never really understood the shadow questions. thanks very much

Honestly, I don't think this question makes sense either. Is the rod attached vertically? Horizontally? The question is very ambiguous. Where are these questions from?

anotherworld2b · « **Reply #2522 on:** July 27, 2017, 10:06:06 pm »

I'm not quite sure what I did wrong.
Can I have some help please?

Shadowxo · « **Reply #2523 on:** July 27, 2017, 10:09:29 pm »

Quote from: anotherworld2b on July 27, 2017, 10:06:06 pm

I'm not quite sure what I did wrong.
Can I have some help please?

It's just the last part (x^-2)
You add one to the power then divide by the resulting power

junzhang · « **Reply #2524 on:** July 27, 2017, 11:33:25 pm »

hello,
how do i do part ii? the answer states to add the two volumes squared. but i subtracted.
thanks heaps

RuiAce · « **Reply #2525 on:** July 28, 2017, 12:04:07 am »

Quote from: junzhang on July 27, 2017, 11:33:25 pm

hello,
how do i do part ii? the answer states to add the two volumes squared. but i subtracted.
thanks heaps

You are certainly meant to subtract here.
$V=\pi \int_{-2}^2 \left((4-x^2)^2-(2-x)^2\right)dx$

K9810 · « **Reply #2526 on:** July 28, 2017, 05:11:32 pm »

Hi, how do you do these questions:
1. Use mathematical induction to show that 3^n-2n-1 is divisible by 4 for all positive integers greater or equal to 3.

2. At time t hours an oil spills occurs, a circular oil slick has radius r km, where r=(t+1)^1/2-1. Find the rate at which the area of the slick is increasing when its radius is 1km.

Thank you!

RuiAce · « **Reply #2527 on:** July 28, 2017, 05:21:30 pm »

Quote from: K9810 on July 28, 2017, 05:11:32 pm

Hi, how do you do these questions:
1. Use mathematical induction to show that 3^n-2n-1 is divisible by 4 for all positive integers greater or equal to 3.

2. At time t hours an oil spills occurs, a circular oil slick has radius r km, where r=(t+1)^1/2-1. Find the rate at which the area of the slick is increasing when its radius is 1km.

Thank you!

$\text{When }n=3, 3^n-2n-1=20=4\times 5\\\text{which is divisible by 4}$
$\text{Assuming the truth when }n=k\text{ gives }3^k-2k-1=4M\\ \text{i.e. }\boxed{3^k=4M+2k+1}\text{ where }M\text{ is an integer}$
$\text{Considering when }n=k+1:\\ \begin{align*}3^{k+1}-2(k+1)-1&=3\left(3^k\right) - 2k - 3\\ &=3(4M+2k+1)-2k-3\tag{assumption}\\ &= 12M+4k\\ &= 4(3M+k)\end{align*}\\ \text{which is divisible by 4, so the truth when }n=k\\ \text{implies the truth when }n=k+1$
$\text{Hence true by induction}$

RuiAce · « **Reply #2528 on:** July 28, 2017, 05:36:14 pm »

Quote from: K9810 on July 28, 2017, 05:11:32 pm

Hi, how do you do these questions:
1. Use mathematical induction to show that 3^n-2n-1 is divisible by 4 for all positive integers greater or equal to 3.

2. At time t hours an oil spills occurs, a circular oil slick has radius r km, where r=(t+1)^1/2-1. Find the rate at which the area of the slick is increasing when its radius is 1km.

Thank you!

$\text{The second one is a less-standard related rates question}\\ \text{We need to gather more information}\\ r=\sqrt{t+1}-1\implies \boxed{\frac{dr}{dt}=\frac1{2\sqrt{r+1}}}$
$\text{Also, }A=\pi r^2 \implies \boxed{\frac{dA}{dr}=2\pi r}$
$\text{When }r=1,\\ \begin{align*}1&=\sqrt{t+1}-1\\ 4&=t+1\\ t&=5\end{align*}$
$\text{So since }\frac{dA}{dt}=\frac{dA}{dr}\times \frac{dr}{dt}\\ \boxed{\frac{dA}{dt}=\frac{\pi r}{\sqrt{t+1}}}\\ \text{Then sub in }r=1,t=5$

inescelic · « **Reply #2529 on:** July 28, 2017, 07:29:29 pm »

Hi Jake, just a general question. I'm doing my trials atm, and my maths ext exam is on tuesday (rip). What do you find is the best way to balance content heavy subjects (i.e. biology and legal) with maths which requires so much time and practice?

RuiAce · « **Reply #2530 on:** July 28, 2017, 08:26:31 pm »

Quote from: inescelic on July 28, 2017, 07:29:29 pm

Hi Jake, just a general question. I'm doing my trials atm, and my maths ext exam is on tuesday (rip). What do you find is the best way to balance content heavy subjects (i.e. biology and legal) with maths which requires so much time and practice?

What works best for one person will not necessarily work best for another person. You need to find the most optimal balance for you.

Content heavy subjects involve gathering (and preferably also understanding) a pile of information/data/etc. and then regurgitating it as some kind of response on paper. And like said, maths is practice. So it may help to think about which you handle better than the other. If you find it easier to store content then maybe you can shift the emphasis towards maths, and vice versa. Only you know what you're most well versed with.

At the same time, time management is obviously of the essence. That speaks in a comparative level, but on an absolute level you should also be doing enough good quality study to ensure you don't prepare too little for a subject.

Sukakadonkadonk · « **Reply #2531 on:** July 28, 2017, 09:16:19 pm »

Hey again,

Could I get some help with part ii of this question?

raymatar · « **Reply #2532 on:** July 28, 2017, 09:22:22 pm »

Can someone please help me with e.ii

RuiAce · « **Reply #2533 on:** July 28, 2017, 09:45:46 pm »

Quote from: raymatar on July 28, 2017, 09:22:22 pm

Can someone please help me with e.ii

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

RuiAce · « **Reply #2534 on:** July 28, 2017, 09:51:04 pm »

Quote from: Sukakadonkadonk on July 28, 2017, 09:16:19 pm

Hey again,

Could I get some help with part ii of this question?

As always, you should be aware of the fact that any papers from 2000 or earlier do NOT necessarily reflect the difficulty of the current HSC.
$\text{From trigonometry, }T\text{ is the point }(r\cos \alpha, r\sin \alpha)$
$\text{Since this point lies on the trajectory, substituting into the Cartesian equation of motion:}\\ \begin{align*}r\sin \alpha&=r\cos \alpha\tan\theta - r^2\cos^2\alpha\sec^2\theta\\ r\cos^2\alpha\sec^2\theta&=\cos\alpha\tan\theta-\sin \alpha\\ r&=\frac{\cos^2\theta(\cos\alpha\tan\theta-\sin\alpha)}{\cos^2\alpha}\\ &= \frac{\cos\theta(\cos\alpha\sin\theta-\sin\alpha\cos\theta)}{\cos^2\alpha}\\ &= \frac{\cos\theta \sin (\theta-\alpha)}{\cos^2\alpha}\end{align*}$

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Shadowxo

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