Author Topic: 3U Maths Question Thread (Read 1505081 times) Tweet Share

RuiAce · « **Reply #2445 on:** July 16, 2017, 11:13:01 pm »

Quote from: winstondarmawan on July 16, 2017, 11:09:46 pm

I could do (i). Thank you!

Anyways, I need help with the Q below. TIA
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20134615_1272831189509068_1563889495_n.png?oh=a3006750c9fdc562896e9b13f276e862&oe=596D675F

This involves forces, which are only in 4U. Where did this come from?

winstondarmawan · « **Reply #2446 on:** July 16, 2017, 11:21:36 pm »

Quote from: RuiAce on July 16, 2017, 11:13:01 pm

This involves forces, which are only in 4U. Where did this come from?

James Ruse 2011 3U Trial Paper

RuiAce · « **Reply #2447 on:** July 17, 2017, 10:08:41 am »

Just checking their solutions I can tell that (for some reason unknown) they've decided to throw a 4U question into a 3U paper. This is very unacceptable (although seeing as though it is a school paper it can be somewhat persuaded). Bits of what's covered here is NOT examinable in 3U.
$\text{Since }A\text{ is the origin, denote }x\text{ as the distance the particle is from }A.$
$\text{A force of }4x\text{ pulls the particle }\textit{back}\text{ to the origin}\\ \text{noting }x\text{ is the distance }PA$
$\text{A force of}(1-x)^2\text{ pushes the particle}\textit{ away}\text{ from the origin}\\ \text{noting }1-x\text{ is the distance }PB\\ \text{since }AB=1$
$\text{Hence resolving the forces, using }F=m\ddot{x}\\ 1\ddot{x} = -4x + (1-x)^2\\ \text{giving the desired result upon expanding.}$
________________________________________
$\text{From straight substitution, when }x=\frac12\text{ we have }\ddot{x}=-\frac74 < 0\\ \text{Hence the particle accelerates in the negative direction, i.e. back towards }A\text{ at that acceleration}$
$\text{But the particle starts at rest}\\ \text{So using your intuition, if it starts from rest and accelerates towards }A\\ \text{It'll start heading towards }A.$
The last part is your very classic computation using $\ddot{x}= \frac{d}{dx}\left(\frac12v^2\right)$ and is left as your exercise for now. This part, of course, is still 3U.

J.B · « **Reply #2448 on:** July 17, 2017, 10:55:00 am »

I was wondering if someone would be able to help me with this question, I keep getting the wrong answer. The answer should be 30.1m and 0.5 m/s.
Thank you

RuiAce · « **Reply #2449 on:** July 17, 2017, 03:26:19 pm »

Quote from: J.B on July 17, 2017, 10:55:00 am

I was wondering if someone would be able to help me with this question, I keep getting the wrong answer. The answer should be 30.1m and 0.5 m/s.
Thank you

beau77bro · « **Reply #2450 on:** July 17, 2017, 03:43:28 pm »

could someone post a solution to this from the lecture - my friend asked me how to do it (i didnt get to go but he was wondering how to do it). thankyou

i am very close to the answer with:

range = 2V² tanß /gsec²ß = (a² + b²)/(a+b) + 2V² /gsec²ß

but i cant get any further/ i dont know if im on the right track

junzhang · « **Reply #2451 on:** July 17, 2017, 05:32:53 pm »

Hi,
Could you help me with this 'show that' question? Thank you!

RuiAce · « **Reply #2452 on:** July 17, 2017, 06:18:08 pm »

Quote from: junzhang on July 17, 2017, 05:32:53 pm

Hi,
Could you help me with this 'show that' question? Thank you!

$\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\dots+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n=(1+x)^n$
$\text{Putting }x=2\\ \begin{align*}\binom{n}{0}+2\binom{n}1+4\binom{n}{2} +\dots + 2^{n-1}\binom{n}{n-1}+2^n\binom{n}n&=3^n\\ 1+2\binom{n}1+4\binom{n}{2} +\dots + 2^{n-1}\binom{n}{n-1}&-3^n-2^n\end{align*}$

RuiAce · « **Reply #2453 on:** July 17, 2017, 06:48:33 pm »

Quote from: beau77bro on July 17, 2017, 03:43:28 pm

could someone post a solution to this from the lecture - my friend asked me how to do it (i didnt get to go but he was wondering how to do it). thankyou

i am very close to the answer with:

range = 2V² tanß /gsec²ß = (a² + b²)/(a+b) + 2V² /gsec²ß

but i cant get any further/ i dont know if im on the right track

I had a go at Jamon's question and got it out but ceebs posting my answer. Especially since I used the same bit-unfair-for-3U method Jamon used for part b).

Anyway, found the answer to part a) on Google

beau77bro · « **Reply #2454 on:** July 17, 2017, 07:30:29 pm »

is that a proper hsc textbook or jsut something you found - the questions seem very high level

RuiAce · « **Reply #2455 on:** July 17, 2017, 07:31:09 pm »

Quote from: beau77bro on July 17, 2017, 07:30:29 pm

is that a proper hsc textbook or jsut something you found - the questions seem very high level

Jamon put it as a challenge question for the break during the lecture. You can make a call on that one.

All I did was Google the first sentence for fun.

jamonwindeyer · « **Reply #2456 on:** July 17, 2017, 08:40:25 pm »

Quote from: beau77bro on July 17, 2017, 03:43:28 pm

could someone post a solution to this from the lecture - my friend asked me how to do it (i didnt get to go but he was wondering how to do it). thankyou

i am very close to the answer with:

range = 2V² tanß /gsec²ß = (a² + b²)/(a+b) + 2V² /gsec²ß

but i cant get any further/ i dont know if im on the right track

I'd approach it differently! Setup the Cartesian equation:

$y=x\tan{\theta}-\frac{gx^2}{2V^2\cos^2{\theta}}$

Substitute the coordinates for the two edges of the wall, $(a,b)$ and $(b,a)$:

$b=a\tan{\theta}-\frac{ga^2}{2V^2\cos^2{\theta}}$
$a=b\tan{\theta}-\frac{gb^2}{2V^2\cos^2{\theta}}$

Now we want to link these together - The trick is to multiply the first equation by 'a' and the second by 'b':

$ab=a^2\tan{\theta}-\frac{ga^3}{2V^2\cos^2{\theta}}\\ab=b^2\tan{\theta}-\frac{gb^3}{2V^2\cos^2{\theta}}$

Then you can equate the two RHS, and I'll let you take it from there - You'll be able to find the range expression appearing there with only a's and b's on the other side

RuiAce · « **Reply #2457 on:** July 17, 2017, 09:10:54 pm »

Hmm, I had the same equations for b and a but I continued differently
$\text{Instead of a fudge factor, I added and subtracted the following equations}\\ \begin{align*}\frac{ga^2}{2V^2\cos^2\theta}&=a\tan \theta-b\\ \frac{gb^2}{2V^2\cos^2\theta}&=b\tan \theta-a\end{align*}$

caitlinlddouglas · « **Reply #2458 on:** July 18, 2017, 04:24:39 pm »

Quote from: RuiAce on July 16, 2017, 09:12:45 pm

$\binom{n}k\text{ is clearly the coefficient of }x^k\text{ in }(1+x)^n-1\\ \text{So it falls to determine the coefficient of }x^k\text{ in the LHS}$
$\text{The LHS directly expands to give}\\ x(1-x)^{n-1}+x(1-x)^{n-2}+\dots + x(1+x) + x$
$\text{Now, the coefficient of }x^k\text{ in }x(1-x)^{n-1}\\ \text{will be the }\textbf{same}\text{ as the coefficient of }x^{k-1}\text{ in }(1-x)^{n-1}\\ \text{due to the extra }x\text{ being multiplied. This will be }\binom{n-1}{k-1}$
$\text{Similarly, the coefficient of }x^k\text{ in }x(1-x)^{n-2}\text{ is }\binom{n-2}{k-1}\\ \text{This process goes all the way down until }x(1-x)^{k-1}\\ \text{to which the coefficient on }x^k\text{ is }\binom{k-1}{k-1}$
$\textbf{Why do we stop?}\\ \text{Because the next term is }x(1-x)^{k-2}\\ \text{and the relevant coefficient, is the coefficient on }x^{k-1}\text{ in }(1-x)^{k-2}\\ \text{But this, of course, doesn't exist because our power is too small now.}$
$\text{Hence regrouping, the coefficient on }x^k\text{ in the LHS is}\\ \binom{n-1}{k-1}+\binom{n-2}{k-1}+\dots+\binom{k-1}{k-1}$
Also, part iii should not be done by the binomial theorem, but just by brute force algebra.

Thanks heaps!

abba9554 · « **Reply #2459 on:** July 18, 2017, 06:05:22 pm »

Hi!
I need help with this question. I am having trouble understanding the notation. Thanks guys

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