HSC 2012 MX1 Marathon #2 (archive) (6 Viewers)

asianese · Jul 22, 2012

Re: HSC 2012 Marathon

$\frac{d}{dx} (x\sqrt{r^2-x^2}=x\times \frac{1}{2\sqrt{r^2-x^2}}\times -2x + \sqrt{r^2-x^2}$

$RHS=\frac{x^2-r^2+r^2}{\sqrt{r^2-x^2}}=LHS$ - don't know how to say it lol

$I = \int_{-r}^r \sqrt{r^2-x^2} dx = \int_{-r}^r \frac{d}{dx} x \sqrt{r^2-x^2} + \frac{x^2}{\sqrt{r^2-x^2}} dx \\ = x \sqrt{r^2-x^2}+\int_{-r}^r \frac{-(r^2-x^2)}{\sqrt{r^2-x^2}}+\frac{r^2}{\sqrt{r^2-x^2}}dx \\ = \left[ x\sqrt{r^2-x^2}-I+r^2\sin^{-1} \left(\frac{x}{r}\right) \right]_{-r}^r\\ 2I=0+r^2\times \frac{\pi}{2} - (0-r^2\times \frac{\pi}{2}) \\ I=\frac{r^2\pi}{2} \\ \therefore 2\times \int_{-r}^r \sqrt{r^2-x^2} dx = \pi r^2$

Sy123 · Jul 22, 2012

Re: HSC 2012 Marathon

Nice work, you have just proved the area of a circle. (Im not sure if you would have to prove that a semi-circle is half a full circle first though before saying it)

Either way, nicely done, also for part b, its just there so people dont get lost when they try to integrate it.

Can someone else post a question? Ive run out of ideas at the moment.

SpiralFlex · Jul 22, 2012

Re: HSC 2012 Marathon

What sort of question do you guys want?

Shadowdude · Jul 22, 2012

Re: HSC 2012 Marathon

How many eight letter words can be formed from the letters of PARRAMATTA?

Sy123 · Jul 22, 2012

Re: HSC 2012 Marathon

$\frac{10!}{4! 2! 2!}$

Also Spiral, a Binomial proof one would be nice, (or anything to do with proofs) but anything will do, as long as its not Probability.

SpiralFlex · Jul 22, 2012

Re: HSC 2012 Marathon

I'll make a projectile motion I guess haha.

Timske · Jul 22, 2012

Re: HSC 2012 Marathon

SpiralFlex said:
I'll make a projectile motion I guess haha.

yeeey and shm

RealiseNothing · Jul 22, 2012

Re: HSC 2012 Marathon

Sy123 said:
Nice work, you have just proved the area of a circle. (Im not sure if you would have to prove that a semi-circle is half a full circle first though before saying it)

Either way, nicely done, also for part b, its just there so people dont get lost when they try to integrate it.

Can someone else post a question? Ive run out of ideas at the moment.

If you want a question, prove the area of a circle without the use of calculus. (this should be pretty easy though).

You can assume that the circumference is given by $2\pi r$ .

SpiralFlex · Jul 22, 2012

Re: HSC 2012 Marathon

Okay, I will post up my question tomorrow. It involves penguins + cannons lol

Carrotsticks · Jul 22, 2012

Re: HSC 2012 Marathon

RealiseNothing said:
If you want a question, prove the area of a circle without the use of calculus. (this should be pretty easy though).

You can assume that the circumference is given by $2\pi r$ .

I can imagine three possible proofs.

1. Unfolding infinitely large numbers of sectors with infinitely small angles, and arrange the 'jagged' shape until you get a rectangle, then find the area of the rectangle (area of circle equivalent) using base x height.

2. Consider an infinitely large number of infinitely thing concentric rings. Unfold them and stack them up to form an isosceles triangle, with the base being the circumference. The area of the circle is the area if the triangle, which can be computed using 1/2 base x height.

3. Construct upper and lower bound n-gons and find the area of them. Take the limit as n -> infinity and squeeze the area of the circle between them.

asianese · Jul 23, 2012

Re: HSC 2012 Marathon

Actually I need help (practice) on probability so something 3unity would be nice =)

I like the squeeze theorem proof of a circle's area, it's a nice idea of the limit as n goes to infinity.

Shadowdude · Jul 23, 2012

Re: HSC 2012 Marathon

Sy123 said:
$\frac{10!}{4! 2! 2!}$

Also Spiral, a Binomial proof one would be nice, (or anything to do with proofs) but anything will do, as long as its not Probability.

Is that for the "How many eight letter words in PARRAMATTA" question?

asianese · Jul 23, 2012

Re: HSC 2012 Marathon

Isn't it $\frac{^{10} P_8}{2!4!2!}$ ... May be completely wrong...

Sy123 · Jul 23, 2012

Re: HSC 2012 Marathon

I am pretty sure I am right, I too may be completely wrong, but I thought, because there are 4 'a's 2 'r's and 2 't's. Because we cant distinguish between the same letters, and there are 4! 2! and 2! ways of sorting each one, I thought I could divide the total number by it..

Thats how Ive always done it, and I dont think I have ever really used the Permutation thing before.

Also about proving the area of a circle without calculus. I came to this:

Let a regular polygon with n-sides be inscribed in a circle. We can split the polygon into an n number of triangles. If the radius of the circle is r, then the total area of the n-number of triangles is:

$n\times \frac{r^2}{2} \sin{\theta}$

Where Theta is the angle subtended by each triangle. However because the total angle of revolution is 2pi

$\therefore \theta=\frac{2\pi}{n}$

$\therefore A_{n-gon}=n \frac{r^2}{2} \sin{\frac{2\pi}{n}}$

And if we take the limit of it as $n \to \infty$

$A_{circle}= \lim_{n \to \infty} r^2 \frac{n}{2} \sin{\frac{2\pi}{n}}$

Graphing this in geogebra, I am able to see an asymptote at pi, then when multiplied with r^2 you get the area of the circle pi r^2

However, I dont know how to evaluate the limit, if it is even possible for a 3U student. Is it possible to evaluate this limit with 3U knowledge?

Carrotsticks · Jul 23, 2012

Re: HSC 2012 Marathon

Sy123 said:
Also about proving the area of a circle without calculus. I came to this:

Let a regular polygon with n-sides be inscribed in a circle. We can split the polygon into an n number of triangles. If the radius of the circle is r, then the total area of the n-number of triangles is:

$n\times \frac{r^2}{2} \sin{\theta}$

Where Theta is the angle subtended by each triangle. However because the total angle of revolution is 2pi

$\therefore \theta=\frac{2\pi}{n}$

$\therefore A_{n-gon}=n \frac{r^2}{2} \sin{\frac{2\pi}{n}}$

And if we take the limit of it as $n \to \infty$

$A_{circle}= \lim_{n \to \infty} r^2 \frac{n}{2} \sin{\frac{2\pi}{n}}$

Graphing this in geogebra, I am able to see an asymptote at pi, then when multiplied with r^2 you get the area of the circle pi r^2

However, I dont know how to evaluate the limit, if it is even possible for a 3U student. Is it possible to evaluate this limit with 3U knowledge?

Very nice!

There a couple of ways of evaluating the limits, one of which can be done using a clever substitution (well within the 3U course).

$\\ $We want to prove that $ \lim_{n \to \infty} \frac{n}{2} \sin{\frac{2\pi}{n}} = \pi \\\\ $We make the substitution $ n=\frac{2 \pi}{k} $ so as $ n \rightarrow \infty, k \rightarrow 0. \\\\ \lim_{n \to \infty} \frac{n}{2} \sin{\frac{2\pi}{n}} =\lim_{k \rightarrow 0} \frac{\frac{2 \pi}{k}}{2} \sin \left ( \frac{2 \pi}{\frac{2 \pi}{k}} \right ) = \pi \times \lim_{k \rightarrow 0} \frac{\sin (k)}{k} = \pi$

And Sy123, your answer for the permutation question is correct.

Sy123 · Jul 23, 2012

Re: HSC 2012 Marathon

Oh nice, Ive learnt a new technique for limits. (Of course substitution is something that can be used in almost anything, but it didn't really occur to me to use it for limits)

Thanks

asianese · Jul 23, 2012

Re: HSC 2012 Marathon

Just asking, if you have 10 letters ( P A R R A M A T T A ) , but you only choose 8, why is it still 10! ?

Carrotsticks · Jul 23, 2012

Re: HSC 2012 Marathon

Oh my apologies, I mis-read the question to be 'How many way can you arrange the letters of the word Parramatta', in which case Sy123's answer would have been correct.

This problem is a little tricky because you have to take cases. We can't just just use 10P8 because that includes permutation of the 8 letters, but how do we know which of the 8 are identical to each other? We can only place the whole thing over 4!2!2! is we KNOW for sure that all 10 letters are included but since we are only choosing 8, there is no guarantee that all the repeated letters are in it.

So we would have to take cases (quite tedious I can imagine).

bleakarcher · Jul 23, 2012

Re: HSC 2012 Marathon

^ It all makes sense again..

Sy123 · Jul 23, 2012

Re: HSC 2012 Marathon

Oops, I misread the question aswell :/

I thought it was your standard old word arrangement questions so I just went right ahead and did it..

HSC 2012 MX1 Marathon #2 (archive) (6 Viewers)

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