HSC 2013 MX2 Marathon (archive) (1 Viewer)

SpiralFlex · Nov 11, 2012

Re: HSC 2013 4U Marathon

$(a) The equation of the ellipse E is$\; \frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\; $where a,b,h are real and a,b is greater than zero.$

$(i) By integration find the area enclosed by E.$

$$(ii) If$\; y=mx \;$is a tangent to$\; E\; $show that$\; m^2=\frac{b^2}{h^2-a^2}$

$(b) If n belongs to the set of natural numbers, let$ \;E_{n} \;$be the ellipse given by$

$\frac{(x-h_{n})^2}{(a_{n})^2}+\frac{y^2}{p^2(a_{n})^2}=1\; $where$\; p>0, h_{n}>h_{n+1}, h_{n}>a_{n}>0\;$

$$Supposing for all natural numbers that$\; E_{n}\;$and$\; E_{n+1}\; $touch each other externally and$\; y=mx\;$is the common tangent to all$\; E_{n}.$

$$(i) Express$\; h_{n}-h_{n+1} \;$in terms of$\; a_{n}\;$and$\; a_{n+1}.$

$Hence show,$

$$(ii)$\;\frac{a_{n+1}}{a_{n}}=\frac{h_{1}-a_{1}}{h_{1}+a_{1}}$

$$(iii)$ $If$\; S_{n}\; $is the area enclosed by the ellipse$\; E_{n}. \;$Evaluate$\; \sum_{n=1}^{\infty } S_{n}\; $in terms of$\; a_{1}, h_{1}, p.$

twinklegal19 · Nov 11, 2012

bleakarcher said:
Notice the point (0,0) representing z=0 lies on the circle lz-1l=1, initially we were given (1/z)+(1/conjugate(z))=1 and so z=/=0.

Sy123 said:
Yes it is the 'mindless algebra', think carefully about what z can be

Ahhhhhhhh ok then. Can't believe I fell for that haha, will definitely remember this next time.

So I just have to add z=/=0 at the end right?

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
Ahhhhhhhh ok then. Can't believe I fell for that haha, will definitely remember this next time.

So I just have to add z=/=0 at the end right?

I would say:

'A circle with radius 1, centre (1,0) with an empty circle at the origin due to z=/=0'

I would make a tiny sketch to show what I mean and that is all. The question is actually originally sketch the locus, but since I cant really ask that on the internet I changed it to describe.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
$(a) The equation of the ellipse E is$\; \frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\; $where a,b,h are real and a,b is greater than zero.$

$(i) By integration find the area enclosed by E.$

$y = b\sqrt{1-\frac{(x-h)^2}{a^2}$

Since it is just a translation, I will use "a" and "-a" as my limits of integration. Noting we want area, and the ellipse is symmetrical, I shall use the limits 0 to "a" and multiply by 4. I will also disregard the "-h" due to this translation:

$4b \int_{0}^{a} 1 - \frac{x^2}{a^2} dx$

$4b[x]_0^a - 4b\int_{0}^{a} \frac{x^2}{a^2} dx$

$4ab - 4b[\frac{x^3}{3a^2}]_0^a$

$4ab - \frac{4b}{3a^2} [a^3 - 0^3]$

$4ab - \frac{4a^3b}{3a^2}$

$4ab - \frac{4ab}{3}$

$\frac{12ab - 4ab}{3}$

$\frac{8ab}{3}$

seanieg89 · Nov 11, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$y = b\sqrt{1-\frac{(x-h)^2}{a^2}$

Since it is just a translation, I will use "a" and "-a" as my limits of integration. Noting we want area, and the ellipse is symmetrical, I shall use the limits 0 to "a" and multiply by 4. I will also disregard the "-h" due to this translation:

$4b \int_{0}^{a} 1 - \frac{x^2}{a^2} dx$

$4b[x]_0^a - 4b\int_{0}^{a} \frac{x^2}{a^2} dx$

$4ab - 4b[\frac{x^3}{3a^2}]_0^a$

$4ab - \frac{4b}{3a^2} [\frac{a^3}{3a^2} - \frac{0^3}{3a^2}]$

$4ab - \frac{4a^3b}{3a^2}$

$4ab - \frac{4ab}{3}$

$\frac{12ab - 4ab}{3}$

$\frac{8ab}{3}$

By setting a=b=1 your answer has the revolutionary consequence that the unit circle has area 8/3.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
By setting a=b=1 your answer has the revolutionary consequence that the unit circle has area 8/3.

What did I do wrong then? I was typing it up whilst waiting to respawn in a game, so might have made some random error. (or completely fucked up).

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

Because the area should be $\pi ab$ shouldn't it.....

seanieg89 · Nov 11, 2012

Re: HSC 2013 4U Marathon

Square root magically disappeared.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Square root magically disappeared.

lol shit, massive fail on my behalf.

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

$$Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1)$

$Show that $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right = \frac{n}{6}(n+1)(n+2) \\ \\ $Hence prove that the product of any three consecutive integers is divisible by 6$

SpiralFlex · Nov 11, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
Ahhhhhhhh ok then. Can't believe I fell for that haha, will definitely remember this next time.

So I just have to add z=/=0 at the end right?

Holes, holes, holes

twinklegal19 · Nov 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Holes, holes, holes

*sigh*(123)

geddit? geddit?

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1)$

$$Find in terms of n $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right$

$\sum_{k=1}^r k = t_r$ where $t_p$ denotes the p'th triangular number.

$\sum_{r=1}^{n} t_r =nt_n - 2t_{n-1}$

Note $t_n = t_{n-1} + n$

$nt_n - 2t_{n-1} = n(t_{n-1} + n) - 2t_{n-1}$

$n^2 + t_{n-1}(n-2)$

Noting that $t_k = \frac{1}{2}(k)(k+1)$

$n^2 + t_{n-1}(n-2) = n^2 + \frac{1}{2}(n-2)(n-1)(n)$

$=\frac{n^3 - n^2 + 2n}{2}$

Let's hope I didn't fuck anything up this time.

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\sum_{k=1}^r k = t_r$ where $t_p$ denotes the p'th triangular number.

$\sum_{r=1}^{n} t_r =nt_n - 2t_{n-1}$

Note $t_n = t_{n-1} + n$

$nt_n - 2t_{n-1} = n(t_{n-1} + n) - 2t_{n-1}$

$n^2 + t_{n-1}(n-2)$

Noting that $t_k = \frac{1}{2}(k)(k+1)$

$n^2 + t_{n-1}(n-2) = n^2 + \frac{1}{2}(n-2)(n-1)(n)$

$=\frac{n^3 - n^2 + 2n}{2}$

Let's hope I didn't fuck anything up this time.

Not quite. I am not sure how you arrived at your observation in the second line.

EDIT: I will make it a 'show that' question so you can check the answer.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Not quite. I am not sure how you arrived at your observation in the second line.

Yep I'm going to go through and make sure the observations are right, I was just using the identities I found when I used to mess around with triangular numbers, probs got one of them slightly wrong.

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Yep I'm going to go through and make sure the observations are right, I was just using the identities I found when I used to mess around with triangular numbers, probs got one of them slightly wrong.

I see, well I changed the question so you know if you got it right or not.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

ok found my mistake, will re-post my solution.

RealiseNothing · Nov 11, 2012

Re: HSC 2013 4U Marathon

Ok so far I have:

$\sum_{k=0}^{n-1} (-2)^k(n-k)t_{n-k}$

Sy123 · Nov 11, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Ok so far I have:

$\sum_{k=0}^{n-1} (-2)^k(n-k)t_{n-k}$

Heh, I believe you are over complicating the question a bit. In your original solution you made an observation that you could of just applied into the expression. Apply the given sum and you are done.

seanieg89 · Nov 11, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1)$

$Show that $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right = \frac{n}{6}(n+1)(n+2) \\ \\ $Hence prove that the product of any three consecutive integers is divisible by 6$

So it is obvious how you WANT students to do this question so I won't write that solution up. But just a neat observation, the notion of a sequences "derivative" from http://community.boredofstudies.org/showthread.php?t=292775 can be used to solve this question kind of nicely. It also allows you to generalise the result to an arbitrary number of nested summations!

(The differentiation operator removes sums from the LHS in the same way that differentiation is inverse to integration.)

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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