HSC 2013 MX2 Marathon (archive) (11 Viewers)

twinklegal19 · Feb 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
By change of base law:

$\log_x e = \frac{\log_e e}{\log_e x} = \frac{1}{\ln x}$

And its straightforward graph manipulation from there.

~~if you just blindly draw the reciprocal then it's wrong though~~

edit: derped really badly

RealiseNothing · Feb 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
By change of base law:

$\log_x e = \frac{\log_e e}{\log_e x} = \frac{1}{\ln x}$

And its straightforward graph manipulation from there.

Or $x^y =e$ which... edit: lol

RealiseNothing · Feb 18, 2013

Re: HSC 2013 4U Marathon

From what I got quickly, it's a hyperbola with horizontal asympote at y=0 and vertical asymptote at x=1. Gonna check though as it was a rough thing.

hayabusaboston · Feb 18, 2013

Re: HSC 2013 4U Marathon

Guys please explain, how do you find the motivation to actually write all this stuff up in laTex? I honestly see it and cant believe you guys write so much, doesnt it take up a shitload of time?? putting it into the laTex format?

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
if you just blindly draw the reciprocal then it's wrong though

The only restriction that I can see is x > 0 which comes via default.

am I missing something?

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Guys please explain, how do you find the motivation to actually write all this stuff up in laTex? I honestly see it and cant believe you guys write so much, doesnt it take up a shitload of time?? putting it into the laTex format?

If you learn latex then no it doesn't take much time at all.
If you are a fast typer, know latex well then the big blocks of latex I sometimes post don't take very long to write up.

asianese · Feb 18, 2013

Re: HSC 2013 4U Marathon

LaTeX just comes with practice (and it really helps if you are fast typer).

twinklegal19 · Feb 18, 2013

Sy123 said:
The only restriction that I can see is x > 0 which comes via default.

am I missing something?

my bad, for some reason I accidentally drew it as x>1 oops

you're right

RealiseNothing · Feb 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
The only restriction that I can see is x > 0 which comes via default.

am I missing something?

$x \neq 1$

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$x \neq 1$

Naturally at x=1 1/ln(x) is infinite so it is covered already by drawing the reciprocal

asianese · Feb 18, 2013

Re: HSC 2013 4U Marathon

It's the same curve I'm pretty sure.

x=/= 1 anyway on the x^y=e because if you convert it 1^y = e which is never possible. And also ln1 = 0 so y=1/0 undefined.

Sy123 · Feb 19, 2013

Re: HSC 2013 4U Marathon

$Find the area enclosed by the ellipses$

$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

$b\neq a$

Sy123 · Feb 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find the area enclosed by the ellipses$

$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

$b\neq a$

Find points of intersection:

$\left (\pm \frac{ab}{\sqrt{a^2+b^2}}, \ \pm \frac{ab}{\sqrt{a^2+b^2}} \right )$

$k=\frac{ab}{\sqrt{a^2+b^2}}$

$A=4 \cdot \left (b\int_{0}^k \sqrt{1-\frac{x^2}{a^2}} \ dx + a\int_k^b \sqrt{1-\frac{x^2}{b^2}} \ dx \right )$

Sy123 · Feb 21, 2013

Re: HSC 2013 4U Marathon

$2 points, P and Q lie on a circle with radius$ \ \ r \ \ $centred at the origin, O$

$$The tangents at P and Q intersect at the point R$ \ \ \textbf{they intersect at a right angle}$

$The tangent at P is called p, and the tangent at Q is called q$

$$The angle OP makes with the$ \ \ x$-axis$ \ \ \ $is$ \ \ \theta$

$Find the equations of the circles (in terms of$ \ \ \theta \ \ $) of radius$ \ \ r \ \ $, such that they are tangents to BOTH p and q$

Lieutenant_21 · Feb 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$The ellipse$ \ \ \ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$

$$Has 2 points$ \ \ P(a\cos \theta, b\sin \theta) \ \ $and$ \ \ Q(a\cos \alpha, b\sin \alpha)$

$The tangents at the points P and Q intersect at R at a right angle$

$$i) Prove that the locus of R is a circle with constant radius$ \ \ \ \fbox{1}$

$$ii) Hence find the equation of the locus of R as P and Q vary$ \ \ \ \fbox{2}$

If you are given parametric coordinates can you just ignore them and use the general tangent instead?
Make the equation of the general tangent a quadratic in terms of m then since they are perpendicular the product of roots = -1 and the locus is a circle with radius sqrt{a^2+b^2}.

Sy123 · Feb 22, 2013

Re: HSC 2013 4U Marathon

$The points$

$P(x_1,y_1)$
$P'(x_1, -y_1)$
$Q(-x_1,y_1$
$Q'(-x_1, -y_1)$

$$lie on the ellipse$ \ \ \ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$

$Tangents are constructed at each of these points, the tangents at:$

$P and Q intersect at R$

$P and P' intersect at P''$

$Q and Q' intersect at Q''$

$P' and Q' intersect at R'$

$Find the area of the quadrilateral P'' R Q'' R'$

OMGITzJustin · Feb 22, 2013

Re: HSC 2013 4U Marathon

integral of xlogx

Sy123 · Feb 22, 2013

Re: HSC 2013 4U Marathon

$$A polynomial$ \ \ S(y)=\sum_{k=0}^{2n}s_k y^k \ \ $has roots$ \ \ \alpha_1, \alpha_2, \dots , \alpha_{2n}$

$$Find a polynomial with roots$ \ \ \pm \alpha_1, \pm \alpha_2, \dots, \pm \alpha_{2n}$

HeroicPandas · Feb 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A polynomial$ \ \ S(y)=\sum_{k=0}^{2n}s_k y^k \ \ $has roots$ \ \ \alpha_1, \alpha_2, \dots , \alpha_{2n}$

$$Find a polynomial with roots$ \ \ \pm \alpha_1, \pm \alpha_2, \dots, \pm \alpha_{2n}$

$\sum_{k=0}^{2n}(-1)^k .(s_k)^2y^{2k}$ ?

Sy123 · Feb 22, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\sum_{k=0}^{2n}(-1)^k .(s_k)^2y^{2k}$ ?

Hmm not quite

HSC 2013 MX2 Marathon (archive) (11 Viewers)

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what is that?It is Cowpea

what is that?It is Cowpea

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what is that?It is Cowpea

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Heroic!

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