HSC 2015 MX1 Marathon (archive) (5 Viewers)

InteGrand · Sep 27, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$F'(x) = cosx $ as $ \frac{cos(h)-1}{h} =0$

$$ Because $ cos(h)=1-2sin^2(\frac{h}{2})=0$

$Except the denominator of $\frac{\cos(h)-1}{h}$ is 0 too if $h$ is 0, so we can't use this argument.$

Drsoccerball · Sep 27, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$Except the denominator of $\frac{\cos(h)-1}{h}$ is 0 too if $h$ is 0, so we can't use this argument.$

$\frac{cos(h) -1}{h} = \frac{-2sin^2(\frac{h}{2})}{h} $ As I mentioned $$

$lim_{h \rightarrow 0} -2sin(\frac{h}{2}) = 0$

kawaiipotato · Sep 28, 2015

Re: HSC 2015 3U Marathon

$y = x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + ... } } }$

$Find $ \frac{dy}{dx} $ in terms of y and x$

braintic · Sep 28, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$y = x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + ... } } }$

$Find $ \frac{dy}{dx} $ in terms of y and x$

AND .... show that this curve is a parabola, and find its focal length and the coordinates of its vertex.

InteGrand · Sep 28, 2015

Re: HSC 2015 3U Marathon

braintic said:
AND .... show that this curve is a parabola, and find its focal length and the coordinates of its vertex.

Are you sure it's a parabola? If we play around with that equation, we seem to end up with a polynomial equation in x and y of degree 4, not 2. And I think the curve would only be a portion of an algebraic curve rather than the whole curve, due to restrictions imposed by the square root.

braintic · Sep 28, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Are you sure it's a parabola? If we play around with that equation, we seem to end up with a polynomial equation in x and y of degree 4, not 2. And I think the curve would only be a portion of an algebraic curve rather than the whole curve, due to restrictions imposed by the square root.

Yep ... that's what happens when I try to do it in my head.

Drsoccerball · Sep 28, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$y = x + \sqrt{y + \sqrt{x + \sqrt{y + \sqrt{x + ... } } }$

$Find $ \frac{dy}{dx} $ in terms of y and x$

You get :

$\frac{dy}{dx} = \frac{4(y-x)^3 -4y(y-x)}{-2y+2x-4y(y-x)+4(y-x)^3 +2y-1}$

leehuan · Sep 28, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
You get :

$\frac{dy}{dx} = \frac{4(y-x)^3 -4y(y-x)}{-2y+2x-4y(y-x)+4(y-x)^3 +2y-1}$

What 3U trick do you use to get that

Drsoccerball · Sep 28, 2015

Re: HSC 2015 3U Marathon

leehuan said:
What 3U trick do you use to get that

Its actually year 9, I posted an integral like this once

From the equation square both sides: take y to the other side then square again and you're left with :

$((y-x)^2-y)^2 =y$ And then differentiate.

leehuan · Sep 28, 2015

Re: HSC 2015 3U Marathon

Oh wow... algebra manipulation. I was not taught that in Year 9 LOL

$y=x+\sqrt { y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } } \\ y-x=\sqrt { y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } } \\ { \left( y-x \right) }^{ 2 }=y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } \\ { \left( y-x \right) }^{ 2 }-y=\sqrt { x+\sqrt { y+\sqrt { x+... } } } \\ { \left( { \left( y-x \right) }^{ 2 }-y \right) }^{ 2 }=x+\sqrt { y+\sqrt { x+... } } \\ \therefore { \left( { \left( y-x \right) }^{ 2 }-y \right) }^{ 2 }=y$

Still, I'd argue implicit differentiation is extreme since it's not in the 3U course...

Drsoccerball · Sep 28, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Oh wow... algebra manipulation. I was not taught that in Year 9 LOL

$y=x+\sqrt { y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } } \\ y-x=\sqrt { y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } } \\ { \left( y-x \right) }^{ 2 }=y+\sqrt { x+\sqrt { y+\sqrt { x+... } } } \\ { \left( y-x \right) }^{ 2 }-y=\sqrt { x+\sqrt { y+\sqrt { x+... } } } \\ { \left( { \left( y-x \right) }^{ 2 }-y \right) }^{ 2 }=x+\sqrt { y+\sqrt { x+... } } \\ \therefore { \left( { \left( y-x \right) }^{ 2 }-y \right) }^{ 2 }=y$

Still, I'd argue implicit differentiation is extreme since it's not in the 3U course...

I asked my extension 2 teacher the integral for something similar he got the answer straight away app its common knowledge.
Its similar to evaluation $x= 1+\frac{1}{1+\frac{1}{1+...}}$

leehuan · Sep 28, 2015

Re: HSC 2015 3U Marathon

Is this how?

$x=1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ x-1=\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ \frac { 1 }{ x-1 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } \\ \therefore \frac { 1 }{ x-1 } =x\\ { x }^{ 2 }-x-1=0$

Drsoccerball · Sep 28, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Is this how?

$x=1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ x-1=\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ \frac { 1 }{ x-1 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } \\ \therefore \frac { 1 }{ x-1 } =x\\ { x }^{ 2 }-x-1=0$

yup

InteGrand · Sep 28, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Is this how?

$x=1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ x-1=\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ \frac { 1 }{ x-1 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } \\ \therefore \frac { 1 }{ x-1 } =x\\ { x }^{ 2 }-x-1=0$

$Or you can see straightaway that $x=1+\frac{1}{x}$ ($x>0$). The solution to this is $x=\varphi=\frac{1+\sqrt{5}}{2}$ (the famous Golden Ratio). Another similar thing for $\varphi$ is $\varphi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$.$

leehuan · Sep 28, 2015

Re: HSC 2015 3U Marathon

Honestly the golden ratio appears everywhere... even in situations I would never have thought of

braintic · Sep 28, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Honestly the golden ratio appears everywhere... even in situations I would never have thought of

Including in Pascal's triangle.

davidgoes4wce · Sep 28, 2015

Re: HSC 2015 3U Marathon

braintic · Sep 28, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:

So are you asking for first principles, or induction?

rand_althor · Sep 28, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:

$\begin{align*}f(x)&=x^n,\,f(x+h)=(x+h)^n \\f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\&=\lim_{h\to0}\frac{(x+h)^n-x^n}{h} \\&=\lim_{h\to0}\frac{{\color{Blue} \binom{n}{0}x^nh^0} + \binom{n}{1}x^{n-1}h^1 + \binom{n}{2}x^{n-2}h^2 +...+\binom{n}{n}x^0h^n{-\color{Blue} x^n}}{h} \\&=\lim_{h\to0}\binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2}h^1 + \binom{n}{3}x^{n-3}h^2 +...+\binom{n}{n}x^0h^{n-1} \\&=\binom{n}{1}x^{n-1} \\&=nx^{n-1}\end{align*}$

davidgoes4wce · Sep 28, 2015

Re: HSC 2015 3U Marathon

I understand that by using first principles, the binomial theorem would come into play.

I have not seen how it is done by induction but would be interested to see.

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