HSC 2016 MX1 Marathon (archive) (1 Viewer)

KingOfActing · Apr 5, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Thought about this question just now. Not sure if it can be done.

$$Let ${L}_{n}\left(x\right)=\underbrace{\ln \left({\ln\left({\ln{\left( \dots \ln{x}\right)}}\right)}\right)}_{n\text{ applications of log}}\\ $Use mathematical induction to show that $\\ \frac{d}{dx} {L}_{n}\left(x\right)=\frac{1}{{ L }_{ 0 }\left( x \right) { L }_{ 1 }\left( x \right) { L }_{ 2 }\left( x \right) { L }_{ 3 }\left( x \right) \dots { L }_{ n-1 }\left( x \right)} \\ $ for integers $n\ge 1$

$$Note that we write $L_0(x) = x$.$\\$Base case: $ \frac{d}{dx}L_1(x) = \frac{d}{dx}\ln{x} = \frac{1}{x} = \frac{1}{L_0(x)} \\$Assume the result for $n = k$, then for $n=k+1$ we get:$\\ \frac{d}{dx}L_{k+1}(x) = \frac{d}{dx} \ln(L_k(x)) = \frac{L'_k(x)}{L_k(x)} = \frac{1}{L_0(x)L_1(x)\cdots L_k(x)}\\$Hence the result is true for all $ n \geq 1$

leehuan · Apr 5, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$$Note that we write $L_0(x) = x$.$\\$Base case: $ \frac{d}{dx}L_1(x) = \frac{d}{dx}\ln{x} = \frac{1}{x} = \frac{1}{L_0(x)} \\$Assume the result for $n = k$, then for $n=k+1$ we get:$\\ \frac{d}{dx}L_{k+1}(x) = \frac{d}{dx} \ln(L_k(x)) = \frac{L'_k(x)}{L_k(x)} = \frac{1}{L_0(x)L_1(x)\cdots L_k(x)}\\$Hence the result is true for all $ n \geq 1$

Basically what I was hoping would work.

Paradoxica · Apr 9, 2016

Re: HSC 2016 3U Marathon

$\int \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}$d$x$

KingOfActing · Apr 10, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$\int \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}$d$x$

$\begin{align*}\int \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}} \,dx &= 2\frac{\frac{\pi}{2}}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}\,dx\\&=2 \int \frac{\sin^{-1}{x}+\cos^{-1}{x}}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}\,dx \\ &= 2\int \frac{1}{\cos^{-1}{x}\sqrt{1-x^2}} + \frac{1}{\sin^{-1}{x}\sqrt{1-x^2}}\,dx\\ &= 2(-\ln{|\cos^{-1}{x}|} + \ln{|\sin^{-1}{x}|}) + C \\ &= 2\ln{\left(\frac{|\sin^{-1}{x}|}{\cos^{-1}{x}}\right)} +C \end{align*}$

Paradoxica · Apr 10, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$\begin{align*}\int \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}} \,dx &= 2\frac{\frac{\pi}{2}}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}\,dx\\&=2 \int \frac{\sin^{-1}{x}+\cos^{-1}{x}}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}\,dx \\ &= 2\int \frac{1}{\cos^{-1}{x}\sqrt{1-x^2}} + \frac{1}{\sin^{-1}{x}\sqrt{1-x^2}}\,dx\\ &= 2(-\ln{|\cos^{-1}{x}|} + \ln{|\sin^{-1}{x}|}) + C \\ &= 2\ln{\left(\frac{|\sin^{-1}{x}|}{\cos^{-1}{x}}\right)} +C \end{align*}$

And hence, evaluate

$\int_{-1}^1 \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}$d$x$

KingOfActing · Apr 10, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
And hence, evaluate

$\int_{-1}^1 \frac{\pi}{\sqrt{1-x^2}\sin^{-1}{x}\cos^{-1}{x}}$d$x$

Trick question, the integral diverges

turntaker · Apr 10, 2016

Re: HSC 2016 3U Marathon

if you can do this you will get a rep from me

Nailgun · Apr 10, 2016

Re: HSC 2016 3U Marathon

turntaker said:
if you can do this you will get a rep from me

e^x + C

InteGrand · Apr 10, 2016

Re: HSC 2016 3U Marathon

turntaker said:
if you can do this you will get a rep from me

Most of us can do this; will we all get a rep from you?

porcupinetree · Apr 10, 2016

Re: HSC 2016 3U Marathon

turntaker said:
if you can do this you will get a rep from me

$\frac{e^{x+1}}{x+1} + C$

hedgehog_7 · Apr 10, 2016

Re: HSC 2016 3U Marathon

Two particles A and B are moving along a horizontal line with their distances X(A) and X(B) to the right of the origin O at time t given by
X(A) = 4te^(-t) and X(B) = -4(t^2)(e^-t). The particles are joined by a piece of elastic, whose midpoint M has position X(M) at time t.

When are A and B furthest apart?
Ans: t = 0.5(1+root 5)

ive tried adding X(A) and X(B) and then differentiating it to find the max tp and i dont seem to get the answer. Thx

leehuan · Apr 10, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
Two particles A and B are moving along a horizontal line with their distances X(A) and X(B) to the right of the origin O at time t given by
X(A) = 4te^(-t) and X(B) = -4(t^2)(e^-t). The particles are joined by a piece of elastic, whose midpoint M has position X(M) at time t.

When are A and B furthest apart?
Ans: t = 0.5(1+root 5)

ive tried adding X(A) and X(B) and then differentiating it to find the max tp and i dont seem to get the answer. Thx

I'm tired so I'm not too sure but don't you want to subtract X(B) from X(A) instead of adding them? If you want to find when they're the furthest apart then wouldn't you be wanting to maximise the difference between their distances?

InteGrand · Apr 10, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
Two particles A and B are moving along a horizontal line with their distances X(A) and X(B) to the right of the origin O at time t given by
X(A) = 4te^(-t) and X(B) = -4(t^2)(e^-t). The particles are joined by a piece of elastic, whose midpoint M has position X(M) at time t.

When are A and B furthest apart?
Ans: t = 0.5(1+root 5)

ive tried adding X(A) and X(B) and then differentiating it to find the max tp and i dont seem to get the answer. Thx

Maximise the squared difference between the two particles' distance functions.

Nailgun · Apr 10, 2016

Re: HSC 2016 3U Marathon

porcupinetree said:
$\frac{e^{x+1}}{x+1} + C$

i dont get it lol

InteGrand · Apr 10, 2016

Re: HSC 2016 3U Marathon

Nailgun said:
i dont get it lol

Troll power rule. (It'd be the answer if the integral was de instead of dx, where e is a variable rather than the number e, and x is a parameter independent of this variable e, and assuming x is not equal to -1.)

Nailgun · Apr 10, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Troll power rule. (It'd be the answer if the integral was de instead of dx, where e is a variable rather than the number e, and x is a parameter independent of this variable e, and assuming x is not equal to -1.)

OHHHH lol
i thought it was some smartly derived equivalent expression
then i tried checking it by equating it to e^x and it didn't work lol

hedgehog_7 · Apr 10, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Maximise the squared difference between the two particles' distance functions.

wait so ive tried what leehuan said and i subtracted it and i end up with t = -0.5(1+ root 5) as the max value instead of t = 0.5(1 + root5) ive doubled checked and i dont see any errors. Thoughts?

InteGrand · Apr 10, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
wait so ive tried what leehuan said and i subtracted it and i end up with t = -0.5(1+ root 5) as the max value instead of t = 0.5(1 + root5) ive doubled checked and i dont see any errors. Thoughts?

What was the function you tried to maximise?

hedgehog_7 · Apr 10, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
What was the function you tried to maximise?

4(e^-t) (-t^2 + t +1)

InteGrand · Apr 10, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
4(e^-t) (-t^2 + t +1)

This is not the difference between the two functions you gave.

The two position functions were:

x_A(t) = 4te^(-t) and x_B(t) = -4(t^2)e^(-t).

The squared-difference is: s(t) := (x_A(t) – x_B(t))².

Try maximising that. (The distance isn't simply xA – xB, it is |xA – xB|. Since absolute values are less nice to work with than squared-distance, and optimising the squared-distance will be equivalent to optimised the distance, it is easier to work with the squared-distance instead.)

Actually, in this case, it is easy to see that xA is always greater than or equal to xA is non-negative always, whilst xB is non-positive always. So in this case, the distance actually is xA – xB, and we can just optimise this. What I said above with the squared-distance stuff was for general distance functions, which may change which one is greater than the other over time, in which case the squared-distance trick is handy.

So for this one, we need to optimise the function

f(t) := x_A(t) – x_B(t)

= 4te^(-t) – [-4(t^2)e^(-t)]

= 4te^(-t) [1+t].

I think you posted the distance functions wrong the first time?

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