HSC 2015 MX2 Marathon (archive) (1 Viewer)

asdfghjklqwerty · Sep 27, 2015

Re: HSC 2015 4U Marathon

ahahhahah gg thanks

seanieg89 · Sep 28, 2015

Re: HSC 2015 4U Marathon

Haven't seen mechanics questions here for a while.

$Suppose a particle moves in simple harmonic motion with amplitude $A$ and angular frequency $\omega$.\\ \\ Find a function $F(x)$ such that $\int_a^b F(x)\, dx$ is equal to the proportion of time that the particle spends in the interval $a\leq x\leq b$ for any $a < b$, with both $a$ and $b$ smaller in absolute value than $A$.\\ \\ For an arbitrary interval $I=[a,b]$ inside $[-A,A]$, define $D(I)=\frac{1}{b-a}\int_a^b F(x)$. ($D(I)$ tells us how much time the particle spends in an interval relative to the size of the interval.) How large can $D(I)$ be?$

braintic · Sep 28, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Haven't seen mechanics questions here for a while.

$Suppose a particle moves in simple harmonic motion with amplitude $A$ and angular frequency $\omega$.\\ \\ Find a function $F(x)$ such that $\int_a^b F(x)\, dx$ is equal to the proportion of time that the particle spends in the interval $a\leq x\leq b$ for any $a < b$, with both $a$ and $b$ smaller in absolute value than $A$.\\ \\ For an arbitrary interval $I=[a,b]$ inside $[-A,A]$, define $D(I)=\frac{1}{b-a}\int_a^b F(x)$. ($D(I)$ tells us how much time the particle spends in an interval relative to the size of the interval.) How large can $D(I)$ be?$

I think this belongs in the advanced marathon. I'm not sure of the difficulty because I haven't tried it, but the wording and notation are going to confuse people, and it is not the sort of question you will see in an HSC exam.

Librah · Sep 28, 2015

Re: HSC 2015 4U Marathon

Solve for k, (8-3/k)(1-e^(-k))=1. Not sure where this belonged.

InteGrand · Sep 28, 2015

Re: HSC 2015 4U Marathon

Librah said:
Solve for k, (8-3/k)(1-e^(-k))=1. Not sure where this belonged.

Don't think you can find a closed form solution.

braintic · Sep 28, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Don't think you can find a closed form solution.

k = 0.536 516 425 697 171 ...

InteGrand · Sep 28, 2015

Re: HSC 2015 4U Marathon

braintic said:
k = 0.536 516 425 697 171 ...

How is that closed-form?

braintic · Sep 28, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
How is that closed-form?

It's not. I was just giving the only answer I could give.
(I replied to the wrong person)

InteGrand · Sep 28, 2015

Re: HSC 2015 4U Marathon

braintic said:
It's not. I was just giving the only answer I could give.
(I replied to the wrong person)

Oh OK.

braintic · Sep 28, 2015

Re: HSC 2015 4U Marathon

Prove that in any triangle ABC:

$\cos \frac {A-B} 2 > \sin \frac C 2$

and

$\sin \frac {A-B} 2 < \cos \frac C 2$

kawaiipotato · Sep 28, 2015

Re: HSC 2015 4U Marathon

braintic said:
Prove that in any triangle ABC:

$\cos \frac {A-B} 2 > \sin \frac C 2$

and

$\sin \frac {A-B} 2 < \cos \frac C 2$

$$ Consider, $ \cos \frac{A-B}{2} - \sin \frac{C}{2}$

$= \cos \frac{A-B}{2} - \sin \frac{180 - A - B}{2} ( $ since $ A+B+C = 180 ($ angle sum of a triangle $ ) )$

$= \cos \frac{A-B}{2} - \sin(90 - \frac{A+B}{2})$

$= \cos \frac{A-B}{2} - \cos \frac{A+B}{2}$

$= 2 \sin \frac{A}{2} \sin \frac{B}{2}$

$$ Since $ 0 < A,B < 180 ,$

$> 0$

...

$$ Consider, $ \sin \frac{A-B}{2} - \cos \frac{C}{2}$

$= \sin \frac{A-B}{2} - \cos(90 - \frac{A+B}{2})$

$= \sin \frac{A-B}{2} - \sin \frac{A+B}{2}$

$= - \cos \frac{A}{2} \cos \frac{B}{2}$

$$ Since, $ 0 < A,B < 180 ,$

$< 0$

asdfghjklqwerty · Sep 28, 2015

Re: HSC 2015 4U Marathon

find the volume of the solid of revolution formed when the are between the hyperbola x^2-2y^2=1 and the ellipse (x^2)/4 + y^2=1 (for x>0) is rotated about the x-axis.

Drsoccerball · Sep 28, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
find the volume of the solid of revolution formed when the are between the hyperbola x^2-2y^2=1 and the ellipse (x^2)/4 + y^2=1 (for x>0) is rotated about the x-axis.

Just from my head they're both at the origin and therefore the area bounded between the two regions can be expressed as 2 times one of them. You use seperate integrals one of the ellipse and one of the hyperbola and evaluate.

braintic · Sep 28, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
the area bounded between the two regions can be expressed as 2 times one of them

Not really relevant since we're doing a volume question.

You don't double for volume.

Edit: Oops - I thought we were rotating about the y-axis.

InteGrand · Sep 28, 2015

Re: HSC 2015 4U Marathon

It's only asking for x > 0 anyway, so we shouldn't double.

Drsoccerball · Sep 28, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
It's only asking for x > 0 anyway, so we shouldn't double.

Didnt see that

asdfghjklqwerty · Sep 29, 2015

Re: HSC 2015 4U Marathon

i was wondering if someone could do this q using shell method? i can do it via slicing but i couldnt get it via shells and idk why

Drsoccerball · Sep 29, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
View attachment 32471
i was wondering if someone could do this q using shell method? i can do it via slicing but i couldnt get it via shells and idk why

I don't know if cylindric shells would work for this. IS the answer 16 $\pi$ ?

flavurr · Sep 29, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I don't know if cylindric shells would work for this. IS the answer 16 $\pi$ ?

I got 7pi, care to show working out?

InteGrand · Sep 29, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
View attachment 32471
i was wondering if someone could do this q using shell method? i can do it via slicing but i couldnt get it via shells and idk why

$The diagram is incorrect, it shows the hyperbola's $x$-intercept being 3, when it should actually be $\sqrt{12}$, which is to the right of 3 (as $\sqrt{12}>3$). So the question should also say that the area is bounded by the $x$-axis, or change the hyperbola to having $9$ on the R.H.S. I haven't checked whether either of these result in their answer of $7\pi$.$

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