Maths 1996 P1 Q7 c (ii)

Tomlowe

Could someone give me a hand with this question? it just face-raped me, i was able to get x in terms of r and theta and all that but the differentiation is killing me... do i actually have to chain rule an inverse sine? i doubt it. It's the one about an astronaut..

MathsManiac

You don't need to diff the inverse sine.

Find dx/d(theta) instead, which is -rcos(theta)/sin^2(theta).

That means that d(theta)/dx is 1 over that, which is -sin^2(theta)/rcos(theta).

d(theta)/dt = [d(theta)/dx] * [dx/dt], which gives -sin^2(theta)/[5cos(theta)]

When x=r, then sin(theta) = 1/2 and theta = 30 and cos(theta) = sqrt(3)/2, so d(theta)/dt evaluates to -1/10sqrt(3) km/hr.

Captain Smiggy

Papers beyond '00 are from the old syllabus though aren't they? You(that is to say we) don't need to know them so much do you? Especially since they are absolute wankas of questions

MathsManiac

Captain Smiggy wrote:

Papers beyond '00 are from the old syllabus though aren't they? You(that is to say we) don't need to know them so much do you? Especially since they are absolute wankas of questions

Nope!

As mentioned on another thread, 94 is the first year of the present syllabus.

mp3guy

I'm keeping away from those old questions, they're way harder than the ones in the last 6 years, and there's no solutions.