Another question i cant do :( (1 Viewer)

[Constip8edSkunk]

something from a past 3U assessment:

A curve has parametric eqations of x=at-sint and y=a-acost} (for 0<t<cost)
Prove dy/dx =cot(t/2) and d²y/dx² = -1/(4x) cosec²(t/2)

Hence show that y regarded as a function of x has a stationary point where t = pi and determine its nature.


i keep getting dy/dx=2/t sin²t/2
i hate anything with parametrics

 

[Newbie]

theres no parametrics in exam tomorrow

 

[Constip8edSkunk]

might be included in calculus of trig functions?... well it was in the '96 paper...

 

[Newbie]

oh shit

 

[wogboy]

A curve has parametric eqations of x=at-sint and y=a-acost

I think there's a mistake in the question, it should be x=at - asint, not x=at - sint.

To find dy/dx, first find dy/dt then find dx/dt. Then divide dy/dt by dx/dt to get dy/dx. (since (dy/dt)/(dx/dt) = dy/dx).

dy/dt = asint
dx/dt = a - acost (with the correction applied)

dy/dx = asint / (a - acost)
= sint / (1-cost) (cancelling off the a's)

now sint = 2tan(t/2) / (1 + tan^2(t/2))
and cost = (1 - tan^2(t/2)) / (1 + tan^2(t/2))
(by the tan@/2 rules)

so 1 - cost = 1 - (1 - tan^2(t/2)) / (1 + tan^2(t/2))
= 2tan^2(t/2) / (1 + tan^2(t/2))
so sint/(1 - cost) = 2tan(t/2) / 2tan^2(t/2) = 1/tan(t/2) = cot(t/2)
therefore dy/dx = cot(t/2)

therefore d^2(y)/dx^2 = d(dy/dx)/dx
= d(dy/dx)/dt * dt/dx
= (1/2)*cosec^2(t/2) / (dx/dt)
= (1/2)*cosec^2(t/2) / (a - acost)
= (1/2)*cosec^2(t/2) / y
= cosec^2(t/2) / 2y
(maybe another mistake here in the question?)

at x=pi, dy/dx = cot(pi/2) = 0 and also y = 2a
and d^2(y)/dx^2 = cosec^2(pi/2) / 2y
= 1/2y
= 1/4a

so for a>0 it is a min pt and for a<0 it is a max pt.

 

[Constip8edSkunk]

sint/(1 - cost) = 2tan^2(t/2) / 2tan(t/2) = 1/tan(t/2) = cot(t/2)

it should be 2tan^2(t/2) / 2tan(t/2) = tan(t/2) shouldnt it?


anyway i think ill skip this question as it seem to have too many errors, thanks for your help.

 

[wogboy]

it should be 2tan^2(t/2) / 2tan(t/2) = tan(t/2) shouldnt it?

Whoops I did a typo in the working out, but the answer is still cot(t/2). I just corrected it.