parametric equations

Question

Evaluate d^2y/dx^2 of the parametric equation x=cos^2t, y=sin^2t, 0<=t<=180.

Answer 1

Note :

Parametric differentiation :

if x = x(t) and y = y(t) then,

dy / dx = [dy  /dt] / [dx / dt], and

d^2y / dx^2 = d /dx(dy / dx) = [ d /dt(dy / dx) ] / [ dx / dt], provided dx / dt is not equals to zero.

The parametric euation is x = cos^2 (t), y = sin^2 (t), 0 <= t <= 180.

dx / dt = 2 cos t(- sin t) = - 2sin tcos t.

dy / dt = 2sin t(cos t) = 2sin tcos t.

dy / dx = [dy  /dt] / [dx / dt]

           = 2sin tcos t / - 2sin tcos t

dy / dx = - 1.

d^2y / dx^2 = d /dx(dy / dx) = [ d /dt(dy / dx) ] / [ dx / dt]

                 = d /dx(- 1)

d^2y / dx^2 = 0.