Question

if x=t^2-1 and y=t^4-2t^3, then when t=1 what's d^2y/dx^2? if x=t^2-1 and y=t^4-2t^3, then when t=1 what's d^2y/dx^2? @MIT 18.01 Single …

Answer 1

dx/dt = 2t and dy/dt = 4t^3 - 6t^2

thus dy/dx = 4t^3 - 6t^2 / 2t = 2t^2 - 3t

Answer 2

d^2y/dx^2 = d/dt(dy/dx) * dt/dx
= d/dt (2t^2 - 3t) * dt/dx
= ( 4t-3) * 1 / 2t = (4t-3)/2t
when t = 1 d^2y/dx^2 = 1/2