Please help with math question

Question

A curve is defined by parametric functions x(t) and y(t), where dx/dt = 3t^2 and dy/dt = t^7-2t^4. For what positive value of t is d^2y/dx^2 = 1?

a) 1.202

b) 1.149

c) 1.079

d) 1.525

Answer 1

The correct answer is d) t = 1.525

The route to this result involves a number of steps.

First dy/dx = is easily obtained from the ratio of dy/dt and dx/dt as

dy/dt = (1/3) t⁵ - (2/3) t² define Q as dy/dt

What is needed is dQ/dx (the second derivative of y wrt x)

dQ/dx = (dQ/dt) ( dt/dx) but dt/dx = 1/( 3 t²) so

dQ/dx = (5/9) t² -(4/3) /t

Setting this equal to 1 results in a cubic equation with solution t = 1.525