a curve is defined by parametric functions x(t) and y(t), where dxdt=3t2 and dydt=t7−2t4. for what positive value of t is d2ydx2=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

Maybe there’s a bug, mine is doing the same thing