Question

A point moves along the curve y= x^3 -3x + 5 so that x= 1/2 square root t + 3 where t is the time. At what rate is y changing when t = 4?

Need some help!!!

Answer 1

dy/dt=dy/dx * dx/dt

but x=1/2 sqrt(t+3)
dx/dt=1/4 sqrt(1/(t+3)

dy/dx=3x^2-3

at t=4, x=sqrt(7) /2
and dy/dx=3*7/4
and dx/dt=1/4 (sqrt(1/7))

so dy/dt=3/16 *sqrt7 at t=4
check that algebra.