Rates of Change

Question

The height of the cylinder is twice the radius, x cm,of its base. If x is decreasing at a rate of 0.1cm s^-1, find the rate of change of the volume of the cylinder when x=2.

Answer 1

Volume, V, of a cylinder = Πr²h = Π(x)²(2x), V = 2Πx³

 

Taking the derivative wrt time, t, we get: dV/dt = 6Πx²dx/dt

 

Substituting x =2, and dx/dt = -0.1, we get: dV/dt = 6Π(2)²(-0.1)

 

So dV/dt = -2.4Π cm³/s (about -7.54 cm³/s)