Question

The radius of a right circular cone is increasing at a rate of 2 inches per second and its height is decreasing at a rate of 4 inches per second. At what rate is the volume of the cone changing when the radius is 10 inches and the height is 20 inches

Answer 1

Answer:

The volume of cone changing at rate $\frac{400}{3}\pi in^3/s$

Step-by-step explanation:

Let r be the radius of cone and h be the height of cone

$\frac{dr}{dt}=2 in/s$

$\frac{dh}{dt}=-4 in/s$

We have to find the rate at which the volume of cone changing when r=10 in and h=20 in

Volume of cone

$V=\frac{1}{3}\pi r^2 h$

Differentiate w.r.t t

$\frac{dV}{dt}=\frac{1}{3}\pi (2rh\frac{dr}{dt}+r^2\frac{dh}{dt})$

Substitute the values

$\frac{dV}{dt}=\frac{1}{3}\pi(2\times 10\times 20\times 2+(10)^2\times (-4))$

$\frac{dV}{dt}=\frac{1}{3}\pi(400)=\frac{400}{3}\pi in^3/s$

Hence, the volume of cone changing at rate $\frac{400}{3}\pi in^3/s$