Question

A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 inches per minute while the height of the cone is decreasing at a rate of 1/3 inches per minute. At the moment the radius of the cone is 2 in and the height of the cone is 1/3 in, find the rate of change of the volume of the cone. Simplify your answer and be sure to include units with your answer.

Answer 1

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

$\frac{dr}{dt} =\frac{1}{2}$ --- rate of the radius

$\frac{dh}{dt} =-\frac{1}{3}$ --- rate of the height

$r = 2$

$h = \frac{1}{3}$

Required

Determine the rate of change of the cone volume

The volume of a cone is:

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

Differentiate with respect to time (t)

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

Substitute values for the known variables

$\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)$

$\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{88}{63}$

$\frac{dV}{dt} =-1.396in^3/min$

The volume is decreasing at the rate of 1.396 cubic inches per minute