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:

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

Step-by-step explanation:

Given

Shape: Cone

[tex]\frac{dr}{dt} =\frac{1}{2}[/tex] --- rate of the radius

[tex]\frac{dh}{dt} =-\frac{1}{3}[/tex] --- rate of the height

[tex]r = 2[/tex]

[tex]h = \frac{1}{3}[/tex]

Required

Determine the rate of change of the cone volume

The volume of a cone is:

[tex]V = \frac{\pi}{3}r^2h[/tex]

Differentiate with respect to time (t)

[tex]\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})[/tex]

Substitute values for the known variables

[tex]\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})[/tex]

[tex]\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})[/tex]

[tex]\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))[/tex]

[tex]\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)[/tex]

[tex]\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}[/tex]

[tex]\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}[/tex]

[tex]\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}[/tex]

[tex]\frac{dV}{dt} = -\frac{88}{63}[/tex]

[tex]\frac{dV}{dt} =-1.396in^3/min[/tex]

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

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