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
