Answer:
Volume of the pile is increasing at the rate = [tex]54\pi\ in^{3}/min[/tex]
Step-by-step explanation:
Given:
The height of the pile is always twice the radius of the base.
Radius of the conical pile r = 6 inches.
When r = 6
The increasing rate of the radius of pile [tex]\frac{dr}{dt} = 0.75\ in/min[/tex]
We need to find the volume of the pile when the radius of the base is 6 inches.
Solution:
We know the volume of the cone.
[tex]V = \frac{\pi }{3}r^{2} h[/tex]
Where:
r = radius of the cone
h = Height of the cone
Substitute h = 2r in the above formula because the height of the pile is always twice the radius of the base.
[tex]V = \frac{\pi }{3}r^{2} (2r)[/tex]
[tex]V = \frac{2}{3}\pi r^{3}[/tex]
Now, differentiate both side of the equation with respect to t.
[tex]\frac{dV}{dt} = \frac{d}{dt}( \frac{2\pi r^{3} }{3} )[/tex]
[tex]\frac{dV}{dt} = \frac{2\pi }{3}. \frac{dr^{3} }{dt}[/tex]
[tex]\frac{dV}{dt} = \frac{2\pi }{3}. 3r^{2}\frac{dr }{dt}[/tex]
[tex]\frac{dV}{dt} = \frac{6\pi r^{2}}{3}.\frac{dr }{dt}[/tex]
Substitute r = 6 and [tex]\frac{dr}{dt} = 0.75\ in/min[/tex].
[tex]\frac{dV}{dt} = \frac{6\pi (6^{2})}{3}\times 0.75[/tex]
[tex]\frac{dV}{dt} = 2\pi\times 36\times 0.75[/tex]
[tex]\frac{dV}{dt} = 54\pi\ in^{3}/min[/tex]
Therefore, volume of the pile is increasing at [tex]54\pi\ in^{3}/min[/tex] in a 6 inches radius of the base.
