Answer:
0.13 ft/min
Step-by-step explanation:
We are given that
[tex]\frac{dV}{dt}=15ft^3/min[/tex]
We have to find the increasing rate of change of height of pile when the pile is 12 ft high.
Let d be the diameter of pile
Height of pile=h
d=h
Radius of pile,r=[tex]\frac{d}{2}=\frac{h}{2}[/tex]
Volume of pile=[tex]\frac{1}{3}\pi r^2 h=\frac{1}{12}\pi h^3[/tex]
[tex]\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}[/tex]
h=12 ft
Substitute the values
[tex]15=\frac{1}{4}\pi(12)^2\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}=\frac{15\times 4}{\pi(12)^2}[/tex]
[tex]\frac{dh}{dt}=0.13ft/min[/tex]
