Answer:
The rate of change of the volume of the cylinder when the radius is 10 ft is
[tex]\frac{dV}{dt}=400\pi \:{\frac{ft^3}{min} }[/tex]
Step-by-step explanation:
This is a related rates problem. A related rates problem is a problem in which we know the rate of change of one of the quantities (the height of a grain) and want to find the rate of change of the other quantity (the volume of grain in the cylinder).
The volume of a cylinder is given by
[tex]V=\pi r^2 h[/tex]
V and h both vary with time so you can differentiate both sides with respect to time, t, to get
[tex]\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}[/tex]
Now use the fact that [tex]\frac{dh}{dt} = 4 \:{\frac{ft}{min} }[/tex] and [tex]r = 10 \:ft[/tex]
[tex]\frac{dV}{dt}=\pi (10)^2 \cdot 4\\\\\frac{dV}{dt}=100\cdot \:4\pi \\\\\frac{dV}{dt}=400\pi[/tex]
