Question

The height of a grain of a cylindrical silo is is increasing at a constant rate of 4 feet per minute At what rate is the volume of grain in the cylinder if the radius of the silo is 10 feet?

Answer 1

Volume of cylinder

`V = \pi h r^2`

radius is given as 10

`V = \pi h (10)^2 = 100\pi h`

Differentiate with respect to 't'

`(dV)/(dt) = 100 \pi (dh)/(dt)`

dh/dt = 4


`(dV)/(dt) = 100 \pi (4) = 400 \pi`

Answer 2

The rate of change of the volume of the cylinder when the radius is 10 ft is

`(dV)/(dt)=400\pi \:{(ft^3)/(min) }`

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

`V=\pi r^2 h`

V and h both vary with time so you can differentiate both sides with respect to time, t, to get

`(dV)/(dt)=\pi r^2 (dh)/(dt)`

Now use the fact that
`(dh)/(dt) = 4 \:{(ft)/(min) }` and
`r = 10 \:ft`

`(dV)/(dt)=\pi (10)^2 \cdot 4\\\\(dV)/(dt)=100\cdot \:4\pi \\\\(dV)/(dt)=400\pi`