Question

The radius of the base of a cylinder is increasing at a rate of 777 millimeters per hour. The height of the cylinder is fixed at 1.51.51, point, 5 millimeters. At a certain instant, the radius is 121212 millimeters. What is the rate of change of the volume of the cylinder at that instant (in cubic millimeters per hour)

Answer 1

252π or 791.7 mm³/h

Explanation:

The volume of a cylinder is given by

`V = \pi r^2h`

We desire to find the volume rate, that is,
`(dV)/(dt)`

`(dV)/(dt) = (dV)/(dr)\cdot(dr)/(dt)`

dr/dt is the rate of change of the radius which is 7 mm/h.

dV/dr is derived by differentiating the volume equation, yielding

`(dV)/(dt) = 2\pi rh`

At r = 12 mm and h = 1.5 mm,

`(dV)/(dt) = 2\pi*12*1.5*7 = 252\pi = 791.7`