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

Answer:

252π  or 791.7 mm³/h

Step-by-step explanation:

The volume of a cylinder is given by

$V = \pi r^2h$

We desire to find the volume rate, that is,$\dfrac{dV}{dt}$

$\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$

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

$\dfrac{dV}{dt} = 2\pi rh$

At r = 12 mm and h = 1.5 mm,

$\dfrac{dV}{dt} = 2\pi\times12\times1.5\times7 = 252\pi = 791.7$