Question

The radius of the base of a cylinder is increasing at a rate of 7 millimeters per hour. The height of the cylinder is fixed at 1.51, point, 5 millimeters. At a certain instant, the radius is 12 millimeters.
What is the rate of change of the volume of the cylinder at that instant (in cubic millimeters per hour)?

Answer 1

Answer:

The rate of change of the volume of the cylinder at that instant = $791.28\ mm^3/hr$

Step-by-step explanation:

Given:

Rate of increase of base of radius of base of cylinder = 7 mm/hr

Height of cylinder = 1.5 mm

Radius at a certain instant = 12 mm

To find rate of change of volume of cylinder at that instant.

Solution:

Let $r$ represent radius of base of cylinder at any instant.

Rate of increase of base of radius of base of cylinder can be given as:

$\frac{dr}{dt}=7\ mm/hr$

Volume of cylinder is given by:

$V=\pi\ r^2h$

Finding derivative of the Volume with respect to time.

$\frac{dV}{dt}=\pi\ h\ 2r\frac{dr}{dt}$

Plugging in the values given:

$\frac{dV}{dt}=\pi\ (1.5)\ 2(12)(7)$

$\frac{dV}{dt}=252\pi$

Using $\pi=3.14$

$\frac{dV}{dt}=252(3.14)$

$\frac{dV}{dt}=791.28\ mm^3/hr$ (Answer)

Thus rate of change of the volume of the cylinder at that instant = $791.28\ mm^3/hr$