Question

The radius of a right circular cylinder is increasing at the rate of 6 in./s, while the height is decreasing at the rateof 3 in./s. At what rate is the volume of the cylinder changing when the radius is 5 in. and the height is 11 in.?​

Answer 1

The volume of a cylinder with radius $r$ and height $h$ is

$V=\pi r^2h$

Differentiate both sides with respect to time:

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dr}{\mathrm dt}+\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

We're given that

$\dfrac{\mathrm dr}{\mathrm dt}=6\dfrac{\rm in}{\rm s}$

$\dfrac{\mathrm dh}{\mathrm dt}=-3\dfrac{\rm in}{\rm s}$

so that at the point when $r=5\,\rm in$ and $h=11\,\rm in$, the volume is undergoing a total change of

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi(5\,\mathrm{in})(11\,\mathrm{in})\left(6\dfrac{\rm in}{\rm s}\right)+\pi(5\,\mathrm{in})^2\left(-3\dfrac{\rm in}{\rm s}\right)$

$\boxed{\dfrac{\mathrm dV}{\mathrm dt}=585\pi\dfrac{\mathrm{in}^3}{\rm s}}$

Answer 2

The rate at which the volume of the cylinder changes is 480πin³/s

The formula for calculating the volume of a cylinder is expressed as:

$V=\pi r^2h$

The change with respect to time is expressed as:

$\frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt}+ \frac{dV}{dh} \frac{dh}{dt}$

$\frac{dV}{dt} = 2\pi rh \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}$

Given the following parameters:

$\frac{dr}{dt} =6in/s\\\frac{dh}{dt} =3in/s\\r = 5in\\h=11in$

Substitute the given values into the formula:

$\frac{dV}{dt} = 2\pi (5)(11) (3) + \pi (5)^2 (6)\\\frac{dV}{dt}=2 \pi (165) + 150 \pi\\\frac{dV}{dt}=330 \pi + 150 \pi\\\frac{dV}{dt} = 480 \pi in^3/s$

Hence the rate at which the volume of the cylinder changes is 480πin³/s

Learn more on the volume of cylinder here: https://brainly.com/question/12748872