Question

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.1 in/s. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 175 in.

Answer 1

Given:

$\dfrac{dr}{dt}=1.4\text{ in/s}$

$\dfrac{dh}{dt}=-2.1\text{ in/s}$

To find:

The rate of change in volume at $r=120\text{ in. and }175\text{ in.}$

Solution:

We know that, volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$

Differentiate with respect to t.

$\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]$

Substitute the given values.

$\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]$

$\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]$

$\dfrac{dV}{dt}=\dfrac{22}{21}\times 28560$

$\dfrac{dV}{dt}=29920$

Therefore, the volume of decreased by 29920 cubic inches per second.