Question

The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At what rate is the volume of the cone changing when the radius is 134 in. and the height is 136 in.

Answer 1

Answer:

Volume of the cone is increasing at the rate $9916\pi \frac{in^3}{s}$.

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of $1.9$ in/s while its height is decreasing at a rate of $2.2$ in/s.

To find: The rate at which volume of the cone changing when the radius is $134$ in. and the height is $136$ in.

Solution:

We have,

$\frac{dr}{dt} =1.9 \:\text{in/s}$, $\frac{dh}{dt}=-2.2\:\text{in/s}$, $r=134 \:\text{in}$, $h=136\:\text{in}$

Now, let $V$ be the volume of the cone.

So, $V=\frac{1}{3}\pi r^{2}h$

Differentiate with respect to $t$.

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

Now, on substituting the values, we get

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left ( 136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]$

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ -39503.2+69251.2 \right ]$  

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]$

$\frac{dv}{dt} =9916\pi \frac{in^3}{s}$

Hence, the volume of the cone is increasing at the rate $9916\pi \frac{in^3}{s}$.