Answer:
Volume of the cone is increasing at the rate
.
Step-by-step explanation:
Given: The radius of a right circular cone is increasing at a rate of
in/s while its height is decreasing at a rate of
in/s.
To find: The rate at which volume of the cone changing when the radius is
in. and the height is
in.
Solution:
We have,
,
,
, ![h=136\:\text{in}](https://tex.z-dn.net/?f=h%3D136%5C%3A%5Ctext%7Bin%7D)
Now, let
be the volume of the cone.
So,
Differentiate with respect to
.
![\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5Cleft%20%5B%20r%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%2Bh%5Cleft%20%28%202r%20%5Cright%20%29%5Cfrac%7Bdr%7D%7Bdt%7D%20%5Cright%20%5D)
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 ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Cleft%20%5B%20%5Cleft%20%28%20134%20%5Cright%20%29%5E2%5Cleft%20%28%20-2.2%20%5Cright%20%29%2B%5Cleft%20%28%20%20136%5Cright%20%29%5Cleft%20%28%202%20%5Cright%20%29%5Cleft%20%28%20134%20%5Cright%20%29%5Cleft%20%28%201.9%20%5Cright%20%29%20%5Cright%20%5D)
![\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%5Cleft%20%5B%2029748%20%5Cright%20%5D)
![\frac{dv}{dt} =9916\pi \frac{in^3}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D9916%5Cpi%20%5Cfrac%7Bin%5E3%7D%7Bs%7D)
Hence, the volume of the cone is increasing at the rate
.