Question

The radius of a right circular cone is increasing at a rate of 1.5 in/s while its height is decreasing at a rate of 2.4 in/s. At what rate is the volume of the cone changing when the radius is 150 in. and the height is 144 in.?

Answer 1

Answer:

$11304 \frac{in^{3}}{s}$

Explanation:

r = radius of right circular cone = 150 in

h = height of right circular cone = 144 in

$\frac{dr}{dt}$ = rate at which radius increase = 1.5 in/s

$\frac{dh}{dt}$ = rate at which height decrease = - 2.4 in/s

Volume of the right circular cone is given as

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

Taking derivative both side relative to "t"

$\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})$

$\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))$

$\frac{dV}{dt} = 11304 \frac{in^{3}}{s}$