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.?

Respuesta :

Answer:

[tex]11304 \frac{in^{3}}{s}[/tex]

Explanation:

r = radius of right circular cone = 150 in

h = height of right circular cone = 144 in

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

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

Volume of the right circular cone is given as

[tex]V = \frac{\pi r^{2}h}{3}[/tex]

Taking derivative both side relative to "t"

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

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

[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]