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insens350 [35]
3 years ago
6

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.?
Physics
1 answer:
kumpel [21]3 years ago
6 0

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}

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