The radius of a cone is increasing at a rate of 333 centimeters per second and the height of the cone is decreasing at a rate of 444 centimeters per second. At a certain instant, the radius is 888 centimeters and the height is 101010 centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per second) ?

Question

Milo Montgomery

Physics

14 March, 01:50

The radius of a cone is increasing at a rate of 333 centimeters per second and the height of the cone is decreasing at a rate of 444 centimeters per second. At a certain instant, the radius is 888 centimeters and the height is 101010 centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per second) ?

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Answers (1)

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Benitez · Answer 1

volume inreases at a rate of 6.22*10^10 cm^3/s

Explanation:

dr/dt = + 333 cm/s, dh/dt = - 444 cm/s, r=888cm, h = 101010 cm

volume of cone is given by formula

v = (1/3) π r^2 h

differentiating with respect to t

dv/dt = (1/3) π {r^2 dh/dt + h * 2r dr/dt)

dv/dt = (1/3) π { 888^2 (-444) + 2*888*101010*333)

dv/dt = 6.22*10^10 cm^3/s