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The radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second. The height of the cylinder is fixed at 2.5, point, 5 kilometers. At a certain instant, the radius is 40 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)

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nuhulawal20

Answer:

the rate of change of the volume of the cylinder at that instant is -2400π km³/s

Explanation:

Given that;

Height of cylinder h = 2.5 km

Radis r = 40 km

radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second

that is; dr/dt = (-12) km/s

we know that; volume of a cylinder V = πr²h

now, we differentiate

dv/dt = π2r×dr/dt×h

so we substitute

dv/dt = 2π × 40 × (-12) × 2.5

dv/dt = -2400π km³/s

Therefore, the rate of change of the volume of the cylinder at that instant is -2400π km³/s

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