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)

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

cancer risk hope this

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2