A cylindrical can is partially filled with water. The radius of the cylindrical can is 8 inches, and its height is 9 inches. If the volume of the water is decreasing at a rate of 527 cubic inches per minute, what is the rate, in inches per minute, at which the height of the water is changing when the height of the water is 5 inches?

The volume of the cylindrical can is given by:

V = πr²h

V = volume, r = base radius, h = height

Differentiate both sides of the equation with respect to time t. The radius r doesn't change over time, so we treat it as a constant:

dV/dt = πr² (dh/dt)

Given values:

dV/dt = - 527in³/min

r = 8in

Plug in and solve for dh/dt:

-527 = π (8) ² (dh/dt)

dh/dt = - 2.62in/min

The height of the water is decreasing at a rate of 2.62in/min