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
