Calculus Word Problem

a tank in the shape of an inverted cone is being drained. the radius is 6 feet, and its height 15 feet. At what rate is the depth of the water tank decreasing at the instant when the depth is 7 feet if the volume is decreasing at the rate of 2 feet cubed per hour at that time?

Solution

Volume V of the cone is V= πr2h/3

The ratio of the radius to the height at any time is r/h = 6:15 = 2:5 = 0.4

The Volume V=π(0.4)2h3/3

The derivative of the Volume with respect to the height is

dV/dt =V’ = 2 = 3(0.4)2h2/3 = 0.16h2 dh/dt = 0.16(7)2 dh/dt = 7,84 dh/dt

Then dh/dy is decreasing at the rate of 2/7.84 = 0.255 feet/hour

It’s harder to type these in than to solve them!