Applications of Derivatives: Related Rates

EXAMPLE 3: A conical container with a height of 12 inches and a radius of 6 inches is filled with water at 8 in^{3}/min. What is the rate of change of the height when h = 4 inches?

Start with a sketch. Do both a 3-D and a 2-D side view. Label variables.

NEED SKETCH

What rates are known and unknown?

The rate of change of volume, dV/dt = 8 in^{3}/min

And dh/dt = ? when h = 4 inches

What equation relates V and h?

V = 3.14/3 r^{2} h

This relates V and h, but r is a variable, too, so we look for an equation that relates r and h. Look at the sketch.

Using ratios: r/h = 6/12 so r = h/2.

Substitute this into the volume equation and take the derivative with respect to t.

V = 1.047(h/2)^{2}h = 0.2618h^{3}

dV/dt = 3 (0.2618)h^{2} dh/dt = 0.7854h^{2}dh/dt

Now plug in the known values and solve.

8 in^{3}/min = 0.7854(4in)^{2}dh/dt

dh/dt = 0.637in^{2}/min