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 in3/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.
What rates are known and unknown?
The rate of change of volume, dV/dt = 8 in3/min
And dh/dt = ? when h = 4 inches
What equation relates V and h?
V = 3.14/3 r2 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)2h = 0.2618h3
dV/dt = 3 (0.2618)h2 dh/dt = 0.7854h2dh/dt
Now plug in the known values and solve.
8 in3/min = 0.7854(4in)2dh/dt
dh/dt = 0.637in2/min