when the water is 12 feet deep?
This “cone” problem is also typical. The key point to getting these right is knowing that the ratio of the
height of a right circular cone to its radius is constant. By telling us that the height of the cone is 30 and the
radius is 15, we know that at any level, the height of the water will be twice its radius, or h = 2r.
The volume of a cone is V = π^2 h. (You’ll learn to derive this
formula through integration in Chapter 17.)You must find the rate at which the water is rising (the height is changing), or . Therefore, you want to
eliminate the radius from the volume. By substituting = r into the equation for volume, we get
V = π h = Differentiate both sides with respect to t.
= 3 h^2Now we can plug in and solve for .
18 π = 3(12)^2 = feet/minIn order to solve related rates problems, you have to be good at determining relationships between
variables. Once you figure that out, the rest is a piece of cake. Many of these problems involve geometric
relationships, so review the formulas for the volumes and areas of cones, spheres, boxes, and other
solids. Once you get the hang of setting up the problems, you’ll see that these problems follow the same
predictable patterns. Look through these sample problems.
PROBLEM 1. A circle is increasing in area at the rate of 16π in.^2 /s. How fast is the radius increasing when
the radius is 2 in.?
Answer: Use the expression that relates the area of a circle to its radius: A = πr^2.
Next, take the derivative of the expression with respect to t.