(b) At what rate (in ft/sec) is the radius of the water in the tank shrinking when the radius is 16
feet?
Step 1: This is a related rates question. We now have a formula for the volume of the cone in
terms of its radius, so if we differentiate it in terms of t, we should be able to solve for the rateof change of the radius .We are given that the rate of change of the volume and the radius are, respectively
= −48π and R = 16Differentiating the formula for the volume, we get = 3πR^2.
Now, we plug in and get −48π = 3π16^2 . Finally, if we solve for , we get
= − ft/sec(c) How fast (in ft/sec) is the height of the water in the tank dropping at the instant that the
radius is 16 feet?
Step 1: This is the same idea as the previous problem, except that we want to solve for . In
order to do this, we need to go back to our ratio of height to radius and solve it for the radius. = 3 or = rSubstituting for R in the original equation, we get V = π h = .
Step 2: Now we need to know what H is when R is 16. Using our ratio,
H = 3(16) = 48.Step 3: Now if we differentiate, we get