Cracking The Ap Calculus ab Exam 2018

− in./s.

We are given the rate at which the water is flowing out of the container, = − 35π (Why is it

negative?), and are looking for the rate at which the depth of the water is dropping, . Thus,

we need to find a way to relate the volume of a cone to its height. We know that the volume of a

cone is V = πr^2 h. Notice that we have a problem. We have a third variable, r, in the equation.

We cannot treat it as a constant the way we did in problem 4 because as the volume of a cone

changes, both its height and radius change. But, we also know that in any cone, the ratio of the

radius to the height is a constant. Here, when the radius is 21 (because the diameter is 42), the

height is 15. Thus, . We can now isolate r in this equation: r = . Now we can

plug it into the volume formula to get rid of r: V = h. This simplifies to V =

. Now we can take the derivative of this equation with respect to t:
. Next, we plug = −35π and h = 5 into the derivative and we get

−35π = . Now we can solve for : = − in./s.

− ft/s

We are given the rate at which the length of the rope, R, is changing, = −4 and are looking

for the rate at which the boat, B, is approaching the dock, . The key to this problem is to

realize that the vertical distance from the dock to the bow, the distance from the boat to the

dock, and the length of the rope form a right triangle.