Cracking The Ap Calculus ab Exam 2018

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Example 3: A spherical balloon is expanding at a rate of 60π in.^3 /sec. How fast is the surface area of the
balloon expanding when the radius of the balloon is 4 in.?

Step 1: You’re given the rate at which the volume’s expanding, and you know the equation that relates
volume to radius. But you have to relate radius to surface area as well, because you have to find the
surface area’s rate of change. This means that you’ll need the equations for volume and surface area of a
sphere.

V   =    πr^3

A   =   4πr^2

You’re trying to find , but A is given in terms of r, so you have to get first. Because we know the

volume, if we work with the equation that gives us volume in terms of radius, we can find . From there,

work with the other equation to find . If we take the derivative of the equation with respect to t we get

    =   4πr^2 . Plugging    in  for     and for r,  we  get 60π =   4π(4)^2.

Solving for , we get

    =       in./sec

Step 2: Now we take the derivative of the other equation with respect to t.

    =   8πr

We can plug in for r and from the previous step and we get

    =   8π(4)   =       =   30π in.^2 /sec

One final example.

Example 4: An underground conical tank, standing on its vertex, is being filled with water at the rate of

18 π ft^3 /min. If the tank has a height of 30 feet and a radius of 15 feet, how fast is the water level rising