AP Calculus AB Related Rates Problem with Solution Yesterday I posted a Level 4 related rates problem for the AP Calculus AB exam (although this problem is a pretty standard Calculus problem that you may come across in any Calculus course). You can see that post here: AP Calculus AB Problem – Related Rates Today I would like to provide a full explanation for this problem. Here is the problem one more time: The radius of a spherical balloon is decreasing at a constant rate of 0.5 centimeters per second. At the instant when the volume V becomes 288π cubic centimeters, what is the rate of decrease, in square centimeters per second, of the surface area of the balloon? (A) 24π (B) 48π (C) 64π (D) 72π Solution: Recall that the volume of a sphere with radius r is V = 4/3 πr3, and the surface area of a sphere with radius r is S = 4πr2. We are given that dr/dt = –0.5, and we are being asked to find dS/dt when V = 288π. First note that when V = 288π, we have 288π = 4/3 πr3 3/4 ⋅ 288π = πr3 216 = r3 6 = r Now, dS/dt = 8πr ⋅ dr/dt = 8πr(–0.5) = –4πr. dS/dt (at r = 6) = –4π(6) = –24π. So the surface area of the balloon is decreasing at a rate of 24π cm2/sec. Therefore, the answer is choice (A). Notes: (1) Observe that one can get the formula for the surface area of a sphere by differentiating the formula for the volume of a sphere. This is just a little trick that can be used to reduce the number of formulas to memorize. (2) Remember that the word rate generally indicates a derivative. “Increasing at a rate of” indicates a positive derivative, and “decreasing at a rate of” indicates a negative derivative. (3) A radius is a length and is therefore measured in single units (in this case centimeters). Area (and in particular, surface area) is measured in square units. Volume is measured in cubic units. The type of units being mentioned usually gives a big hint as to what measurement is being given or asked for. (4) This is a related rates problem. In a related rates problem we differentiate the independent and dependent variables with respect to a new variable, usually named t, for time. (5) A related rates problem can be pictured as a dynamic (moving) process that gets fixed at a specific moment in time. For this problem we can picture a sphere shaped balloon deflating. We then freeze time at the moment when the volume is 288π cm3. At this moment in time we want to know what the rate of decrease of the surface area is. The word “rate” indicates that we want the derivative of surface area with respect to time, dS/dt. Don’t forget to apply the chain rule when differentiating the right hand side. In this case, the derivative of r is dr/dt. More AP Calculus Problems with Explanations If you are preparing for one of the AP Calculus exams, you may want to take a look at one of the following books. And if you liked this article, please share it with your Facebook friends: Comments comments