Cracking The Ap Calculus ab Exam 2018

100 km/hr

We are given the rate at which Car A is moving south, = 80, and the rate at which Car B is

moving west, = 60, and we are looking for the rate at which the distance between them is

increasing, which we’ll call . Note that the directions south and west are at right angles to

each other. Thus, the distance that Car A is from the starting point, which we’ll call A, and the

distance that Car B is from the starting point, which we’ll call B, are the legs of a right triangle,

with C as the hypotenuse. We can relate the three distances using the Pythagorean Theorem.

Here, because A and B are the legs and C is the hypotenuse, A^2 + B^2 = C 2 . Now, we take the

derivative of the equation with respect to t: . This simplifies to

. We know that Car A has been driving for 3 hours at 80 km/hr and Car

B has been driving for 3 hours at 60 km/hr, so A = 240 and B = 180. Using the Pythagorean

Theorem, 240^2 + 180^2 = C 2 , so C = 300. Now we can substitute into our derivative equation:

(240)(80) + (180)(60) = (300) . If we solve for C, we get C = 100 km/hr.

243 in.^2 /s

We are given the rate at which the sides of the triangle are increasing, = 27, and are looking

for the rate at which the area is increasing, . Thus, we need to find a way to relate the area

of an equilateral triangle to the length of a side. We know that the area of an equilateral

triangle, in terms of its sides, is A = . (If you don’t know this formula, memorize it! It

will come in very handy in future math problems.) Now we take the derivative of this equation

with respect to t: . Next, we plug = 27 and s = 18 into the derivative

and we get = (36)(27) = 243 in.^2 /s.