AP Calc BC Applications of Differentiation

Question 1

A 13-foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at 2 ft/s. At the instant the bottom is 5 feet from the wall, what is the rate of change of the height of the top of the ladder on the wall?

Accepted Answer

-5/6 ft/s

Answer

With x² + y² = 169, differentiating gives x(dx/dt) + y(dy/dt) = 0. At x = 5, y = 12, so 5(2) + 12(dy/dt) = 0 and dy/dt = -5/6 ft/s.

Question 2

Let f(x) = x³ − 3x on the interval [0, 2]. Which value of c in (0, 2) satisfies the conclusion of the Mean Value Theorem?

Accepted Answer

2/√3

Answer

The average rate of change is (f(2) − f(0))/2 = (2 − 0)/2 = 1. Setting f′(c) = 3c² − 3 = 1 gives c² = 4/3, so c = 2/√3.

Question 3

What is the absolute maximum value of f(x) = x³ − 12x on the closed interval [−3, 5]?

Accepted Answer

65

Answer

f′(x) = 3x² − 12 = 0 gives x = ±2. Comparing f(−3) = 9, f(−2) = 16, f(2) = −16 and f(5) = 65, the largest value is 65.

Question 4

The derivative of a function f is f′(x) = x²(x − 2)³. At which x-values does the graph of f have points of inflection?

Accepted Answer

x = 0 and x = 4/5

Answer

f″(x) = x(x − 2)²(5x − 4), which is zero at x = 0, 4/5 and 2. f″ changes sign at 0 and 4/5 but not at 2, because (x − 2)² never changes sign.

Question 5

A particle moves along a line with position s(t) = t³ − 9t² + 15t for t ≥ 0. At which time is the particle speeding up?

Accepted Answer

t = 2

Answer

v(t) = 3(t − 1)(t − 5) and a(t) = 6t − 18. At t = 2, v = −9 and a = −6 have the same sign, so the speed is increasing.

AP Calc BC Practice Test

AP Calc BC Practice Test

AP Calc BC Applications of Differentiation