AP Calc AB Applications of Differentiation — Questions and Answers
Question 1: A 13-foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant 2 ft/s. How fast is the top of the ladder sliding down the wall at the instant the bottom is 5 feet from the wall?
- 2/5 ft/s
- 6/5 ft/s
- 5/6 ft/s (Correct answer)
- 12/5 ft/s
Correct answer: 5/6 ft/s
From x² + y² = 169, x(dx/dt) + y(dy/dt) = 0. When x = 5, y = 12, so 5(2) + 12(dy/dt) = 0 and dy/dt = -5/6 ft/s, so the top slides down at 5/6 ft/s.
Related-rates problems start with an equation that links the changing quantities. Here the ladder is the hypotenuse, so x² + y² = 13² holds at every moment. Differentiating implicitly with respect to t gives 2x(dx/dt) + 2y(dy/dt) = 0. Substitute only after differentiating: x = 5, y = 12 (from the Pythagorean triple), dx/dt = 2. Solving gives dy/dt = -5/6 ft/s. The negative sign means y is decreasing, so the top moves down at 5/6 ft/s.
Question 2: Evaluate lim (x→0) (e^(2x) − 1 − 2x) / x².
Correct answer: 2
Direct substitution gives 0/0, so apply L'Hospital's Rule twice. The first pass gives (2e^(2x) − 2)/(2x), still 0/0. The second gives 4e^(2x)/2, which tends to 2.
At x = 0 the numerator is e⁰ − 1 − 0 = 0 and the denominator is 0, so the form is 0/0 and L'Hospital's Rule applies. Differentiating gives (2e^(2x) − 2)/(2x), which is again 0/0 at x = 0. Differentiating a second time gives 4e^(2x)/2. As x → 0 this approaches 4/2 = 2. A common mistake is to stop after one application, or to apply the quotient rule instead of differentiating the numerator and denominator separately.
Question 3: Let f(x) = x³ − 3x on the interval [0, 2]. At which value of c in (0, 2) does the Mean Value Theorem guarantee that f'(c) equals the average rate of change of f on [0, 2]?
- 1
- √2
- 4/3
- 2/√3 (Correct answer)
Correct answer: 2/√3
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, which lies in (0, 2).
The Mean Value Theorem applies because f is a polynomial, so it is continuous on [0, 2] and differentiable on (0, 2). It guarantees some c with f'(c) = (f(2) − f(0))/(2 − 0). Here f(2) = 8 − 6 = 2 and f(0) = 0, so the secant slope is 1. Since f'(x) = 3x² − 3, solve 3c² − 3 = 1 to get c² = 4/3. Only the positive root, 2/√3 ≈ 1.155, lies in the open interval. Note that 4/3 is c², not c.
Question 4: What is the absolute maximum value of f(x) = x³ − 6x² + 9x + 1 on the closed interval [0, 5]?
Correct answer: 21
f'(x) = 3(x − 1)(x − 3), so the critical points are x = 1 and x = 3. Evaluating gives f(0) = 1, f(1) = 5, f(3) = 1, f(5) = 21, so the maximum is 21.
By the Extreme Value Theorem, a continuous function on a closed interval attains an absolute maximum and minimum. Use the Candidates Test: find critical points by solving f'(x) = 0, then compare f at those points and at the endpoints. Here f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). The values are f(0) = 1, f(1) = 5, f(3) = 1, and f(5) = 125 − 150 + 45 + 1 = 21. The local maximum at x = 1 (value 5) is not the absolute maximum, because the endpoint x = 5 gives a larger value.
Question 5: A function f is twice differentiable with f''(x) = x(x − 2)². At which x-values does the graph of f have a point of inflection?
- x = 0 and x = 2
- x = 0 only (Correct answer)
- x = 2 only
- f has no inflection points
Correct answer: x = 0 only
An inflection point requires f'' to change sign. f'' is zero at x = 0 and x = 2, but (x − 2)² is never negative, so the sign of f'' changes only at x = 0.
Inflection points occur where concavity changes, meaning f'' changes sign. The zeros of f'' are x = 0 and x = 2. For x < 0, f'' is negative (x is negative and the squared factor is positive). For 0 < x < 2 and for x > 2, f'' is positive. So f'' changes from negative to positive at x = 0, but keeps the same sign on both sides of x = 2. The graph is therefore concave up on both sides of x = 2, with no inflection there.
Question 6: A farmer has 200 meters of fencing and will enclose a rectangular pen along a straight river. No fence is needed along the river side. What is the maximum possible area of the pen?
- 2500 m²
- 10000 m²
- 5000 m² (Correct answer)
- 4000 m²
Correct answer: 5000 m²
With two sides of length x perpendicular to the river and one side of length 200 − 2x parallel to it, A(x) = x(200 − 2x). A'(x) = 200 − 4x = 0 gives x = 50, so A = 50 · 100 = 5000 m².
Let x be the length of each side perpendicular to the river. The fencing constraint 2x + y = 200 gives y = 200 − 2x for the side parallel to the river. The area is A(x) = 200x − 2x² on 0 ≤ x ≤ 100. Then A'(x) = 200 − 4x = 0 at x = 50, and A'' = −4 < 0 confirms a maximum. The dimensions are 50 by 100, so the area is 5000 m². The value 2500 m² comes from wrongly using a 50 by 50 square.
Question 7: Let f(x) = x⁴ − 4x³. Which statement about f at its critical points is correct?
- f has a local minimum at x = 3 and no local extremum at x = 0. (Correct answer)
- f has a local maximum at x = 0 and a local minimum at x = 3.
- f has a local minimum at both x = 0 and x = 3.
- f has a local minimum at x = 0 and a local maximum at x = 3.
Correct answer: f has a local minimum at x = 3 and no local extremum at x = 0.
f'(x) = 4x²(x − 3), so the critical points are x = 0 and x = 3. f' is negative on both sides of 0, so there is no extremum there. f' changes from negative to positive at 3, so that is a local minimum.
Factor f'(x) = 4x³ − 12x² = 4x²(x − 3), giving critical points x = 0 and x = 3. Since f''(x) = 12x² − 24x, f''(3) = 36 > 0, so x = 3 is a local minimum. But f''(0) = 0, so the second derivative test is inconclusive. Use the first derivative test instead: 4x² is nonnegative and (x − 3) is negative near 0, so f' < 0 on both sides of 0. The sign does not change, so f has no local extremum at x = 0.
Question 8: A particle moves along a line with position s(t) = t³ − 6t² + 9t for t ≥ 0. At which time is the particle speeding up?
- t = 0.5
- t = 2
- t = 2.5
- t = 4 (Correct answer)
Correct answer: t = 4
v(t) = 3(t − 1)(t − 3) and a(t) = 6t − 12. At t = 4, v = 9 and a = 12 have the same sign, so the speed is increasing. At t = 0.5 and t = 2.5 the signs differ, and at t = 2 the acceleration is 0.
A particle speeds up when velocity and acceleration have the same sign, and slows down when they have opposite signs. Here v(t) = 3t² − 12t + 9 and a(t) = 6t − 12. At t = 0.5: v = 3.75 > 0 and a = −9 < 0, so it is slowing down. At t = 2: a = 0, so the speed is momentarily not changing. At t = 2.5: v = −2.25 < 0 and a = 3 > 0, so it is slowing down. At t = 4: v = 9 > 0 and a = 12 > 0, so it is speeding up.
Question 9: A spherical balloon is inflated so that its volume increases at 36π cm³/s. At the instant its radius is 3 cm, how fast is its surface area increasing? (V = (4/3)πr³, S = 4πr²)
- 12π cm²/s
- 24π cm²/s (Correct answer)
- 36π cm²/s
- 8π cm²/s
Correct answer: 24π cm²/s
From dV/dt = 4πr²(dr/dt), 36π = 4π(9)(dr/dt) gives dr/dt = 1 cm/s. Then dS/dt = 8πr(dr/dt) = 8π(3)(1) = 24π cm²/s.
This related-rates problem has two linked equations that share the radius r. Differentiating V = (4/3)πr³ gives dV/dt = 4πr²(dr/dt). Substituting dV/dt = 36π and r = 3 gives 36π = 36π(dr/dt), so dr/dt = 1 cm/s. Differentiating S = 4πr² gives dS/dt = 8πr(dr/dt). At r = 3 this is 24π cm²/s. Substituting r = 3 before differentiating would be an error, since r is changing with time.
Question 10: The derivative of a function f is f'(x) = (x + 2)(x − 1)². Which statement describes f at x = 1?
- f has a local maximum at x = 1.
- f has a local minimum at x = 1.
- f has a vertical tangent at x = 1.
- f has neither a local maximum nor a local minimum at x = 1. (Correct answer)
Correct answer: f has neither a local maximum nor a local minimum at x = 1.
f'(1) = 0, so the tangent is horizontal, but (x − 1)² is nonnegative, so f' stays positive on both sides of x = 1. Because f' does not change sign, f has no local extremum there.
A critical point is a local extremum only if f' changes sign there (first derivative test). Near x = 1, the factor (x + 2) is positive, and (x − 1)² is positive on both sides of 1. So f' > 0 just left and just right of x = 1, and f is increasing through that point with a momentarily horizontal tangent. In contrast, at x = −2 the derivative changes from negative to positive, so f has a local minimum there. A vertical tangent is ruled out because f' is defined and equals 0 at x = 1.
