AP Calc BC 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 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?
- -12/5 ft/s
- 5/6 ft/s
- -5/6 ft/s (Correct answer)
- -6/5 ft/s
Correct answer: -5/6 ft/s
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.
Related-rates problems link quantities through an equation that holds at all times. Here x² + y² = 13² relates the base distance x and the height y. Differentiate implicitly with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0. Only after differentiating do you substitute the instant's values: x = 5, so y = 12, and dx/dt = 2. Solving gives dy/dt = -5/6 ft/s. The negative sign shows the top is sliding down the wall.
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?
- 1
- 2/√3 (Correct answer)
- √3
- 2/3
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.
The Mean Value Theorem applies because f is continuous on [0, 2] and differentiable on (0, 2). It guarantees some c where f′(c) equals the average rate of change (f(b) − f(a))/(b − a). Here f(2) = 8 − 6 = 2 and f(0) = 0, so the secant slope is 1. Solving 3c² − 3 = 1 gives c = ±2/√3, and only the positive root, about 1.155, lies in (0, 2).
Question 3: What is the absolute maximum value of f(x) = x³ − 12x on the closed interval [−3, 5]?
- 65 (Correct answer)
- 16
- 9
- −16
Correct answer: 65
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.
By the Extreme Value Theorem, a continuous function on a closed interval attains its absolute extrema at critical points or endpoints. The critical points are x = −2 and x = 2, both inside [−3, 5]. Evaluating f at x = −3, −2, 2 and 5 gives 9, 16, −16 and 65. The local maximum of 16 at x = −2 is not the absolute maximum, because the endpoint x = 5 gives a larger value.
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?
- x = 2 only
- x = 0 and x = 2
- x = 0 and x = 4/5 (Correct answer)
- x = 0, x = 4/5 and x = 2
Correct answer: x = 0 and x = 4/5
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.
Differentiate f′ with the product rule: f″ = 2x(x − 2)³ + 3x²(x − 2)² = x(x − 2)²[2(x − 2) + 3x] = x(x − 2)²(5x − 4). The candidates are x = 0, 4/5 and 2. At x = 0 and x = 4/5 the sign of f″ changes, so concavity changes there. At x = 2 the factor (x − 2)² is nonnegative on both sides, so f″ keeps the same sign and there is no inflection point.
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?
- t = 0.5
- t = 3
- t = 4
- t = 2 (Correct answer)
Correct answer: t = 2
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.
A particle speeds up when velocity and acceleration have the same sign, and slows down when the signs differ. Here v = 3t² − 18t + 15 = 3(t − 1)(t − 5) and a = 6t − 18. At t = 0.5, v > 0 and a < 0, so it slows. At t = 3, a = 0, so the speed is momentarily not changing. At t = 4, v < 0 and a > 0, so it slows. At t = 2, both are negative, so it speeds up.
Question 6: Air is pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing at the instant the radius is 5 cm? (V = (4/3)πr³)
- 4/π cm/s
- 1/π cm/s (Correct answer)
- 1 cm/s
- π/4 cm/s
Correct answer: 1/π cm/s
Differentiating gives dV/dt = 4πr²(dr/dt). With r = 5, 100 = 100π(dr/dt), so dr/dt = 1/π cm/s.
The volume and radius are both functions of time, so differentiate V = (4/3)πr³ with respect to t to get dV/dt = 4πr²(dr/dt). Substituting dV/dt = 100 and r = 5 gives 100 = 4π(25)(dr/dt) = 100π(dr/dt). Therefore dr/dt = 1/π ≈ 0.318 cm/s. Substituting r = 5 before differentiating is a common mistake. It would treat the radius as constant.
Question 7: A farmer has 200 feet of fencing to enclose a rectangular pen along a straight river. No fence is needed along the river, so only three sides are fenced. What is the maximum possible area of the pen?
- 10,000 ft²
- 2,500 ft²
- 4,000 ft²
- 5,000 ft² (Correct answer)
Correct answer: 5,000 ft²
With width x and length y along the river, 2x + y = 200, so A = x(200 − 2x). A′ = 200 − 4x = 0 gives x = 50, y = 100, and A = 5,000 ft².
Let x be each side perpendicular to the river and y the side parallel to it. The constraint 2x + y = 200 gives y = 200 − 2x, so A(x) = 200x − 2x². Setting A′(x) = 200 − 4x = 0 gives x = 50. A″ = −4 < 0 confirms a maximum. Then y = 100 and A = 50 · 100 = 5,000. Treating all four sides as fenced would give a 50 × 50 square of 2,500 ft², which is a wrong setup.
Question 8: A function f has derivative f′(x) = (x − 1)(x + 2)². Which statement about the relative extrema of f is correct?
- f has a relative minimum at x = 1 and no other relative extrema. (Correct answer)
- f has a relative maximum at x = −2 and a relative minimum at x = 1.
- f has a relative maximum at x = 1 only.
- f has relative minima at both x = −2 and x = 1.
Correct answer: f has a relative minimum at x = 1 and no other relative extrema.
f′ changes from negative to positive at x = 1, giving a relative minimum. At x = −2 the factor (x + 2)² never changes sign, so f′ stays negative and there is no extremum.
The critical points are x = 1 and x = −2. For x < 1 (excluding −2), (x − 1) is negative and (x + 2)² is positive, so f′ < 0. For x > 1, f′ > 0. So f decreases and then increases around x = 1, which is a relative minimum by the First Derivative Test. At x = −2, f′ is negative on both sides, so f is decreasing through that point. It is a horizontal-tangent point, not an extremum.
Question 9: What is the relative maximum value of f(x) = x·e^(−x)?
- e
- 1
- 1/e (Correct answer)
- 1/e²
Correct answer: 1/e
f′(x) = e^(−x)(1 − x) = 0 at x = 1, and f″(1) = e^(−1)(1 − 2) < 0, so x = 1 is a maximum. The value is f(1) = 1/e.
By the product rule, f′(x) = e^(−x) − x·e^(−x) = e^(−x)(1 − x). Since e^(−x) is never zero, the only critical point is x = 1. Differentiating again, f″(x) = e^(−x)(x − 2), so f″(1) = −1/e < 0. The graph is concave down there, so x = 1 is a relative maximum. Substitute into the original function, not the derivative, to get the value f(1) = 1/e.
Question 10: The tangent line to f(x) = √x at x = 25 is used to approximate √26. What is the approximation, and is it an overestimate or an underestimate of the true value?
- 5.1, underestimate
- 5.1, overestimate (Correct answer)
- 5.2, overestimate
- 5.01, underestimate
Correct answer: 5.1, overestimate
L(x) = 5 + (1/10)(x − 25), so L(26) = 5.1. Since f″(x) = −¼x^(−3/2) < 0, f is concave down and tangent lines lie above the graph, so the approximation is an overestimate.
With f(25) = 5 and f′(x) = 1/(2√x), f′(25) = 1/10. The linearization is L(x) = 5 + 0.1(x − 25), so L(26) = 5.1. The second derivative f″(x) = −(1/4)x^(−3/2) is negative for x > 0, so the graph is concave down. For a concave-down function the tangent line lies above the curve, so the linear approximation is larger than the true value. The actual value √26 ≈ 5.099 is slightly below 5.1.
