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AP Calculus AB: Applications of Derivatives Practice Questions

Test yourself on Applications of Derivatives with 12 original AP Exams practice questions. Pick an answer to see instant feedback and a full explanation.

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1. A particle moves along a line with position s(t) = t³ - 6t² + 9t for t ≥ 0. At what time is the particle momentarily at rest and changing direction from moving left to moving right?
Explanation. v(t) = s'(t) = 3t² - 12t + 9 = 3(t-1)(t-3), zero at t=1 and t=3. For 1<t<3, v<0 (moving left); for t>3, v>0 (moving right). So at t=3 the particle changes from left to right.
2. The function f(x) = x³ - 3x + 2 has a local maximum at which x-value?
Explanation. f'(x) = 3x² - 3 = 3(x-1)(x+1), critical points at x = ±1. f''(x) = 6x; f''(-1) = -6 < 0, so x = -1 is a local maximum.
3. A spherical balloon's volume increases at 36π cm³/s. When the radius is 3 cm, how fast is the radius increasing? (V = (4/3)πr³)
Explanation. dV/dt = 4πr² (dr/dt). So 36π = 4π(9)(dr/dt) = 36π(dr/dt), giving dr/dt = 1 cm/s.
4. For f(x) = x⁴ - 4x³, the graph is concave down on which interval?
Explanation. f''(x) = 12x² - 24x = 12x(x-2), which is negative when 0 < x < 2. So the graph is concave down on (0, 2).
5. Use a linear approximation of f(x) = √x at x = 25 to estimate √26.
Explanation. f(25) = 5, f'(x) = 1/(2√x), f'(25) = 1/10. L(x) = 5 + 0.1(x-25). L(26) = 5 + 0.1(1) = 5.1.
6. A rectangle is inscribed under the curve y = 9 - x² (x ≥ 0) with one side on the x-axis. What x-value maximizes the rectangle's area A = x(9 - x²)?
Explanation. A(x) = 9x - x³, A'(x) = 9 - 3x² = 0 gives x² = 3, x = √3. A''(x) = -6x < 0 confirms a maximum.
7. Given f is continuous on [1,5] and differentiable on (1,5) with f(1)=2 and f(5)=14, the Mean Value Theorem guarantees some c where f'(c) equals what?
Explanation. MVT guarantees f'(c) = (f(5)-f(1))/(5-1) = (14-2)/4 = 12/4 = 3.
8. A 13-ft ladder leans against a wall. The bottom slides away at 2 ft/s. When the bottom is 5 ft from the wall, how fast is the top sliding down?
Explanation. x² + y² = 169. At x=5, y=12. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. So 2(5)(2) + 2(12)(dy/dt) = 0, giving dy/dt = -20/24 = -5/6 ft/s.
9. On what interval is f(x) = xe^(-x) increasing?
Explanation. f'(x) = e^(-x) - xe^(-x) = e^(-x)(1 - x). Since e^(-x) > 0, f'(x) > 0 when 1 - x > 0, i.e., x < 1. So f increases on (-∞, 1).
10. The position of an object is s(t) = -16t² + 64t + 80 ft. What is its maximum height?
Explanation. v(t) = -32t + 64 = 0 at t = 2. s(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = 144 ft.
11. For f(x) = 2x³ - 9x² + 12x, which x-value is an inflection point?
Explanation. f''(x) = 12x - 18 = 0 gives x = 3/2. Since f'' changes sign there, it is an inflection point.
12. By the Candidates Test, find the absolute minimum of f(x) = x³ - 3x on [0, 2].
Explanation. f'(x) = 3x² - 3 = 0 at x = 1 (in [0,2]). Evaluate: f(0)=0, f(1)=1-3=-2, f(2)=8-6=2. The absolute minimum is -2 at x=1.
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FAQ

What's the difference between a critical point and an inflection point?

A critical point is where f'(x)=0 or is undefined—candidates for local extrema. An inflection point is where f''(x) changes sign—the concavity flips. They are tested using the first derivative (extrema) versus the second derivative (concavity).

How do I approach related rates problems on the AP exam?

Identify all variables and the given/wanted rates, write an equation relating the quantities, differentiate both sides with respect to time (using chain rule), then substitute known values—substitute numbers only AFTER differentiating, never before.

When should I use the first vs second derivative test?

Use the first derivative test by checking sign changes of f' around a critical point—it always works including where f''=0. Use the second derivative test (f''>0 means min, f''<0 means max) when f'' is easy to compute and nonzero; it's faster but inconclusive when f''=0.

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