Applications of Differentiation Flashcards

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?

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.

Evaluate lim (x→0) (e^(2x) − 1 − 2x) / x².

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.

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]?

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).

What is the absolute maximum value of f(x) = x³ − 6x² + 9x + 1 on the closed interval [0, 5]?

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.

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?

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.

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?

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 f(x) = x⁴ − 4x³. Which statement about f at its critical points is 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.

A particle moves along a line with position s(t) = t³ − 6t² + 9t for t ≥ 0. At which time is the particle speeding up?

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 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)

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.

The derivative of a function f is f'(x) = (x + 2)(x − 1)². Which statement describes f at x = 1?

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.