Applications of Differentiation Flashcards

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?

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.

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?

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.

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

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.

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?

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.

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

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.

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)

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.

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?

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.

A function f has derivative f′(x) = (x − 1)(x + 2)². Which statement about the relative extrema of f is 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.

What is the relative maximum value of f(x) = x·e^(−x)?

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.

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?

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.