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Apply derivatives to solve problems involving optimization, related rates, and curve sketching.

Putting Derivatives to Work: Applications of Derivatives

Derivatives are not just abstract calculations — they solve real problems. Finding maximum and minimum values, determining where a function increases or decreases, analyzing the shape of curves, and modeling motion are all powered by derivatives. These applications appear in physics, economics, biology, and engineering.

Components of Applications of Derivatives

This section covers the major application areas:

  • Finding Extrema: Use the first derivative to locate critical points (where f'(x) = 0 or undefined), then classify them as local maxima, local minima, or neither.
  • Increasing & Decreasing Intervals: A function increases where f'(x) > 0 and decreases where f'(x) < 0.
  • Optimization: Set up a function from a real-world scenario, take its derivative, set f'(x) = 0, and solve to find the optimal value.
  • Related Rates: When two or more quantities change over time, use implicit differentiation and the chain rule to find how their rates relate.

Examples of Applications of Derivatives

Finding Extrema Examples

  • For f(x) = x² - 4x + 3, f'(x) = 2x - 4 = 0 gives x = 2. Since f''(2) = 2 > 0, x = 2 is a local minimum with f(2) = -1.
  • For f(x) = -x² + 6x, f'(x) = -2x + 6 = 0 gives x = 3. Since f''(3) = -2 < 0, x = 3 is a local maximum with f(3) = 9.
  • For f(x) = x³ - 3x, f'(x) = 3x² - 3 = 0 gives x = ±1. f(-1) = 2 is a local max and f(1) = -2 is a local min.

Increasing & Decreasing Examples

  • For f(x) = x² - 2x, f'(x) = 2x - 2. Setting f'(x) = 0 gives x = 1. The function decreases on (-∞, 1) and increases on (1, ∞).
  • For f(x) = x³, f'(x) = 3x². Since 3x² ≥ 0 for all x (and equals 0 only at x = 0), the function is increasing everywhere.
  • For f(x) = -x³ + 12x, f'(x) = -3x² + 12 = 0 gives x = ±2. The function increases on (-2, 2) and decreases outside that interval.

Optimization Examples

  • Maximize the area of a rectangle with perimeter 40 cm: If width = x, length = 20 - x, area A = x(20 - x) = 20x - x². A'(x) = 20 - 2x = 0 gives x = 10. Maximum area is 100 cm² (a square).
  • A farmer has 200 meters of fencing for a rectangular pen against a barn wall. With three sides fenced: A = x(200 - 2x), A'(x) = 200 - 4x = 0, x = 50 m. Maximum area is 5,000 m².
  • Find the minimum sum of a positive number and its reciprocal: f(x) = x + 1/x, f'(x) = 1 - 1/x² = 0, x = 1. Minimum value is 2.
  • A balloon's radius grows at 2 cm/s. When r = 5, how fast does volume change? V = 4/3 Ï€r³, dV/dt = 4Ï€r² × dr/dt = 4Ï€(25)(2) = 200Ï€ ≈ 628.3 cm³/s.
  • A 10-foot ladder slides down a wall. The base moves out at 1 ft/s. When the base is 6 ft from the wall, the top slides down: using x² + y² = 100, 2x(dx/dt) + 2y(dy/dt) = 0, y = 8, so dy/dt = -6/8 × 1 = -0.75 ft/s.
  • Water fills a cone (radius 3 m, height 6 m) at 2 m³/min. When h = 4, r = 2, dh/dt = 2/(Ï€ × 4) = 1/(2Ï€) ≈ 0.159 m/min.