Application of Derivatives

Derivatives answer the question: "How fast is something changing?" This chapter transforms derivatives from abstract mathematical objects into practical tools for solving real-world problems. Engineers use derivatives to optimize designs, economists use them to maximize profit, biologists track population growth rates, and physicists describe motion. Whether you're finding the maximum volume of a box cut from a sheet of material, determining when a bacteria culture stops growing, or calculating the optimal angle for a projectile, derivatives provide the systematic approach. This chapter bridges the theory of chapter-05-continuity-and-differentiability with tangible applications.

Rate of Change: The Core Idea

The derivative f'(x) measures the instantaneous rate of change of f at point x. If f(t) represents distance traveled over time, then f'(t) is the instantaneous velocity. If f(p) represents revenue as a function of price p, then f'(p) is the marginal revenue.

Average vs. Instantaneous:

• Average rate of change from a to b: [f(b) - f(a)]/(b - a) (slope of secant line)
• Instantaneous rate of change at a: f'(a) = lim(h→0) [f(a+h) - f(a)]/h (slope of tangent line)

Tangent and Normal Lines

The tangent line to a curve at point (a, f(a)) has slope m = f'(a) and equation:

y - f(a) = f'(a)(x - a)

The normal line is perpendicular to the tangent line, with slope m = -1/f'(a) (negative reciprocal):

y - f(a) = [-1/f'(a)](x - a)

Application: Finding where light reflects off a curved mirror or where a ladder slips against a wall.

Increasing and Decreasing Functions

• If f'(x) > 0 on interval I, then f is increasing on I
• If f'(x) < 0 on interval I, then f is decreasing on I
• If f'(x) = 0 at x = c, then c is a critical point (possible local extremum)

Strategy: Find critical points by solving f'(x) = 0, then test the sign of f'(x) on either side.

Local and Absolute Extrema

Local extremum: A point where f has a local maximum or minimum (extreme value in a neighborhood).

Tests for local extrema:

- If f'(x) changes from + to - as x passes through c: local maximum - If f'(x) changes from - to + as x passes through c: local minimum - If f'(x) doesn't change sign: neither (inflection point)

1. First Derivative Test: At critical point c:

- If f''(c) > 0: local minimum (concave up) - If f''(c) < 0: local maximum (concave down) - If f''(c) = 0: test inconclusive

1. Second Derivative Test: At critical point c where f'(c) = 0:

Absolute extrema: On a closed interval [a, b], the absolute maximum and minimum occur at critical points or endpoints. Always evaluate f at all critical points and endpoints.

Concavity and Inflection Points

Concavity:

• If f''(x) > 0 on interval I, then f is concave up (bowl-shaped) on I
• If f''(x) < 0 on interval I, then f is concave down (hill-shaped) on I

Inflection point: Where concavity changes. At an inflection point, f''(x) = 0 (or undefined) and f'' changes sign.

Intuition: Concave up means f' is increasing (curve getting steeper). Concave down means f' is decreasing (curve getting less steep). At inflection points, the rate of change of the rate of change switches direction.

Optimization Problems: Systematic Approach

Many real-world problems reduce to: maximize or minimize some quantity subject to constraints.

Steps:

1. Identify the quantity to maximize/minimize; express as function f(x)
2. Identify constraints; express relationships algebraically
3. Express f as function of single variable using constraints
4. Find domain of f (where the problem makes sense)
5. Find critical points by solving f'(x) = 0
6. Evaluate f at critical points and endpoints
7. Interpret the result in context

Example: A rectangular field has fixed perimeter 100 m. What dimensions maximize area?

• Quantity: Area A = length × width = lw
• Constraint: Perimeter 2l + 2w = 100, so w = 50 - l
• Express A as function of l: A(l) = l(50 - l) = 50l - l²
• Domain: (0, 50)
• Critical points: dA/dl = 50 - 2l = 0 → l = 25
• Maximum area: A(25) = 25 × 25 = 625 m²

Motion and Related Rates

When quantities change with time, related rates problems connect their rates of change through implicit differentiation.

Example: A ladder 5 m long leans against a wall. If the base slides away at 2 m/s, how fast is the top sliding down when the base is 3 m from the wall?

- 2(3)(2) + 2(4)(dy/dt) = 0 - dy/dt = -12/8 = -1.5 m/s (negative means sliding down)

• Constraint: x² + y² = 25 (Pythagorean theorem)
• Differentiate with respect to time: 2x(dx/dt) + 2y(dy/dt) = 0
• Solve for dy/dt: When x = 3, y = 4, dx/dt = 2 m/s:

Connections to Other Topics

• chapter-05-continuity-and-differentiability: Foundation of differentiability and derivatives
• chapter-07-integrals: Inverse operation; antiderivatives and integration
• chapter-08-application-of-integrals: Using derivatives and integrals together for area/volume
• chapter-03-matrices and chapter-04-determinants: Hessian matrices for multivariable optimization
• chapter-12-linear-programming: Alternative optimization method for linear constraints

Socratic Questions

1. Explain why a function can have a critical point where f'(x) = 0 but no local extremum exists. What is such a point called, and how can the second derivative test help identify it?

1. Consider a function whose second derivative f''(x) is always positive on an interval. What does this guarantee about the function's behavior? Why are businesses interested in functions whose second derivative (rate of change of marginal cost) is positive?

1. In optimization problems, we always check critical points and endpoints. Why is checking endpoints essential? Can the absolute maximum on a closed interval occur at a point where f'(x) ≠ 0?

1. For a related rates problem involving a ladder sliding down a wall, why is the answer (rate at the top) negative? What does the negative sign represent physically, and how would you interpret it to someone unfamiliar with calculus?

1. If you're designing a cylindrical can to minimize material used (surface area) while holding fixed volume V, how do the first and second derivatives guide your optimization? Why is the second derivative test essential here?