Applications of Derivatives
1. Introduction
Derivatives have wide-ranging applications in physics, economics, engineering, and geometry. This chapter explores how derivatives describe rates of change, slopes of curves, monotonicity, and optimization.
2. Rate of Change
If y = f(x), then dy/dx represents the instantaneous rate of change of y with respect to x.
If s = f(t) is the position function, v = ds/dt is velocity and a = dv/dt = d²s/dt² is acceleration.
'In related rates problems, identify all variables, write an equation relating them, then differentiate with respect to time.'
3. Tangents and Normals
3.1 Tangent
The slope of the tangent to y = f(x) at (x₀, y₀) is m = f'(x₀). Equation: y - y₀ = m(x - x₀).
3.2 Normal
The normal is perpendicular to the tangent. Slope m_n = -1/m (if m ≠ 0). Equation: y - y₀ = -(1/m)(x - x₀).
3.3 Special Cases
- If m = 0, tangent is horizontal (y = y₀), normal is vertical (x = x₀).
- If m → ∞ (vertical tangent), equation is x = x₀, normal is y = y₀.
4. Increasing and Decreasing Functions
f is increasing on (a, b) if f'(x) ≥ 0 for all x ∈ (a, b). f is decreasing on (a, b) if f'(x) ≤ 0 for all x ∈ (a, b).
'Strictly increasing means f'(x) > 0 (not just ≥ 0) almost everywhere. Points where f'(x) = 0 are critical points.'
5. Maxima and Minima
5.1 Critical Points
Points where f'(x) = 0 or f'(x) does not exist.
5.2 First Derivative Test
If f' changes sign from + to - at c, c is a local maximum. If - to +, local minimum. If no sign change, neither.
5.3 Second Derivative Test
If f'(c) = 0 and f''(c) < 0, c is a local maximum. If f'(c) = 0 and f''(c) > 0, c is a local minimum. If f''(c) = 0, the test fails.
5.4 Absolute Maxima and Minima on [a, b]
Evaluate f at all critical points in (a, b) and at endpoints. The largest is the absolute maximum, the smallest is the absolute minimum.
6. Optimization Problems
Steps: (1) Identify variable to optimize. (2) Express it in terms of one variable. (3) Use constraints. (4) Differentiate and find critical points. (5) Verify using second derivative. (6) State the result.
7. Worked Problems
Problem 1: Find the equation of tangent and normal to y = x³ - 2x + 1 at x = 1. Solution: At x = 1, y = 0. dy/dx = 3x² - 2 = 1. Tangent: y - 0 = 1(x - 1) ⇒ y = x - 1. Normal: y - 0 = -1(x - 1) ⇒ y = -x + 1.
Problem 2: Show that f(x) = x³ - 3x² + 3x + 2 is increasing for all real x. Solution: f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1) = 3(x - 1)² ≥ 0. Hence always increasing.
Problem 3: Find the dimensions of a rectangle with maximum area that can be inscribed in a semicircle of radius r. Solution: Let rectangle have width 2x and height y, with x² + y² = r². Area A = 2xy = 2x√(r² - x²). dA/dx = 0 gives x = r/√2, y = r/√2. Rectangle has sides r√2 and r/√2.
Problem 4: Show that the height of a closed cylinder of maximum volume for a given surface area S is equal to its diameter. Solution: S = 2πrh + 2πr², V = πr²h. Eliminate h, differentiate, get h = 2r.
8. Common Mistakes
'Students often forget to verify that a critical point is indeed a maximum or minimum using the second derivative test.'
'When solving optimization problems, always check the feasibility of the answer — negative dimensions are not allowed.'
9. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Rate of change | 3 | 2 |
| Tangents and normals | 3 | 2 |
| Increasing/decreasing | 3 | 1 |
| Maxima and minima | 4 | 3 |
| Optimization | 4 | 3 |
10. Self-Test Questions
- Find the points on y = 3x³ - 5x² + 3x - 1 where the tangent is parallel to the x-axis.
- Prove that f(x) = 2x³ - 9x² + 12x + 1 is decreasing in (1, 2) and increasing elsewhere.
- Find all local maxima and minima of f(x) = x⁴ - 4x³ + 4x².
- A wire of length L is cut into two parts to form a circle and a square. How should it be cut to minimize total area?
- Find the shortest distance from (4, 2) to the curve y² = x.
11. Quick Revision Points
- The slope of the tangent at a point equals the derivative at that point.
- Normal is perpendicular to tangent: m₁ × m₂ = -1.
- f'(x) > 0 ⇒ function is increasing; f'(x) < 0 ⇒ function is decreasing.
- For local maxima: f'(c) = 0 and f''(c) < 0.
- For local minima: f'(c) = 0 and f''(c) > 0.
- Absolute maxima/minima on [a, b]: Check critical points AND endpoints.
- In optimization, ensure the second derivative confirms the nature of the critical point.
12. Additional Worked Problems
Problem A: Find the equation of the tangent to the curve y = x³ - 4x + 1 at the point where x = 2.
Solution: At x = 2, y = 8 - 8 + 1 = 1. dy/dx = 3x² - 4 = 12 - 4 = 8. Tangent: y - 1 = 8(x - 2) ⇒ y = 8x - 15. Normal: y - 1 = (-1/8)(x - 2) ⇒ 8y - 8 = -x + 2 ⇒ x + 8y = 10.
Problem B: A farmer has 200 m of fencing to enclose a rectangular field along a river (no fence needed on the river side). Find the dimensions that maximize the area.
Solution: Let width = x (perpendicular to river), length = y (parallel to river). Fencing: 2x + y = 200 ⇒ y = 200 - 2x. Area A = xy = x(200 - 2x) = 200x - 2x². dA/dx = 200 - 4x = 0 ⇒ x = 50. d²A/dx² = -4 < 0 (maximum). y = 200 - 100 = 100. Maximum area = 50 × 100 = 5000 m². Dimensions: 50 m × 100 m.
Problem C: Show that the function f(x) = x³ - 6x² + 12x - 8 is increasing for all real x.
Solution: f'(x) = 3x² - 12x + 12 = 3(x² - 4x + 4) = 3(x - 2)² ≥ 0 for all x. Since f'(x) ≥ 0 everywhere and equals 0 only at x = 2 (an isolated point), f is increasing on R.
