How to Solve: Applications of Differentiation (Tangents, Normals, Turning Points, Optimisation, Rates of Change)

How to Solve: Applications of Differentiation (Tangents, Normals, Turning Points, Optimisation, Rates of Change)

Complete Guide for GCSE/A-Level Maths

Introduction

"Mastering differentiation applications doesn’t just get you marks—it unlocks real-world problems like designing the fastest race car, minimising costs in business, or even predicting how fast a balloon inflates. On your exam, this topic is worth 15-20% of your calculus questions, so nailing it could boost your grade by a full level. Let’s break it down step by step."

What You Need To Know First

Before diving in, you must already understand: 1. Basic differentiation rules (power rule, chain rule, product/quotient rules). 2. Equation of a straight line (gradient-intercept form, point-slope form). 3. Second derivatives (how to find and interpret them).

If any of these feel shaky, pause and review them first.

Key Vocabulary

Formulas To Know

Step-by-Step Method

1. Finding the Equation of a Tangent

Steps: 1. Differentiate the function to find dy/dx. 2. Substitute the x-coordinate of the point into dy/dx to find the gradient (m). 3. Use the point-slope form: y – y₁ = m(x – x₁). 4. Rearrange into y = mx + c if needed.

Example: Find the tangent to y = x³ – 3x at x = 2. 1. dy/dx = 3x² – 3 2. At x = 2: m = 3(2)² – 3 = 9 3. Point: (2, 2³ – 3(2)) = (2, 2) 4. Equation: y – 2 = 9(x – 2) → y = 9x – 16

2. Finding the Equation of a Normal

Steps: 1. Find the gradient of the tangent (m₁) using dy/dx. 2. The normal’s gradient (m₂) is -1/m₁. 3. Use the point-slope form with m₂.

Example: Find the normal to y = x² at x = 1. 1. dy/dx = 2x → At x = 1: m₁ = 2 2. m₂ = -1/2 3. Point: (1, 1) 4. Equation: y – 1 = -½(x – 1) → y = -½x + 1.5

3. Finding Turning Points

Steps: 1. Differentiate the function to get dy/dx. 2. Set dy/dx = 0 and solve for x. 3. Find the y-coordinate by substituting x back into the original function. 4. Use the second derivative test:
- If d²y/dx² > 0 → minimum.
- If d²y/dx² < 0 → maximum.
- If d²y/dx² = 0 → test fails (use first derivative test).

Example: Find and classify the turning points of y = x³ – 3x². 1. dy/dx = 3x² – 6x 2. Set 3x² – 6x = 0 → x = 0 or x = 2 3. y-values:
- At x = 0: y = 0
- At x = 2: y = 8 – 12 = -4 4. d²y/dx² = 6x – 6
- At x = 0: d²y/dx² = -6 → maximum at (0, 0)
- At x = 2: d²y/dx² = 6 → minimum at (2, -4)

4. Optimisation Problems

Steps: 1. Define variables (e.g., let x = length, y = width). 2. Write the function to optimise (e.g., area A = xy). 3. Express in one variable (e.g., if perimeter is fixed, 2x + 2y = 20 → y = 10 – x). 4. Differentiate the function and set dy/dx = 0. 5. Check nature (max/min) using second derivative or first derivative test. 6. Answer the question (e.g., "The maximum area is...").

Example: A farmer has 100m of fencing to make a rectangular pen. Find the dimensions for maximum area. 1. Let x = length, y = width. 2. Perimeter: 2x + 2y = 100 → y = 50 – x 3. Area: A = xy = x(50 – x) = 50x – x² 4. dA/dx = 50 – 2x → Set 50 – 2x = 0 → x = 25 5. d²A/dx² = -2 → maximum (since d²A/dx² < 0) 6. y = 50 – 25 = 25 → Dimensions: 25m × 25m

5. Rates of Change (Related Rates)

Steps: 1. Identify the given rate (e.g., dV/dt = 5 cm³/s). 2. Write the relationship between variables (e.g., volume V = (4/3)πr³). 3. Differentiate implicitly with respect to time (t). 4. Substitute known values and solve for the unknown rate.

Example: A spherical balloon inflates at 10 cm³/s. How fast is the radius increasing when r = 2 cm? 1. Given: dV/dt = 10 cm³/s 2. V = (4/3)πr³ 3. Differentiate: dV/dt = 4πr² (dr/dt) 4. Substitute r = 2, dV/dt = 10:
10 = 4π(2)² (dr/dt) → 10 = 16π (dr/dt) → dr/dt = 10/(16π) ≈ 0.2 cm/s

Worked Examples

Example 1 – Basic: Tangent to a Curve

Question: Find the equation of the tangent to y = x² + 3x at x = 1. Solution: 1. dy/dx = 2x + 3 2. At x = 1: m = 2(1) + 3 = 5 3. Point: (1, 1² + 3(1)) = (1, 4) 4. Equation: y – 4 = 5(x – 1) → y = 5x – 1 What we did and why: We found the gradient at the point using differentiation, then used the point-slope form to write the tangent’s equation.

Example 2 – Medium: Optimisation with Constraints

Question: A box with a square base has volume 500 cm³. Find the dimensions that minimise the surface area. Solution: 1. Let x = side of base, h = height. 2. Volume: x²h = 500 → h = 500/x² 3. Surface area: S = x² + 4xh = x² + 4x(500/x²) = x² + 2000/x 4. dS/dx = 2x – 2000/x² → Set 2x – 2000/x² = 0 → 2x³ = 2000 → x = 10 5. d²S/dx² = 2 + 4000/x³ → At x = 10: d²S/dx² = 6 > 0 → minimum 6. h = 500/10² = 5 → Dimensions: 10 cm × 10 cm × 5 cm What we did and why: We expressed the surface area in one variable using the volume constraint, then found the minimum by setting the derivative to zero.

Example 3 – Exam-Style: Rates of Change

Question: A ladder 5m long leans against a wall. The bottom slides away at 0.5 m/s. How fast is the top sliding down when the bottom is 3m from the wall? Solution: 1. Let x = distance from wall, y = height on wall. 2. x² + y² = 5² (Pythagoras) 3. Differentiate: 2x (dx/dt) + 2y (dy/dt) = 0 → x (dx/dt) + y (dy/dt) = 0 4. Given: dx/dt = 0.5 m/s, x = 3 m 5. Find y: 3² + y² = 25 → y = 4 m 6. Substitute: 3(0.5) + 4(dy/dt) = 0 → 1.5 + 4(dy/dt) = 0 → dy/dt = -0.375 m/s What we did and why: We used implicit differentiation to relate the rates, then substituted the known values to find the unknown rate.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet for differentiation applications: 1. Tangents/Normals: Differentiate, find gradient, use y – y₁ = m(x – x₁). Normals are perpendicular (m₂ = -1/m₁). 2. Turning Points: Set dy/dx = 0, find y, then use d²y/dx² to classify (positive = min, negative = max). 3. Optimisation: Write the function, express in one variable, differentiate, set to zero, check nature. 4. Rates of Change: Differentiate implicitly, substitute known rates, solve for the unknown. Common traps? Forgetting units, mixing up tangents/normals, and not checking turning point nature. Now go smash that exam!