Library › Pure Mathematics 4 WMA14 › Further Parametric Equations

A2 Level · Pure Mathematics 4 WMA14

Further Parametric Equations

Revise Further Parametric Equations for Pure Mathematics 4 WMA14 (A2 Level) — revision notes and instant AI marking.

📖 Revision notes · preview

What You Need to Know

The Big Idea

Instead of defining a curve as y = f(x), parametric equations describe both x and y separately using a third variable called a parameter (usually t):

x = f(t) and y = g(t)
This parameter acts as a bridge — you use it to find gradients, areas, and volumes without having to convert back to y = f(x).
The three main skills are: differentiation (finding gradients), integration (finding areas), and volumes of revolution.

Parametric Differentiation

Why Do We Need a Special Method?

When a curve is defined parametrically, we can't directly find dy/dx by differentiating y with respect to x. Instead, we use the chain rule with the parameter as a stepping stone.

The Core Formula

dy/dx = (dy/dt) ÷ (dx/dt)

Or equivalently: dy/dx = (dy/dt) × (dt/dx), where dt/dx = 1 ÷ (dx/dt)

Why This Works

Think of the parameter t as a clock. As time (t) changes by a tiny amount, both x and y change. The rate at which y changes per unit time is dy/dt, and the rate x changes per unit time is dx/dt. To find how fast y is changing with respect to x, you divide: (change in y) ÷ (change in x) = (dy/dt) ÷ (dx/dt).

Key Point

dy/dx will be expressed in terms of t — this is perfectly fine. You don't need to eliminate the parameter.

Finding Gradients at a Point

To find the gradient of a parametric curve at a specific point, follow these 4 steps:

Step 1: Find dx/dt and dy/dt

Differentiate both equations separately with respect to t.

Step 2: Find dy/dx in terms of t

Use the formula: dy/dx = (dy/dt) ÷ (dx/dt). Leave it in terms of t.

Step 3: Find the value of t at the required point

Use the condition given (e.g., "where x = 5") to find t. Substitute into x = f(t) and solve.

Step 4: Substitute into dy/dx

Plug this value of t into your expression for dy/dx to get the numerical gradient.

Worked Example: Finding a Gradient

Problem: A curve is defined parametrically by x = 2t − 3 and y = t³ − 4t + 1. Find the gradient at the point where x = 5.

Step 1: Differentiate

dx/dt = 2
dy/dt = 3t² − 4

Step 2: Find dy/dx

dy/dx = (3t² − 4) ÷ 2 = ½(3t² − 4)

Step 3: Find t when x = 5

5 = 2t − 3
8 = 2t
t = 4

Step 4: Substitute t = 4

dy/dx = ½(3(4)² − 4) = ½(48 − 4) = ½(44) = 22

The gradient at x = 5 is 22.

Tangent and Normal Lines

Once you have the gradient (m) and the coordinates (x₁, y₁) of a point, you use the standard straight-line formula:

Tangent Line Equation

y − y₁ = m(x − x₁)

For a normal (perpendicular to the tangent), the gradient is the negative reciprocal:

Normal Gradient (Perpendicular)

m_normal = −1/m_tangent

Or: m₁ × m₂ = −1

Common Mistake

y-coordinate both

🔓 Read the full Further Parametric Equations note → You're seeing the preview · sign in to read it all

Also in the full note

Parametric Integration
Parametric Volumes of Revolution
Key Terms to Memorise
Concepts Checklist
Exam Tips & Common Mistakes
Quick Reference — All Formulas
Stationary Points (dy/dx = 0)
Finding Areas Under Parametric Curves

What's inside

📖 Revision notes 🎯 Learn mode ✦ AI flashcards ✓ Instant AI marking 🧊 3D explorers 🧪 Experiments & simulations 📈 Progress tracking

📄 Practise Further Parametric Equations with Pure Mathematics 4 WMA14 past papers Every paper with its mark scheme — answer online, marked instantly. Open →

Read the full Further Parametric Equations notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →

Further Parametric Equations