Parametric differentiation

Parametric differentiation finds the gradient of a curve given by parametric equations, where x and y are both written in terms of a third variable, the parameter t. Differentiate each with respect to t, then divide: dy/dx = (dy/dt) ÷ (dx/dt). The answer is usually in terms of t, so substitute the value of t at the point you want.

The parametric differentiation formula

This is the chain rule rearranged. Think of y as a function of x, and x as a function of t; the chain rule links the three rates of change, and dividing through by dx/dt gives the formula. It works whenever dx/dt is not zero.

AQA's A-level Maths specification asks you to differentiate simple functions and relations defined implicitly or parametrically, for the first derivative only, and to use derivatives to find gradients, tangents, normals and stationary points. The parametric equations guide covers what the equations mean and how to convert them to Cartesian form.

How to differentiate parametric equations

1. Differentiate x with respect to t to get dx/dt.
2. Differentiate y with respect to t to get dy/dt.
3. Divide: dy/dx = (dy/dt) ÷ (dx/dt), and simplify.
4. For a gradient at a point, find the value of t there and substitute it.

Example 1: a polynomial curve

Take the curve x = t² − 3, y = t³ − 12t. Find dy/dx, and the gradient at the point where t = 1.

• dx/dt = 2t and dy/dt = 3t² − 12.
• dy/dx = (3t² − 12) ÷ 2t = (3t² − 12)/(2t).
• At t = 1: dy/dx = (3 − 12) ÷ 2 = −4.5. The point is x = −2, y = −11, so the gradient at (−2, −11) is −4.5.

Example 2: trigonometric parameters

A curve has x = 4 sin θ and y = cos 2θ. Find dy/dx in its simplest form.

• dx/dθ = 4 cos θ and dy/dθ = −2 sin 2θ.
• Use sin 2θ = 2 sin θ cos θ: dy/dθ = −4 sin θ cos θ.
• dy/dx = (−4 sin θ cos θ) ÷ (4 cos θ) = −sin θ.

At θ = π/6, the point is (2, ½) and the gradient is −sin(π/6) = −½. Simplifying with an identity before substituting often makes the answer much neater.

Tangents and normals to a parametric curve

Once you have the gradient, the method is the same as for any curve: find the point, find the gradient, then use y − y₁ = m(x − x₁). The normal meets the tangent at a right angle, so its gradient is −1/m.

Example 3: tangent and normal

Find the equations of the tangent and normal to the curve x = 2t, y = t² + 3/t (t > 0) at the point where t = 1.

1. The point: x = 2 and y = 1 + 3 = 4, so (2, 4).
2. dx/dt = 2 and dy/dt = 2t − 3/t², so dy/dx = (2t − 3/t²) ÷ 2 = t − 3/(2t²).
3. At t = 1 the tangent's gradient is 1 − 3/2 = −½, so y − 4 = −½(x − 2). Multiplying by 2 and tidying gives x + 2y = 10.
4. The normal has gradient −1 ÷ (−½) = 2, so y − 4 = 2(x − 2), which gives y = 2x. It passes through the origin.

Because the axes have the same scale, you can see the tangent and normal meet at a true right angle. This curve is in fact y = x²/4 + 6/x: substituting t = x/2 into y gives it, and differentiating that directly gives x/2 − 6/x², which is 1 − 1.5 = −½ at x = 2, the same answer.

Stationary points and vertical tangents

A fraction is zero when its top is zero, so:

Example 4: finding stationary points

For the curve in example 1, x = t² − 3 and y = t³ − 12t, find the stationary points.

Set dy/dt = 0: 3t² − 12 = 0, so t² = 4 and t = ±2. At both, dx/dt = 2t is not zero, so these are genuine stationary points. When t = 0, dx/dt = 0 while dy/dt = −12, so the curve has a vertical tangent at (−3, 0).

Find the coordinates by putting each t back into both equations. t = 2 gives x = 1, y = 8 − 24 = −16, and t = −2 gives x = 1, y = −8 + 24 = 16, so the stationary points are (1, −16) and (1, 16).

Common mistakes

• Dividing the wrong way. It is dy/dt over dx/dt: the y derivative on top.
• Substituting x instead of t. dy/dx is in terms of t; find the t value for the point first.
• Using one equation only. A point needs both coordinates: put t into x and into y.
• Forgetting the chain rule in the parts. d/dθ(cos 2θ) = −2 sin 2θ, not −sin 2θ.
• Ignoring dx/dt = 0. If both derivatives are zero at the same t, the formula breaks down there, so check before calling it stationary.

The derivative calculator differentiates each equation with working, and the parametric integration guide does the reverse, finding areas under parametric curves. For practice in a short timed paper, try today's A-level Maths: Calculus paper.

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