For a curve given by parametric equations x = f(t) and y = g(t), differentiate each with respect to t and divide: dy/dx = (dy/dt) ÷ (dx/dt). The answer is usually in terms of t. To find a gradient or tangent at a point, substitute the value of t, and use the parametric equations to get the coordinates.
Parametric differentiation
- dy/dx = (dy/dt) ÷ (dx/dt)
- Gradient at t = t₀: substitute t₀ into dy/dx
- Point at t = t₀: (f(t₀), g(t₀))
- Horizontal tangent: dy/dt = 0 (with dx/dt ≠ 0)
- Vertical tangent: dx/dt = 0
How do you differentiate parametric equations?
Differentiate x and y separately with respect to the parameter t, then divide dy/dt by dx/dt. This comes from the chain rule: dy/dx = dy/dt × dt/dx, and dt/dx = 1/(dx/dt). The answer usually still contains t, which is expected.
Dividing the wrong way round, dx/dt over dy/dt, is the most common slip.
Finding the gradient at a given value of t
Find dy/dx in terms of t first, then substitute the given value of t. Never substitute the value before differentiating, because that loses the function you need to differentiate.
Equations of tangents and normals
A tangent needs a gradient and a point. Find the gradient by substituting t₀ into dy/dx, and find the point by substituting t₀ into both parametric equations. Then use y − y₁ = m(x − x₁), or the negative reciprocal gradient for a normal. Forgetting to find the actual coordinates is a frequent loss of marks.
Stationary points and vertical tangents
dy/dx is zero when its numerator, dy/dt, is zero, provided dx/dt is not also zero there. So to find horizontal tangents, solve dy/dt = 0 for t, then find the coordinates. A vertical tangent occurs where dx/dt = 0, because the gradient becomes undefined.
Exponential, log and trig curves
The method does not change when the parametric equations use eᵗ, ln t or trig functions. Differentiate each part with the standard results, divide, and simplify; common factors such as eᵗ often cancel. For trig parameters, identities can simplify the final expression.
Worked examples
Example 1. Given x = t² and y = t³, find dy/dx.
- dx/dt = 2t and dy/dt = 3t².
- dy/dx = 3t² / 2t.
Answer: dy/dx = 3t/2.
Marks: M1 for dividing the derivatives, A1 for 3t/2.
Example 2. For x = t³ and y = t², find dy/dx at t = 2.
- dy/dx = 2t / 3t² = 2/(3t).
- At t = 2: 2/6 = 1/3.
Answer: 1/3.
Marks: M1 for 2/(3t), A1 for 1/3.
Example 3. Find the equation of the tangent to x = t², y = 2t at t = 2.
- dx/dt = 2t and dy/dt = 2, so dy/dx = 1/t. At t = 2 the gradient is ½.
- Point: x = 4, y = 4.
- Tangent: y − 4 = ½(x − 4).
Answer: y = ½x + 2.
Marks: M1 for the gradient ½, M1 for the point (4, 4), A1 for the equation.
Example 4. Find the values of t for which x = t², y = t³ − 3t has a horizontal tangent.
- dy/dt = 3t² − 3. Set it to zero: 3(t² − 1) = 0, so t = ±1.
- Check dx/dt = 2t is not zero at t = ±1.
Answer: t = 1 and t = −1.
Marks: M1 for 3t² − 3 = 0, A1 for t = ±1.
Example 5. Find dy/dx for x = eᵗ, y = teᵗ.
- dx/dt = eᵗ. Product rule: dy/dt = eᵗ + teᵗ = eᵗ(1 + t).
- dy/dx = eᵗ(1 + t) / eᵗ.
Answer: dy/dx = 1 + t.
Marks: M1 for the product rule on teᵗ, A1 for 1 + t.
Common mistakes
- Dividing the wrong way round. dy/dx = (dy/dt)/(dx/dt), not (dx/dt)/(dy/dt).
- Substituting t before differentiating. Find dy/dx in terms of t first, then substitute.
- Not finding the coordinates. A tangent needs the point (x, y), found from the parametric equations.
- Mixing up horizontal and vertical tangents. Horizontal: dy/dt = 0. Vertical: dx/dt = 0.
- Forgetting the product rule. A term like teᵗ needs the product rule when differentiating with respect to t.
How parametric differentiation is examined
Parametric differentiation is tested on the Pure 4 paper (WMA14) in Edexcel International A-Level, and in the second year of Edexcel A-Level (9MA0). Common question parts:
- show that dy/dx can be written in a given form
- find the gradient and coordinates at a given value of t
- find a tangent or normal, often in the form ax + by + c = 0
- find points where the tangent is horizontal or vertical, or explain why there are none
Full questions with mark schemes are among the Pure 4 differentiation exam questions. Then build fluency with the Pure 4 differentiation worksheet, or generate fresh parametric curves in the revision engine for A-Level differentiation.
For converting parametric equations to Cartesian form, see parametric equations. A related technique is implicit differentiation.
Frequently asked questions
What is the formula for parametric differentiation?
dy/dx = (dy/dt) ÷ (dx/dt).
Why does the answer contain t?
Because the gradient changes as you move along the curve, and t tells you where you are. Substitute a value of t to get a number.
How do you find a stationary point on a parametric curve?
Set dy/dt = 0, check dx/dt is not zero there, solve for t, then substitute into x and y for the coordinates.