Parametric equations

Parametric equations describe a curve by giving x and y separately in terms of a third variable, the parameter, usually t or θ: for example x = t − 2 and y = t² + 1. Each value of t gives one point (x, y), and as t changes the point traces out the curve. To find the Cartesian equation, the ordinary equation linking x and y, eliminate the parameter: here t = x + 2, so y = (x + 2)² + 1. The gradient of a parametric curve is dy/dx = (dy/dt) ÷ (dx/dt).

What parametric equations are

Think of t as time: x(t) and y(t) tell you where a moving point is at each moment. That makes parametric equations natural for motion, such as a projectile or a planet in orbit, and for curves like circles that cannot be written as a single function y = f(x). Pearson Edexcel's A-level specification asks students to understand and use parametric equations of curves, to convert between Cartesian and parametric forms, and to use parametric equations in modelling.

Plotting from a table of values

Choose values of t, work out x and y for each, and plot the points in order of t. For x = t − 2, y = t² + 1:

The points lie on a parabola with its lowest point at (−2, 1), where t = 0.

Here x and y are the usual coordinates; each labelled point shows the value of t that produces it.

Converting parametric equations to a Cartesian equation

There are two main methods. Which to use depends on how t appears.

Method 1: rearrange and substitute

If one equation can be rearranged for t, substitute it into the other.

Example 1. Find the Cartesian equation of x = t − 2, y = t² + 1.

From the first equation, t = x + 2. Substituting:

Method 2: use a trig identity

If the equations involve sin and cos of the parameter, use an identity such as sin²θ + cos²θ = 1 to get rid of θ.

Example 2: the parametric equations of a circle. Find the Cartesian equation of x = 3 cos t, y = 3 sin t.

Divide by 3 to get cos t = x ÷ 3 and sin t = y ÷ 3, then use cos²t + sin²t = 1:

This is a circle of radius 3 centred on the origin. The parameter t is the angle from the positive x-axis to the point, as the diagram shows.

In general, x = a + r cos t, y = b + r sin t is a circle of radius r centred at (a, b).

Example 3: an ellipse

Find the Cartesian equation of x = 1 + 2 cos θ, y = 2 − sin θ.

Rearrange: cos θ = (x − 1) ÷ 2 and sin θ = 2 − y. Then cos²θ + sin²θ = 1 gives:

Example 4: watch the domain

Find the Cartesian equation of x = sin θ, y = cos 2θ, and state the possible values of x.

Use the double angle identity cos 2θ = 1 − 2 sin²θ:

But x = sin θ can only take values from −1 to 1, so the curve is only the part of the parabola with −1 ≤ x ≤ 1. Edexcel's specification tells students to pay attention to the domain of the parameter, because a specific section of a curve may be described.

Here is a quick check.

Differentiating parametric equations

To find dy/dx without converting first, differentiate x and y with respect to t, then divide:

This is the chain rule. Edexcel's specification includes differentiation of functions defined parametrically, for the first derivative, and finding tangents and normals to curves given parametrically.

Example 5: a tangent and a stationary point

A curve has parametric equations x = 2t + 1, y = t² − 6t + 2. Find the equation of the tangent where t = 1, and the coordinates of the stationary point.

dx/dt = 2 and dy/dt = 2t − 6, so:

At t = 1, the point is (3, −3) and the gradient is 1 − 3 = −2. The tangent is y + 3 = −2(x − 3), which is y = −2x + 3.

A stationary point has dy/dx = 0, so t = 3, giving the point (7, −7).

Parametric equations in modelling

Parametric equations describe motion directly, with t as time. Edexcel's specification gives modelling as an application, and says it may also appear in kinematics.

Example 6: constant velocity

A boat sails at a steady velocity from (2, 3) at t = 0 to (10, 9) at t = 4, with distances in km and t in hours. Find parametric equations for its position, and its speed.

x rises by 8 in 4 hours and y by 6, so:

Its speed is √(2² + 1.5²) = √6.25 = 2.5 km/h. For a projectile launched at speed u and angle α, the same idea gives x = (u cos α)t and y = (u sin α)t − ½gt².

The Custom Graph plots the Cartesian form of a curve so you can check its shape, and the Derivative Calculator differentiates x(t) and y(t) step by step. For the area under a parametric curve, see our guide to parametric integration.

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Common mistakes

• Forgetting the domain. If x = sin θ, the curve only exists for −1 ≤ x ≤ 1, even if the Cartesian equation suggests more.
• Dividing the wrong way round. dy/dx = (dy/dt) ÷ (dx/dt): y on top.
• Leaving t in the answer when asked for a Cartesian equation, or not substituting t when asked for a gradient at a point.
• Using the wrong identity. Look at the trig functions present: sin and cos suggest sin² + cos² = 1; sin θ and cos 2θ suggest a double angle formula; sec and tan suggest 1 + tan² = sec².
• Squaring and losing information. Squaring can add points that the original curve does not have, so check the range of x and y.

Practise calculus in a short timed paper

A-level Maths: Calculus gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so parametric curves turn up regularly but not in every paper.

Read next

Maths · CalculusParametric differentiation: dy/dx, tangents and normalsHow to differentiate parametric equations: dy/dx = (dy/dt) ÷ (dx/dt), with worked examples for gradients, tangents, normals and stationary points.