Maths › Coordinate geometry › Parametric equations

Parametric equations

Instead of relating y to x directly, both can be given in terms of a third variable, so a curve is traced out as that parameter increases. Parametric equations describe circles, parabolas and flight paths naturally, and eliminating the parameter converts them back to Cartesian form when a question calls for it.

Year 12-13EDEXCEL 9MA0 3.3, 3.4

Builds on Circles and Trigonometric graphs and equations.

IN THIS TOPIC

  • Trace and sketch a curve given parametrically, using values of the parameter.
  • Convert between parametric and Cartesian forms, and say which part of the Cartesian curve you actually get.
  • Use parametric equations as models, projectile motion included.

COMMON MISCONCEPTION

Eliminating the parameter changes nothing.

The Cartesian equation gives the curve the points lie on; the range of the parameter decides which part of it is covered. For x = t2, y = t4 the equation is y = x2, but x = t2 ≥ 0 for every real t, so the curve is only the right-hand half of that parabola. State the restriction as part of the answer.

A point with a clock

A parametric curve gives x and y separately as functions of a parameter t, and the curve is the set of points (x(t), y(t)) for t in the stated range. In a motion model the parameter is time.

A sideways parabola swept out by the computed points of x equals t squared, y equals 2t, with amber dots labelled by their t values from minus 2 up to 2: the parameter is a clock, and the curve is where the point has been.
FIG. 1x = t², y = 2t for −2 ≤ t ≤ 2. The dots mark the plotted values of t, and the curve through them is a parabola with a horizontal axis.

WORKED EXAMPLE

Plotting, then converting

Sketch the curve x = t2, y = 2t for −2 ≤ t ≤ 2, and find its Cartesian equation.

A short table of t-values gives (4, −4), (1, −2), (0, 0), (1, 2) and (4, 4), sweeping up the right-opening curve in the figure.

To convert, make t the subject of the simpler equation. t = y/2, and substituting gives x = (y/2)2.

The Cartesian equation is y2 = 4x, a parabola on its side.

Make t the subject of the simpler of the two equations, then substitute into the other.

Converting, with care

A pair defined by sin and cos is converted using an identity rather than by making t the subject. For x = 3 cos t, y = 3 sin t, squaring and adding uses sin2 t + cos2 t = 1 to give x2 + y2 = 9, a circle of radius 3. For x = 5t, y = 5/t, multiplying gives xy = 25 straight away. The domain of t must be stated: the second pair excludes t = 0, and a parametrisation may cover only part of the Cartesian curve.

x equals 5t, y equals 5 over t with t positive traces only the first-quadrant branch of the hyperbola x y equals 25: the third-quadrant branch belongs to negative t, and t 0 is excluded
Describe the drawing

Axes with the hyperbola x y equals 25 ghosted in dashed grey in both the first and third quadrants, and the branch actually traced by x equals 5t, y equals 5 over t with t positive drawn solid in cyan through marked points at t equals one half, 1, 2 and 5: the parameter's domain decides which part of the Cartesian curve exists, and t equals 0 never happens.

FIG. 2Both branches of the Cartesian curve xy = 25 are drawn dashed. The parametrisation x = 5t, y = 5/t with t > 0 gives only the solid first-quadrant branch, through the marked points at t = ½, 1, 2 and 5. Negative t gives the other branch, and t = 0 is excluded.

WORKED EXAMPLE

Which part of the Cartesian curve

The curve C has parametric equations x = t2, y = t4. Find its Cartesian equation, and state which part of that Cartesian curve C actually is.

Since y = (t2)2, the Cartesian equation is y = x2.

But x = t2 ≥ 0 for every real t, so C is only the right-hand half of the parabola, x ≥ 0.

The Cartesian equation gives the curve on which the points lie; the range of the parameter determines which part of it the parametrisation covers. State that restriction as part of the answer.

GUIDED PRACTICE

Two classic pairs

Find Cartesian equations for (a) x = 3 cos t, y = 3 sin t, and (b) x = 5t, y = 5/t with t ≠ 0, before opening the working.

Show the working

(a) Square and add. x2 + y2 = 9 cos2 t + 9 sin2 t = 9, the circle of radius 3 about the origin, fully traced as t runs through a period.

(b) Multiply. xy = (5t)(5/t) = 25, the reciprocal curve; t ≠ 0 matches the fact that the curve never meets the axes.

One conversion used an identity and the other used cancellation, each determined by the form of the pair.

Parametric models

In a motion model both coordinates are functions of the same time variable, which is why projectile motion appears in this topic and in mechanics.

A computed projectile arc rising to its marked peak of 11.25 metres at one and a half seconds and returning to the ground 60 metres downrange at three seconds: the two parametric equations run the horizontal and vertical motions on the same clock.
FIG. 3x = 20t, y = 15t − 5t²: horizontal and vertical motion in terms of the same t. Greatest height 11.25 m at t = 1.5 s, landing 60 m away at t = 3 s.

INDEPENDENT PRACTICE

A flight, fully read

A ball's flight is modelled by x = 20t, y = 15t − 5t2, in metres and seconds. Find when and where it lands, its greatest height, and the Cartesian equation of its path.

Show the working

Landing. y = 0 gives 5t(3 − t) = 0, so t = 3, and x = 20 × 3 = 60 m.

Greatest height. y peaks midway between the roots, at t = 1.5, giving y = 11.25 m.

Substituting t = x/20 into y gives y = 3x/4 − x2/80, the parabola of the figure. The model only runs from launch to landing, 0 ≤ t ≤ 3, so the flight is the piece of that parabola with 0 ≤ x ≤ 60; the unrestricted curve carries on into negative x and below the ground, points the ball never visits.

The parametric form gives the times, and the Cartesian form gives the shape of the path, with the domain stating which part of it the flight covers. Edexcel 9MA0 3.3 requires conversion in both directions.

ASSESSMENT FOCUS

  • Sketch parametric curves from a t-table, with arrows on the curve showing the direction of increasing t.
  • Convert by making t the subject of the simpler equation, or by a trig identity when sin and cos both appear.
  • State the domain of t, and any part of the Cartesian curve the parametrisation does not cover.
  • In motion models, landing is given by y = 0 and the greatest height by the vertex in t. Keep values exact until the final step, since a rounded value of t affects the later parts.

CHECK YOURSELF

The curve C is given by x = 2 cos t, y = 2 sin t for 0 ≤ t ≤ π. Find the Cartesian equation of C, and state precisely which points of that Cartesian curve C consists of.

Show a hint

Square and add; then think about what y does for t in the top half of a period.

Show the answer

Squaring and adding gives x2 + y2 = 4, the circle of radius 2.

For 0 ≤ t ≤ π the sine is never negative, so y ≥ 0 throughout.

C is the upper semicircle, from (2, 0) round through (0, 2) to (−2, 0), and not the full circle the Cartesian equation alone would suggest.

A parametric curve is the set of points (x(t), y(t)) for t in a stated range.

Eliminate t by substitution or by an identity, then state which part of the Cartesian curve the parameter range covers.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

PDF · 76 KB · screen edition · navy

Answer book

PDF · 79 KB · fully worked · navy
15 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the parametric equations questions page.

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CHECK YOUR PROGRESS

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  • Trace and sketch a curve given parametrically, using values of the parameter.
  • Convert between parametric and Cartesian forms, and say which part of the Cartesian curve you actually get.
  • Use parametric equations as models, projectile motion included.

Open the full revision checklist to see every objective in the course in one place.