Maths › Coordinate geometry › Parametric equations

Parametric equations

Instead of tying y to x directly, let both answer to a third variable, and a curve becomes a moving point with a clock. Parametric equations trace circles, parabolas and flight paths one instant at a time, and eliminating the parameter translates them back into Cartesian when needed.

Year 12-13EDEXCEL 9MA0 3.3, 3.4

Builds on Circles and Trigonometric graphs and equations.

IN THIS TOPIC

  • Trace and sketch curves given parametrically, treating the parameter as a clock.
  • Convert between parametric and Cartesian forms, watching the parameter's domain.
  • Use parametric equations as models, including projectile-style motion.

WHAT YOU PROBABLY THINK

Eliminating the parameter changes nothing.

A point with a clock

A parametric curve gives x and y separately as functions of a parameter t, and the curve is everywhere the point (x(t), y(t)) visits as t runs. The parameter often reads naturally as time, and the curve as a trail.

The curve with parametric equations x equals t squared and y equals 2t, traced as t runs from minus 2 to 2: the points sweep up the left-opening parabola y squared equals 4x, with t labelling the journey ordert = 2t = 1t = −1t = −2x = t², y = 2t
FIG. 1x = t², y = 2t traced from t = −2 to 2. The dots mark the clock readings; the trail they leave is a sideways parabola.

WORKED EXAMPLE

Plot first, name later

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), (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 substitute: x = (y/2)2.

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

The simpler of the two equations is nearly always the door: solve it for t and feed the other one.

Converting, with care

Trig pairs convert by identity rather than substitution. 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. And for x = 5t, y = 5/t, multiplying gives xy = 25 directly. The care comes with the domain of t, which is what the opening lie ignores. The parametric point never has t = 0 in the second example, and a parametrisation can cover only part of its Cartesian curve.

WORKED EXAMPLE

The half that is really there

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 describes where the point is allowed; the parameter's range says where it actually goes. Stating the restriction is the mark most often dropped in this topic.

YOUR TURN

Two spec-book classics

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, with t ≠ 0 matching the curve's own refusal to touch the axes.

One conversion ran on an identity, the other on cancellation. Which tool fits is usually visible at a glance from the pair.

Parametric models

Motion is parametric by nature, one clock driving two coordinates, which is why projectiles land in this topic and again in the applied paper.

A projectile modelled parametrically by x equals 20t and y equals 15t minus 5 t squared: the flight peaks at 11.25 metres after 1.5 seconds and lands 60 metres away after 3 secondst = 1.5 s: peak 11.25 mt = 3 s: lands at 60 mx = 20t, y = 15t − 5t²
FIG. 2x = 20t, y = 15t − 5t²: horizontal and vertical motion on one clock. Peak of 11.25 m at t = 1.5 s, landing 60 m away at t = 3 s.

TRY IT UNSEEN

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, at t = 1.5, giving y = 11.25 m.

Substituting t = x/20 into y: y = 3x/4 − x2/80, the parabola of the figure.

The parametric form answered the when questions and the Cartesian form describes the shape; each earns its keep, which is why the spec asks for fluency in both directions.

THE EXAM BIT

  • Sketch parametric curves from a t-table, 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 misses; the restriction is a mark on its own.
  • In motion models, landing means y = 0, greatest height means the vertex in t, and each is a one-line solve.
  • Keep answers exact until the last step, particularly with trig parameters, where decimal t-values wreck 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 a point with a clock; the trail is the graph.

Eliminate t by substitution or identity, then say which part of the curve the parameter really visits.

CHECK YOUR PROGRESS

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  • Trace and sketch curves given parametrically, treating the parameter as a clock.
  • Convert between parametric and Cartesian forms, watching the parameter's domain.
  • Use parametric equations as models, including projectile-style motion.

No animated video for this topic yet; these notes stand alone.