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 a question wants that.
Builds on Circles and Trigonometric graphs and equations.
IN THIS TOPIC
- Trace and sketch a curve given parametrically, treating the parameter as a clock.
- 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.
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 as time, and then the curve is a trail.
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) 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.
The simpler of the two equations is nearly always the door. Solve that one for t and feed the other.
Converting, with care
A pair built from sin and cos wants an identity, and making t the subject is the wrong door. 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 care comes with the domain of t. The second pair never has t = 0, 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 says where the point is allowed to be. The parameter's range says where it actually goes, and stating that restriction is the mark most often dropped in this topic.
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, with t ≠ 0 matching the curve's own refusal to touch the axes.
One conversion ran on an identity and 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. That is how projectiles end up in this topic and again on the applied paper.
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 parametric form answered the when questions and the Cartesian form describes the shape. Each earned its keep, and Edexcel asks you to move between them 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 misses. That restriction is a mark on its own.
- In motion models, landing means y = 0 and greatest height means the vertex in t, each a one-line solve. Keep the values exact until the last step, because decimal t-values wreck 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 a point with a clock, and the trail it leaves is the graph.
Eliminate t by substitution or by identity, then say which part of the curve the parameter really visits.
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.
Or read them with their worked answers on the parametric equations questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Trace and sketch a curve given parametrically, treating the parameter as a clock.
- 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.