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Parametric Equations

Parametric Equations – definition and example, Precalculus math glossary – Mathovore

Key takeaway: Parametric equations describe a curve by writing both x and y as functions of a third variable, the parameter t, often representing time as the point moves.

Grade 12 – PrecalculusF-IF.C.7Functions and Graphs

Definition of Parametric Equations

Definition

Parametric equations give a curve a clock. Rather than a single rule linking y to x, two separate rules tell you where a moving point is at each moment: one rule for the horizontal position and one for the vertical position, both fed by the parameter \(t\).

This approach handles motion naturally, since a thrown ball, a rolling wheel or a planet all have positions that depend on time. It also describes curves that fail the vertical line test, such as a full circle. To see the shape without the clock, solve one equation for \(t\) and substitute into the other, which is called eliminating the parameter.

Formula and notation

A parametric curve is a pair of rules over an interval of the parameter:

\[x=f(t),\quad y=g(t),\qquad a\le t\le b.\]

The curve is traced in the direction of increasing \(t\). Starting and ending points come from \(t=a\) and \(t=b\).

t = 00.511.522.5peak 30 ft at t = 1.25x = 18ty = −16t² + 40t + 5
Each marked dot is the ball’s position at a different time, half a second apart.

Key facts

  • Each value of t produces exactly one point (x, y) on the plane.
  • Eliminating the parameter gives an ordinary equation in x and y, though it may hide the direction.
  • The same path can be traced faster or slower by changing the rules for t.
  • Lines, projectile parabolas and circles (with cosine and sine) are common cases of parametric equations.

Parametric Equations example

Problem

A ball is thrown so that its horizontal position is \(x=18t\) feet and its height is \(y=-16t^2+40t+5\) feet, with \(t\) in seconds. Find its maximum height, and write y as a function of x.

  1. The height rule is a downward parabola in \(t\) with vertex at \(t=\frac{-40}{2(-16)}=1.25\).

  2. Height at that moment: \(-16(1.5625)+50+5=-25+55=30\) feet.

  3. Eliminate \(t\): from \(x=18t\) we get \(t=\frac{x}{18}\), so \(y=-\frac{4}{81}x^2+\frac{20}{9}x+5\).

The maximum height is 30 feet at 1.25 seconds, and the path is \(y=-\frac{4}{81}x^2+\frac{20}{9}x+5\).

Common mistake

Watch out

Noelle graphs the ball’s height \(y\) against \(t\) and says the picture shows the flight path through the air.

Why it is wrong

That graph shows height versus time. The flight path plots \(y\) against \(x\), so each point uses both rules at the same \(t\). With parametric equations, the path always means y against x, not either one against t.

Related terms and lessons

Browse the glossary: A–Z list · Precalculus vocabulary · Functions and Graphs

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