PrecalculusParametric Equations

Parametric Equations

x and y both depend on a third variable t.

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A graph like $y = x^2$ shows a path but not how something travels along it. To describe motion you want to know where an object is at each moment. So let both coordinates depend on a third variable, usually time $t$. You already know the shapes of basic functions and the unit circle; both reappear here.

Key idea

Parametric equations $x = f(t)$, $y = g(t)$ give a point for each value of the parameter $t$. As $t$ increases, the point traces a curve in a definite direction, starting and stopping where the $t$-interval says.

Why it works

Think of a bug crawling across graph paper. At every moment it has an $x$-coordinate and a $y$-coordinate, each a function of time. Listing $(x(t), y(t))$ for increasing $t$ plots its trail in order. Two different pairs of equations can trace the same trail in different directions or at different speeds, and parametric equations keep that information.
Drag the point or press Play. The green arrowheads show which way the point moves as $t$ increases.

Worked examples

Worked example: A table of points

Plot $x = 2t + 1$, $y = t^2 - 3$ for $-2 \le t \le 2$.
  1. Make a table: $t = -2$: $(-3, 1)$; $t = -1$: $(-1, -2)$; $t = 0$: $(1, -3)$; $t = 1$: $(3, -2)$; $t = 2$: $(5, 1)$.
  2. Plot the points in order of $t$ and join them.
  3. The path is part of a parabola, traced from left to right as $t$ increases.

Answer: A parabolic arc from $(-3, 1)$ to $(5, 1)$, moving rightward

Worked example: Same path, different motion

Compare $x = \cos t$, $y = \sin t$ with $x = \sin t$, $y = \cos t$ for $0 \le t \le 2\pi$.
  1. Both satisfy $x^2 + y^2 = 1$: the same circle.
  2. The first starts at $(1, 0)$ and at $t = \frac{\pi}{2}$ is at $(0, 1)$: counterclockwise.
  3. The second starts at $(0, 1)$ and at $t = \frac{\pi}{2}$ is at $(1, 0)$: clockwise.

Answer: Same circle, opposite directions and different starting points

Worked example: When is it there?

For $x = t^2 - 1$, $y = 2t$, at what time is the point at $(3, 4)$?
  1. From $y$: $2t = 4$, so $t = 2$.
  2. Check $x$: $2^2 - 1 = 3$. ✓ (From $x$ alone, $t = \pm 2$; the $y$-equation picks $t = 2$.)

Answer: $t = 2$

Common mistake

Losing the direction of motion. The equation $x^2 + y^2 = 1$ describes the circle but cannot tell you which way the point travels, where it starts, or how many times it goes around. To find the direction, plot the points for two or three increasing values of $t$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Builds on

Leads to

Also used in other ladders:

Sources and further reading

  • OpenStax Precalculus 2e, Section 8.6 Parametric Equations
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