Parametric Equations and Curves

Main idea

Instead of tying $y$ to $x$ , let both coordinates depend on a third variable $t$ , usually time. Then a curve is the path of a moving point, so it can loop, cross itself and double back, which no graph $y = f(x)$ can do.

x = {\htmlData{sym=1}{f(t)}}, \qquad y = {\htmlData{sym=2}{g(t)}}, \qquad {\htmlData{sym=3}{\alpha}} \le {\htmlData{sym=0}{t}} \le {\htmlData{sym=4}{\beta}}

What the symbols mean

$t$
Parameter
Think of it as time: each value of t gives one point on the curve.
$f(t)$
Horizontal position
The x-coordinate at time t.
$g(t)$
Vertical position
The y-coordinate at time t.
$\alpha$
Start
The first value of t.
$\beta$
End
The last value of t.

Explanation

A parametric curve is simply the path of a moving point, described by saying where the point is at each moment. The name comes from the parameter $t$ , an extra helper variable (usually time) that both coordinates depend on, instead of $y$ depending on $x$ . It's good for any curve a graph can't handle, like loops, circles and paths that double back, and for anything that actually moves, like a thrown ball or a robot arm. The pair $x = f(t)$ , $y = g(t)$ looks like twice the work, but it's just two simple pieces: $f(t)$ (how far right the point is at time $t$ ) and $g(t)$ (how high up it is). As $t$ runs from $\alpha$ to $\beta$ , the point moves like a bug crawling across the page: the trail is the curve, and the equations also tell you which direction it went and how fast.

Why bother? Up to now, every curve you've done calculus on was a graph $y = f(x)$ : pick an $x$ , get one $y$ . That rules out a lot. A circle fails the vertical line test, and so does the path of a ball someone throws in a loop. So what can we do? Stop making $y$ depend on $x$ , and let both depend on time instead.

Eliminating the parameter. Sometimes you want to know what shape you're looking at. Getting rid of $t$ turns the pair of equations into one equation in $x$ and $y$ , and often you'll recognize it:

$x = 3\cos t$ , $y = 2\sin t$ : since $\cos^2 t + \sin^2 t = 1$ , we get $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ , an ellipse traced counterclockwise.
$x = t^2 - 2t$ , $y = t + 1$ : solve the second equation for $t = y - 1$ and substitute.

Notice what you lose when you do this, though. The equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$ describes the ellipse, but it has no idea where the point starts or which way it goes.

The same curve, different trips. $x = \cos t, y = \sin t$ and $x = \cos 2t, y = \sin 2t$ both trace the unit circle. The second one goes twice as fast, so on $0 \le t \le 2\pi$ it goes around twice. Same trail, different trip, and that difference will matter when we measure length later in this unit.

The cycloid is the path of a point on the rim of a rolling wheel of radius $r$ :

x = r(t - \sin t), \qquad y = r(1 - \cos t).

It makes sense if you read it in pieces: $rt$ is how far the wheel's center has rolled, and $-r\sin t$ , $-r\cos t$ is where the point sits on the rim relative to the center. The curve has a sharp cusp every time the point touches the ground, because right at that instant the point is momentarily standing still.

See it

A wheel rolls along the ground, and one point on its rim traces a cycloid: x = r(t − sin t), y = r(1 − cos t). Notice the point slows to a stop every time it touches down, which is why each arch ends in a sharp cusp instead of a smooth bottom.

A kicked ball, written as two separate rules: x = 15t and y = 20t − 4.9t² (meters and seconds). Notice the sideways part of the velocity never changes, while the upward part shrinks, hits zero and turns negative. That's why the ball peaks at about 20.4 m after about 2.04 s, halfway across at x ≈ 30.6 m, and lands about 61.2 m away after 4.08 s. The dashed path alone couldn't tell you any of those times.

Worked examples

Example 1

Eliminate the parameter: $x = t^2 - 2t$ , $y = t + 1$ .

1
The second equation is the easy one to solve for $t$ : $t = y - 1$ .

Example 2

Describe the curve $x = 3\cos t$ , $y = 2\sin t$ , $0 \le t \le 2\pi$ .

1
Divide out the 3 and the 2 so the Pythagorean identity can do the work: $\left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = \cos^2 t + \sin^2 t = 1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Eliminate the parameter from $x = 2t + 1$ , $y = 4t - 3$ . Write $y$ in terms of $x$ .

Problem 2+5 XP

What curve is $x = \sec t$ , $y = \tan t$ ?

Problem 3+5 XP

On the cycloid with $r = 1$ , find the x-coordinate of the point at $t = \pi/2$ .

Problem 4+5 XP

How many times does $x = \cos 2t$ , $y = \sin 2t$ , $0 \le t \le \pi$ , go around the unit circle?

Problem 5+5 XP

Does $x = \cos t$ , $y = -\sin t$ trace the circle clockwise or counterclockwise as $t$ increases?

Problem 6+5 XP

A kicked ball follows $x(t) = 15t$ , $y(t) = 20t - 4.9t^2$ (meters and seconds). How far away does it land?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Fonts and vector graphics draw curves as Bézier curves. The quadratic Bézier from $P_0 = (0, 0)$ to $P_2 = (4, 0)$ with control point $P_1 = (2, 4)$ is $B(t) = (1 - t)^2P_0 + 2t(1 - t)P_1 + t^2P_2$ , for $0 \le t \le 1$ . Find its highest point.

Where it's used

Every animation path, robot trajectory and orbit is a parametric curve: position as a function of time. The projectile lab plots $x = v_0 t\cos\theta$ , $y = h_0 + v_0 t\sin\theta - \tfrac12 g t^2$ , and fonts are drawn with parametric Bézier curves.

Projectile MotionAlgebra · Physics Bézier CurvesCalc II · Computer Science

Probability Tangents with Parametric Equations