Parametric Curves, calculus lesson

A fly crawling on a window traces a path, but the path alone does not say where the fly was at each moment or which way it went. Giving its position as two functions of time, $x(t)$ and $y(t)$ , records all of that. This also lets you describe curves like circles and loops that are not the graph of any single function $y = f(x)$ .

Key terms

Parameter and parametric equations

Definition Let $I$ be an interval and let $f$ and $g$ be functions defined on $I$ . The equations

x = f(t), \qquad y = g(t), \qquad t \in I,

are parametric equations, and the variable $t$ is the parameter. Each value of $t$ gives one point $(f(t), g(t))$ .

In plain words Instead of tying $y$ directly to $x$ , both coordinates are tied to a third variable. Often $t$ is time and $(x, y)$ is where something is at that time.

Example For $x = t^2$ and $y = t + 1$ ,

$t$	$-2$	$-1$	$0$	$1$	$2$
$(x, y)$	$(4, -1)$	$(1, 0)$	$(0, 1)$	$(1, 2)$	$(4, 3)$

The points lie on a parabola opening to the right.

Plane curve

Definition If $f$ and $g$ are continuous on the interval $I$ , the set of points

C = \{ (f(t), g(t)) : t \in I \}

is a plane curve, and the pair of equations is a parametrization of $C$ . The same curve has many different parametrizations.

In plain words The curve is the trail of points. The parametrization is one particular way of walking along it.

Example The equations $x = \cos t$ , $y = \sin t$ for $0 \le t \le 2\pi$ and $x = \cos 3t$ , $y = \sin 3t$ for $0 \le t \le \frac{2\pi}{3}$ are two parametrizations of the same curve, the unit circle, since in both cases $x^2 + y^2 = 1$ and every angle from $0$ to $2\pi$ is reached.

Orientation

Definition The orientation of a parametrized curve is the direction in which the point $(f(t), g(t))$ moves as $t$ increases.

In plain words Orientation is the direction of travel. Two parametrizations can trace the same set of points in opposite directions.

Example The parametrization $x = \cos t$ , $y = \sin t$ starts at $(1, 0)$ when $t = 0$ and reaches $(0, 1)$ at $t = \frac{\pi}{2}$ , so it goes counterclockwise. Swapping the two, $x = \sin t$ , $y = \cos t$ starts at $(0, 1)$ and reaches $(1, 0)$ at $t = \frac{\pi}{2}$ , so it goes clockwise around the same circle.

Repeated tracing

Definition A parametrization traces part of its curve more than once if there are two intervals $[a, b]$ and $[c, d]$ in $I$ with $a < b \le c < d$ that are carried onto the same arc of the curve, an arc being a piece of the curve with more than one point. Crossing itself at isolated points, as a figure eight does, is not retracing. A parametrization that is one to one on $I$ passes through each point of the curve exactly once, so it never retraces.

In plain words A point can run around the same path several times, or go back and forth over it. The picture looks the same, but the motion is different.

Example The point $x = \cos t$ , $y = \sin t$ for $0 \le t \le 4\pi$ goes around the unit circle twice. And $x = \sin t$ , $y = 0$ for $0 \le t \le 2\pi$ starts at the origin, moves right to $(1, 0)$ , left to $(-1, 0)$ and right again to the origin, so every point of the segment other than its endpoints is passed at least twice.

Eliminating the parameter

Definition Eliminating the parameter means finding an equation in $x$ and $y$ alone that every point of the curve satisfies, for example by solving $x = f(t)$ for $t$ and substituting into $y = g(t)$ , or by using an identity such as $\cos^2 t + \sin^2 t = 1$ . The resulting equation may describe a larger set than the curve, so the range of $x$ and $y$ over $I$ must be recorded too.

In plain words Get rid of $t$ to see which familiar curve you are on. Then check which part of that curve is actually traced.

Example For $x = 2t - 1$ , $y = t^2$ , solve $t = \frac{x + 1}{2}$ and substitute, giving

y = \frac{(x + 1)^2}{4}.

For $x = 3\cos t$ , $y = 2\sin t$ , the identity $\cos^2 t + \sin^2 t = 1$ gives

\frac{x^2}{9} + \frac{y^2}{4} = 1,

an ellipse, all of which is traced when $t$ runs over $[0, 2\pi]$ . For $x = \sqrt{t}$ , $y = t$ with $t \ge 0$ , the result is $y = x^2$ with only $x \ge 0$ , the right half of the parabola.

Cycloid

Definition A cycloid is the curve traced by a point on the rim of a circle of radius $r > 0$ that rolls without slipping along a straight line. If the circle rolls along the $x$ axis and the point starts at the origin, then after the circle has turned through the angle $\theta$ the point is at

x = r(\theta - \sin\theta), \qquad y = r(1 - \cos\theta), \qquad \theta \in \mathbb{R}.

The center is then at $(r\theta, r)$ , because rolling without slipping moves it forward by the arc length $r\theta$ .

In plain words Paint a spot on a bicycle tire and watch it as the bike rolls. It rises in arches and touches the ground once per turn.

Example With $r = 1$ , at $\theta = \pi$ the point is at $(\pi, 2)$ , the top of the first arch, and at $\theta = 2\pi$ it is back on the ground at $(2\pi, 0)$ . One arch spans the length $2\pi r$ of the circumference.

Smooth and regular curves

Definition A parametrization $x = f(t)$ , $y = g(t)$ , $t \in I$ , is smooth (also called regular) if $f'$ and $g'$ exist and are continuous on $I$ , using one-sided derivatives at any endpoint that belongs to $I$ , and

(f'(t), g'(t)) \ne (0, 0) \quad \text{for every } t \in I \text{ that is not an endpoint of } I.

In words, $f'(t)$ and $g'(t)$ are never zero at the same time inside the interval. A curve is smooth if it has a smooth parametrization. Some books reserve the word regular for the condition $(f', g') \ne (0, 0)$ alone.

In plain words The moving point never stops, and its velocity changes continuously. Such a curve has no corners or cusps.

Example For the cycloid, $x'(\theta) = r(1 - \cos\theta)$ and $y'(\theta) = r\sin\theta$ . Both are zero exactly when $\theta = 2\pi k$ for an integer $k$ , the moments the point touches the ground. There the cycloid has sharp cusps. On each open interval $(2\pi k, 2\pi(k + 1))$ the parametrization is smooth.

Check yourself

Eliminate the parameter from $x = t + 2$ , $y = 3t - 1$ .

Show the answer

Solving the first equation gives $t = x - 2$ , so $y = 3(x - 2) - 1 = 3x - 7$ , a line.

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

Show the answer

The points satisfy $x^2 + y^2 = 4$ with $y \ge 0$ , the upper half of the circle of radius $2$ , traced counterclockwise from $(2, 0)$ to $(-2, 0)$ .

Where is the point on a cycloid with $r = 2$ when $\theta = \frac{\pi}{2}$ ?

Show the answer

The point is at $(\pi - 2, 2) \approx (1.14, 2)$ , since

x = 2\left(\frac{\pi}{2} - \sin\frac{\pi}{2}\right) = \pi - 2, \qquad y = 2\left(1 - \cos\frac{\pi}{2}\right) = 2.

Parametric Curves

Key terms

Parameter and parametric equations

Plane curve

Orientation

Repeated tracing

Eliminating the parameter

Cycloid

Smooth and regular curves

Check yourself

How this lesson connects

Used later in