Sketching Parametric Curves, calculus lesson

A table of points is a start, but connecting dots can mislead. Between two plotted points the curve might double back, loop or turn a corner. A few targeted questions about where the curve starts and ends, which way it moves, and where it crosses itself give a sketch you can trust. One curve runs through the whole page,

x = t^2, \qquad y = t^3 - 3t, \qquad -2 \le t \le 2.

Key terms

Parameter interval

Definition The parameter interval is the set $I$ of values of $t$ for which the parametric equations are used. The curve is the set $\{ (x(t), y(t)) : t \in I \}$ , so changing $I$ changes the curve, even with the same formulas.

In plain words The formulas say how to move, and the interval says when to start and stop.

Example The pair $x = \cos t$ , $y = \sin t$ gives a quarter of the unit circle for $0 \le t \le \frac{\pi}{2}$ and the whole circle for $0 \le t \le 2\pi$ . For the running curve, a table already shows the outline.

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

Initial and terminal points

Definition If $I = [a, b]$ is a closed interval, the initial point of the curve is $(x(a), y(a))$ and the terminal point is $(x(b), y(b))$ . The curve is closed if the two points are equal.

In plain words Mark where the moving point starts and where it ends. If they coincide, the path is a loop.

Example The running curve starts at $(4, -2)$ when $t = -2$ and ends at $(4, 2)$ when $t = 2$ . They differ, so the curve is not closed. The circle $x = \cos t$ , $y = \sin t$ on $[0, 2\pi]$ starts and ends at $(1, 0)$ , so it is closed.

Direction arrows

Definition If $x$ and $y$ are differentiable on an open interval of $t$ values, then on that interval the point moves right where $x'(t) > 0$ , left where $x'(t) < 0$ , up where $y'(t) > 0$ and down where $y'(t) < 0$ . Arrows drawn along the curve show the direction of increasing $t$ .

In plain words The signs of the two derivatives tell you which way the point is heading, so you can draw arrows without plotting extra points.

Example For the running curve, $x'(t) = 2t$ and $y'(t) = 3t^2 - 3 = 3(t - 1)(t + 1)$ .

Interval of $t$	$x'$	$y'$	Motion
$(-2, -1)$	$-$	$+$	left and up
$(-1, 0)$	$-$	$-$	left and down
$(0, 1)$	$+$	$-$	right and down
$(1, 2)$	$+$	$+$	right and up

Symmetry

Definition Suppose the interval $I$ is symmetric about $0$ . If $x(-t) = x(t)$ and $y(-t) = -y(t)$ for all $t \in I$ , the curve is symmetric about the $x$ axis. If $x(-t) = -x(t)$ and $y(-t) = y(t)$ , it is symmetric about the $y$ axis. If $x(-t) = -x(t)$ and $y(-t) = -y(t)$ , it is symmetric about the origin.

In plain words If the point at $-t$ is the mirror image of the point at $t$ , you only need to sketch half the curve and reflect it.

Example For the running curve, $x(-t) = t^2 = x(t)$ and $y(-t) = -t^3 + 3t = -y(t)$ , and $[-2, 2]$ is symmetric. So the curve is symmetric about the $x$ axis, which matches the table, where $(1, 2)$ and $(1, -2)$ appear in pairs.

Intersections

Definition A curve intersects itself at a point $P$ if there are parameter values $t_1 \ne t_2$ in $I$ with

(x(t_1), y(t_1)) = (x(t_2), y(t_2)) = P.

Two curves $(x_1(t), y_1(t))$ and $(x_2(s), y_2(s))$ intersect where $x_1(t) = x_2(s)$ and $y_1(t) = y_2(s)$ for some $t$ and $s$ . If this happens with $t = s$ , two objects moving along the curves are at the same place at the same time, which is a collision.

In plain words Use different letters for the two parameter values, because the curve can pass through the same spot at different times.

Example For the running curve, $t_1^2 = t_2^2$ with $t_1 \ne t_2$ forces $t_2 = -t_1$ and $t_1 \ne 0$ . Then $y(t_1) = y(-t_1) = -y(t_1)$ means $y(t_1) = 0$ , so $t_1(t_1^2 - 3) = 0$ and therefore $t_1 = \pm\sqrt{3}$ , both inside $[-2, 2]$ . These two values give the only crossing point, $(3, 0)$ , so the curve has a loop. For a collision, take $(t, 2t)$ and $(2 - s, s + 1)$ . Equal coordinates give $t = 2 - s$ and $2t = s + 1$ , so $s = t = 1$ , and the two objects meet at $(1, 2)$ at the same time.

Eliminating parameters without losing domain restrictions

Definition If eliminating $t$ gives an equation $F(x, y) = 0$ , then every point of the curve satisfies it, but the converse can fail. The curve is exactly the set of points $(x, y)$ with $F(x, y) = 0$ that are actually reached, so the elimination is complete only once you state the range of $x$ and $y$ over $I$ , and any other restriction, alongside the equation.

In plain words The Cartesian equation often describes more than was traced. Write down which part is really there.

Example For $x = e^t$ , $y = e^{2t}$ , $t \in \mathbb{R}$ , we get $y = x^2$ , but only with $x > 0$ . For $x = \sin t$ , $y = \sin^2 t$ , the curve is $y = x^2$ only for $-1 \le x \le 1$ , traced back and forth. For the running curve,

y^2 = t^2(t^2 - 3)^2 = x(x - 3)^2,

and the curve is the part of that graph with $0 \le x \le 4$ . Without the restriction the equation would also include points such as $(9, \pm 18)$ , which the curve never reaches.

Check yourself

Sketch $x = 1 + 2t$ , $y = 3 - t$ , $0 \le t \le 2$ , by finding its initial and terminal points.

Show the answer

It starts at $(1, 3)$ and ends at $(5, 1)$ . Both coordinates are linear in $t$ , so the curve is the line segment between them, traced from left to right.

Show that $x = \cos t$ , $y = \sin 2t$ , $0 \le t \le 2\pi$ , crosses itself at the origin.

Show the answer

At $t = \frac{\pi}{2}$ the point is $(0, \sin \pi) = (0, 0)$ , and at $t = \frac{3\pi}{2}$ it is $(0, \sin 3\pi) = (0, 0)$ . Two different parameter values give the same point.

Eliminate the parameter from $x = \sqrt{t}$ , $y = t + 1$ , $t \ge 0$ , and state the restriction.

Show the answer

Since $x = \sqrt{t} \ge 0$ , we get $t = x^2$ , so $y = x^2 + 1$ with $x \ge 0$ . The curve is the right half of that parabola starting at $(0, 1)$ .

Sketching Parametric Curves

Key terms

Parameter interval

Initial and terminal points

Direction arrows

Symmetry

Intersections

Eliminating parameters without losing domain restrictions

Check yourself

How this lesson connects

Builds on