Parametric Equations and Vectors — AP® Calculus BC Notes

Unit 9 follows points that move in the plane. To describe them you need two tools from precalculus: parametric equations, which give the position at each time, and vectors, which describe a change in position. This page rebuilds both, one step at a time.

A point for each value of t

A graph $y = f(x)$ gives one $y$ for each $x$ . Parametric equations instead give both coordinates from a third variable, the parameter $t$ :

$x = x(t), \qquad y = y(t).$

Think of $t$ as time. At each time the point is at $(x(t), y(t))$ , and as $t$ increases the point moves, tracing a path. The order of the points matters, so a parametric path has a direction (its orientation). Draw it with arrows.

Worked example 1. Sketch the path $x = t^2 - 2t$ , $y = 3 - t$ for $-1 \le t \le 3$ .

Make a table: put each $t$ into both equations, with parentheses around negative inputs.

$t$	$-1$	$0$	$1$	$2$	$3$
$x = t^2 - 2t$	$3$	$0$	$-1$	$0$	$3$
$y = 3 - t$	$4$	$3$	$2$	$1$	$0$

Plot $(3, 4)$ , $(0, 3)$ , $(-1, 2)$ , $(0, 1)$ , $(3, 0)$ in that order and join them. $y$ keeps falling, so the point moves down the page; $x$ falls until $t = 1$ and then rises, so it moves left and then right. The figure shows the path with its arrows.

Left: the path x = t squared minus 2t, y = 3 minus t for t from -1 to 3, with dots at t = -1 (3, 4), t = 0 (0, 3), t = 1 (-1, 2), t = 2 (0, 1) and t = 3 (3, 0), and arrows showing that the point moves down the page, first left and then right. Right: the vector from P(1, 1) to Q(4, 5), with dashed components 3 across and 4 up; it is the vector 3, 4, and its length is 5. — **Figure 1.** A path traced point by point, and a vector with its components. LevlPrep original diagram.

Eliminating the parameter

To see the shape of a path, remove $t$ and get one equation in $x$ and $y$ . Two methods cover almost every case.

Solve and substitute. When one equation is linear in $t$ , solve it for $t$ and put the result into the other. In worked example 1, $y = 3 - t$ gives $t = 3 - y$ , so

$x = (3 - y)^2 - 2(3 - y) = y^2 - 4y + 3,$

a parabola that opens to the right. Check a point: $y = 2$ gives $x = 4 - 8 + 3 = -1$ , as in the table at $t = 1$ .

Use an identity. When the equations hold $\cos t$ and $\sin t$ , square and add: $\cos^2 t + \sin^2 t = 1$ removes $t$ .

Worked example 2. Describe $x = 3\cos t$ , $y = 3\sin t$ for $0 \le t \le 2\pi$ .

$x^2$ $+ y^2$ $= 9\cos^2 t$ $+ 9\sin^2 t$ $= 9(\cos^2 t + \sin^2 t)$ $= 9$ : a circle of radius 3 centered at the origin.

Direction: $t = 0$ gives $(3, 0)$ and $t = \frac{\pi}{2}$ gives $(0, 3)$ . From the right side to the top is counterclockwise, and $t$ from 0 to $2\pi$ is one full turn.

Elimination keeps the shape and throws away the rest. $x = 3\sin t$ , $y = 3\cos t$ satisfies the same equation $x^2 + y^2 = 9$ , yet it starts at the top, $(0, 3)$ , and goes clockwise. And on $0 \le t \le 4\pi$ either path goes around twice. Always find the start and the direction from the parametric equations themselves.

Vectors: a change in position

A vector has a size and a direction. In the plane it is written by its components,

$\left\langle a, b \right\rangle: \quad a \text{ across (right if positive)}, \; b \text{ up (if positive)}.$

The vector from $P(x_1, y_1)$ to $Q(x_2, y_2)$ is end minus start: $\left\langle x_2 - x_1, y_2 - y_1 \right\rangle$ . The vector from the origin to a point is that point's position; the position of $Q(4, 5)$ is $\left\langle 4, 5 \right\rangle$ .

The magnitude (length) of $\left\langle a, b \right\rangle$ is the hypotenuse of a right triangle with legs $|a|$ and $|b|$ :

$\left\| \left\langle a, b \right\rangle \right\| = \sqrt{a^2 + b^2}.$

Add vectors and multiply them by numbers one component at a time:

$\left\langle 1, -3 \right\rangle + \left\langle 4, 5 \right\rangle = \left\langle 5, 2 \right\rangle, \qquad 2\left\langle 1, -3 \right\rangle = \left\langle 2, -6 \right\rangle.$

Worked example 3. A drone flies from $P(1, 1)$ to $Q(4, 5)$ , in meters. Find the vector of the trip and its length.

End minus start: $\left\langle 4 - 1, 5 - 1 \right\rangle = \left\langle 3, 4 \right\rangle$ , so 3 meters east and 4 meters north.

Length: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$ meters. Not $3 + 4 = 7$ : the drone flies the diagonal, not the two legs.

Common slips

Slip	Fix
Writing $(y, x)$ or putting $t$ in the $x$ place	A point is $(x(t), y(t))$ : evaluate both functions at the same $t$
Reading the shape and guessing the direction	Plot two nearby values of $t$ and see which way the point moves
Start minus end for the vector from $P$ to $Q$	End minus start: $Q - P$
Length as $a + b$ , or forgetting the square root	$\sqrt{a^2 + b^2}$ ; a magnitude is never negative

With these two tools, the rest of Unit 9 is calculus you already know, applied to each coordinate: rates of $x$ and $y$ , slopes of paths, and how far a point travels.