Parametric Equations and Vectors

Part 1 · Hook

Why this matters

A graph

y = f(x)

tells you where a curve is, but not when you were there or which way you were going. A drone's flight, a ball's arc, a point on a spinning wheel: each has an

x

and a

y

that change with time. Unit 9 describes motion like that with two equations in

t

, and measures its changes with vectors. This page builds both ideas from the precalculus you already know.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On the unit circle, the point at angle $\theta$ is

$(\cos\theta, \sin\theta)$
$(\sin\theta, \cos\theta)$
$(\theta, \sin\theta)$
$(\cos\theta, \tan\theta)$

Show the answer

Cosine is the $x$ -coordinate and sine the $y$ -coordinate, so $\cos^2\theta + \sin^2\theta = 1$ .

Correct: $(\cos\theta, \sin\theta)$ :
$(\sin\theta, \cos\theta)$ :
$(\theta, \sin\theta)$ :
$(\cos\theta, \tan\theta)$ :

2. The distance between $(1, 2)$ and $(4, 6)$ is

$5$
$7$
$25$
$\sqrt{7}$

Show the answer

The changes are 3 and 4, the legs of a right triangle: $\sqrt{3^2 + 4^2} = 5$ by the Pythagorean theorem.

Correct: $5$ :
$7$ :
$25$ :
$\sqrt{7}$ :

3. If $f(x) = x^2 - 3$ , then $f(t + 1) =$

$t^2 + 2t - 2$
$t^2 - 2$
$t^2 + 1 - 3t$
$(t^2 - 3) + 1$

Show the answer

Replace every $x$ by $(t + 1)$ : $(t + 1)^2 - 3$ $= t^2 + 2t + 1 - 3$ .

Correct: $t^2 + 2t - 2$ :
$t^2 - 2$ :
$t^2 + 1 - 3t$ :
$(t^2 - 3) + 1$ :

Part 4 · See it

See it first

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. — Left: the path $x = t^2 - 2t$ , $y = 3 - t$ ; each value of $t$ gives one point, and the arrows show the direction of increasing $t$ . Right: the vector from $P(1, 1)$ to $Q(4, 5)$ is $\left\langle 3, 4 \right\rangle$ , and its length is $\sqrt{3^2 + 4^2} = 5$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Give each coordinate its own function of a parameter $t$ each $t$ gives one point $(x(t), y(t))$ , and together the points trace a path
Increase $t$ and watch the pointsthe path gets a direction (its orientation), which an equation in $x$ and $y$ cannot show
Solve one equation for $t$ , or use $\cos^2 t + \sin^2 t = 1$ eliminating $t$ gives the shape of the path as an equation in $x$ and $y$
Record a change in position as $\left\langle \Delta x, \Delta y \right\rangle$ a vector, whose length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ comes from the Pythagorean theorem

Part 6 · Key ideas

Key ideas

Parametric equations $x = x(t)$ , $y = y(t)$ give a point for each $t$ . Make a table of $t$ , $x$ , $y$ , plot the points in order of $t$ , and draw arrows for the direction.
Eliminate the parameter to see the shape: solve a linear equation for $t$ and substitute, or square and add sines and cosines. The direction and the timing are lost when you do.
$x = r\cos t$ , $y = r\sin t$ is a circle of radius $r$ traced counterclockwise from $(r, 0)$ ; swapping sine and cosine starts at the top and goes clockwise.
A vector $\left\langle a, b \right\rangle$ is a change: $a$ across and $b$ up. From $P$ to $Q$ it is end minus start; its magnitude is $\sqrt{a^2 + b^2}$ . Add vectors and multiply by numbers one component at a time.

Part 7 · Misconception

A common mistake

The wrong idea: The vector from $P(1, -2)$ to $Q(4, 2)$ is $\left\langle 4, 2 \right\rangle$ , and its length is $4 + 2 = 6$ .

What actually happens: $\left\langle 4, 2 \right\rangle$ is the position of $Q$ , the vector from the origin. From $P$ to $Q$ it is end minus start, $\left\langle 3, 4 \right\rangle$ , and its length is $\sqrt{3^2 + 4^2} = 5$ , not the sum of the components.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A path given by two equations in $t$

A point moves so that

x(t) = 2t - 1

and

y(t) = t^2 - 3

for

-2 \le t \le 2

.

Coordinates of the point at five values of $t$
$t$	$x(t)$	$y(t)$
$-2$	$-5$	$1$
$-1$	$-3$	$-2$
$0$	$-1$	$-3$
$1$	$1$	$-2$
$2$	$3$	$1$

1. Where is the point when $t = -1$ ?

$(-2, -3)$
$(-3, -2)$
$(-1, -2)$
$(-3, -4)$

Show the answer

$x(-1) = 2(-1) - 1 = -3$ and $y(-1) = (-1)^2 - 3 = -2$ , so the point is $(-3, -2)$ : the $x$ -value comes first.

$(-2, -3)$ : These are the right numbers in the wrong order: the point is $(x, y) = (-3, -2)$ .
Correct: $(-3, -2)$ : Right: $(x(-1), y(-1))$ $= (-3, -2)$ .
$(-1, -2)$ : This puts $t = -1$ in the $x$ place. The $x$ -coordinate is $x(-1) = -3$ , not $t$ itself.
$(-3, -4)$ : $(-1)^2 = 1$ , not $-1$ : $y(-1) = 1 - 3 = -2$ . Put parentheses around a negative input before squaring.

2. Eliminate the parameter: write $y$ as a function of $x$ for this path.

Type an expression for $y =$ .

Show the answer

From $x = 2t - 1$ , $t = \frac{x + 1}{2}$ . Substitute into $y = t^2 - 3$ : $y = \frac{(x + 1)^2}{4} - 3$ . Check with the table: $x = 3$ gives $\frac{16}{4} - 3 = 1$ , as at $t = 2$ .

Answer: $y =$ $\frac{(x + 1)^{2}}{4} - 3$

3. As $t$ increases from $-2$ to $2$ , how does the point move?

Left the whole time, first down and then up
Right the whole time, first up and then down
Right the whole time, first down and then up
Down and then up, first right and then left

Show the answer

$x$ goes $-5, -3, -1, 1, 3$ : it keeps increasing, so the point moves right. $y$ goes $1, -2, -3, -2, 1$ : down until $t = 0$ , then up. The path is a parabola traced from left to right.

Left the whole time, first down and then up: The $x$ -values increase ( $-5$ up to $3$ ), so the motion is to the right, not the left.
Right the whole time, first up and then down: The $y$ -values fall from 1 to $-3$ before rising: down first, then up.
Correct: Right the whole time, first down and then up: Right: $x$ increases throughout; $y$ falls to $-3$ at $t = 0$ , then rises.
Down and then up, first right and then left: Read the $x$ column: it increases at every step, so the point never turns back to the left.

4. Find the magnitude of the vector from the point where $t = 0$ to the point where $t = 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From $(-1, -3)$ to $(3, 1)$ the vector is $\left\langle 3 - (-1), 1 - (-3) \right\rangle$ $= \left\langle 4, 4 \right\rangle$ . Its magnitude is $\sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$ .

Answer: $4\,\sqrt{2}$ ≈ 5.657

5. The path $x = 2\sin t$ , $y = 2\cos t$ , $0 \le t \le 2\pi$ , is a circle of radius 2. Where does it start, and which way does it go around?

It starts at $(2, 0)$ and goes counterclockwise
It starts at $(0, 2)$ and goes counterclockwise
It starts at $(2, 0)$ and goes clockwise
It starts at $(0, 2)$ and goes clockwise

Show the answer

At $t = 0$ : $(2\sin 0, 2\cos 0)$ $= (0, 2)$ , the top. At $t = \frac{\pi}{2}$ : $(2, 0)$ , the right side. Going from the top to the right side is clockwise.

It starts at $(2, 0)$ and goes counterclockwise: $x(0) = 2\sin 0 = 0$ , so the start is $(0, 2)$ ; swapping sine and cosine moves the start to the top.
It starts at $(0, 2)$ and goes counterclockwise: The start $(0, 2)$ is right, but the next quarter turn reaches $(2, 0)$ , on the right: that is clockwise.
It starts at $(2, 0)$ and goes clockwise: Plug in $t = 0$ : $x = 0$ and $y = 2$ , so the start is the top of the circle, not $(2, 0)$ .
Correct: It starts at $(0, 2)$ and goes clockwise: Right: from $(0, 2)$ to $(2, 0)$ to $(0, -2)$ is clockwise.

6. What is the vector from $P(1, -2)$ to $Q(4, 2)$ ?

$\left\langle -3, -4 \right\rangle$
$\left\langle 5, 0 \right\rangle$
$\left\langle 3, 4 \right\rangle$
$\left\langle 4, 2 \right\rangle$

Show the answer

End minus start: $\left\langle 4 - 1, 2 - (-2) \right\rangle$ $= \left\langle 3, 4 \right\rangle$ . From $P$ , go 3 right and 4 up to reach $Q$ .

$\left\langle -3, -4 \right\rangle$ : That is start minus end, the vector from $Q$ back to $P$ . Subtract the start from the end.
$\left\langle 5, 0 \right\rangle$ : That adds the coordinates of $P$ and $Q$ . A vector from $P$ to $Q$ is the change: end minus start.
Correct: $\left\langle 3, 4 \right\rangle$ : Right: $\left\langle 4 - 1, 2 - (-2) \right\rangle$ .
$\left\langle 4, 2 \right\rangle$ : That is the position of $Q$ alone, the vector from the origin to $Q$ , not from $P$ .

Part 9 · Summary

Summary

Parametric equations describe a path by giving

x

and

y

as functions of a parameter

t

; increasing

t

traces the path in a direction. Eliminating

t

shows the shape but loses the direction. A vector

\left\langle a, b \right\rangle

records a change in position, from start to end; its magnitude is

\sqrt{a^2 + b^2}

, and vectors add and scale component by component.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

A path given by two equations in tt

Summary

What comes next

Connections

A path given by two equations in $t$