4.4 - Parametrically Defined Circles and Lines

Overview

Circles and line segments are the two paths you'll parametrize most often. Motion around a circle is modeled with cosine and sine, and motion along a straight segment is modeled with linear functions of

t

. Both build on ideas you already know: the unit circle from Unit 3 and transformations of functions.

The unit circle, parametrically

The parametric function

\big(x(t),\, y(t)\big) = (\cos t, \, \sin t), \qquad 0 \le t \le 2\pi

models one complete counterclockwise revolution around the unit circle (center at the origin, radius 1), starting and ending at

(1, 0)

Why it's a circle: for every $t$ , $x^2 + y^2 = \cos^2 t + \sin^2 t = 1$ , so every point is 1 unit from the origin.

What $t$ is: the angle in radians, measured counterclockwise from the positive $x$ -axis. As $t$ increases, the point moves counterclockwise.

Start and end: $t = 0$ gives $(\cos 0, \sin 0) = (1, 0)$ , and $t = 2\pi$ brings the point back to $(1, 0)$ .

$t$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$(\cos t, \sin t)$	$(1, 0)$	$(0, 1)$	$(-1, 0)$	$(0, -1)$	$(1, 0)$

4.4 - Parametrically Defined Circles and Lines

Transforming the circle

Transformations of

(\cos t, \sin t)

can model any circular path. Each change to the formula does one job:

Change	Parametric function	Effect
Radius $r$	$(r\cos t, \, r\sin t)$	dilates the circle to radius $r$
Center $(h, k)$	$(h + r\cos t, \, k + r\sin t)$	translates the center to $(h, k)$
Clockwise	$(h + r\cos t, \, k - r\sin t)$	reflects the motion so it goes clockwise
Speed	$(h + r\cos(bt), \, k + r\sin(bt))$	one revolution takes $\frac{2\pi}{b}$ units of time
Starting point	$(h + r\cos(t + c), \, k + r\sin(t + c))$	starts at angle $c$ instead of angle $0$
Part of a circle	restrict the domain	for example, $0 \le t \le \pi$ traces only the top half

Putting the first two together, the counterclockwise circle with center

(h, k)

and radius

r

x(t) = h + r\cos t, \qquad y(t) = k + r\sin t, \qquad 0 \le t \le 2\pi

Check:

(x - h)^2 + (y - k)^2 = r^2\cos^2 t + r^2\sin^2 t = r^2

, which is the equation of that circle. (Stretching by different amounts in each direction, like

(3\cos t, 2\sin t)

, gives an ellipse instead. Topic 4.6 covers ellipses, and Topic 4.7 parametrizes them.)

Example 1: Writing a circle

Write a parametric function for a particle that travels once around the circle with center

(3, -2)

and radius

4

, (a) counterclockwise, starting at the rightmost point, and (b) clockwise, starting at the top.

(a) Use

h = 3

k = -2

, and

r = 4

x(t) = 3 + 4\cos t, \qquad y(t) = -2 + 4\sin t, \qquad 0 \le t \le 2\pi

Check:

t = 0

gives

(3 + 4, -2 + 0) = (7, -2)

, which is 4 units right of the center, the rightmost point.

(b) Swap cosine and sine:

x(t) = 3 + 4\sin t, \qquad y(t) = -2 + 4\cos t, \qquad 0 \le t \le 2\pi

Check:

t = 0

gives

(3, 2)

, the top. At

t = \frac{\pi}{2}

the particle is at

(7, -2)

, the rightmost point. Going from the top to the right side is clockwise.

Answer: (a)

(3 + 4\cos t, \, -2 + 4\sin t)

and (b)

(3 + 4\sin t, \, -2 + 4\cos t)

, each for

0 \le t \le 2\pi

. Other correct answers exist, since one path can be parametrized many ways.

Example 2: A Ferris wheel

A Ferris wheel has a radius of 25 meters, and its center is 30 meters above the ground. Put the origin on the ground directly below the center, so the center is at

(0, 30)

. The wheel turns counterclockwise in this coordinate system and makes one full revolution every 8 minutes. A rider boards at the bottom at

t = 0

, where

t

is in minutes.

(a) Write a parametric function for the rider's position. (b) Find the rider's position at

t = 3

. (c) When does the rider first reach a height of 42.5 meters?

(a) Set three things. Center and radius:

(0, 30)

and

25

. Speed: one revolution (

2\pi

radians) every 8 minutes gives

b = \frac{2\pi}{8} = \frac{\pi}{4}

. Start: the bottom of the circle is at angle

-\frac{\pi}{2}

, so shift the angle by

-\frac{\pi}{2}

x(t) = 25\cos\left(\frac{\pi}{4}t - \frac{\pi}{2}\right), \qquad y(t) = 30 + 25\sin\left(\frac{\pi}{4}t - \frac{\pi}{2}\right), \qquad 0 \le t \le 8

Using

\cos\left(\theta - \frac{\pi}{2}\right) = \sin\theta

and

\sin\left(\theta - \frac{\pi}{2}\right) = -\cos\theta

, this simplifies to

x(t) = 25\sin\left(\frac{\pi}{4}t\right), \qquad y(t) = 30 - 25\cos\left(\frac{\pi}{4}t\right), \qquad 0 \le t \le 8

Check:

t = 0

gives

(0, 5)

, the bottom.

t = 2

gives

(25, 30)

, the rightmost point, so the rider moves counterclockwise.

t = 8

brings the rider back to

(0, 5)

(b) At

t = 3

, the angle input is

\frac{3\pi}{4}

x(3) = 25\sin\left(\frac{3\pi}{4}\right) = 25 \cdot \frac{\sqrt{2}}{2} \approx 17.68, \qquad y(3) = 30 - 25\cos\left(\frac{3\pi}{4}\right) = 30 + 25 \cdot \frac{\sqrt{2}}{2} \approx 47.68

Answer: After 3 minutes, the rider is at about

(17.68, 47.68)

: about 47.68 meters above the ground and 17.68 meters to the right of the wheel's center.

(c) Set the height equal to 42.5:

30 - 25\cos\left(\frac{\pi}{4}t\right) = 42.5 \quad \Rightarrow \quad \cos\left(\frac{\pi}{4}t\right) = -0.5

During the first half turn,

\frac{\pi}{4}t

runs from

0

\pi

, and the first angle in that range with cosine

-0.5

\frac{2\pi}{3}

. So

\frac{\pi}{4}t = \frac{2\pi}{3}

, which gives

t = \frac{8}{3}

Answer: The rider first reaches 42.5 meters

\frac{8}{3} \approx 2.67

minutes (2 minutes 40 seconds) after boarding.

Line segments, parametrically

A linear path along the segment from

(x_1, y_1)

(x_2, y_2)

can be parametrized many ways. The most useful one starts at the initial position and adds a constant rate of change in each direction:

x(t) = x_1 + (x_2 - x_1)\,t, \qquad y(t) = y_1 + (y_2 - y_1)\,t, \qquad 0 \le t \le 1

Initial position: at $t = 0$ you're at $(x_1, y_1)$ , and at $t = 1$ you're at $(x_2, y_2)$ .

Rates of change: $x$ changes at a constant $x_2 - x_1$ per unit of $t$ , and $y$ changes at a constant $y_2 - y_1$ per unit of $t$ . Constant rates in both directions are what make the path straight.

Slope: the slope of the segment is the ratio of the rates, $\frac{y_2 - y_1}{x_2 - x_1}$ , just like in Topic 4.3.

More generally, a particle that starts at

(x_1, y_1)

and moves with constant rates

a

(horizontal) and

b

(vertical) has

x(t) = x_1 + at

and

y(t) = y_1 + bt

. Changing the rates changes how long the trip takes, swapping the endpoints reverses the direction, and dropping the domain restriction gives the whole line.

Example 3: A drone's flight

A drone flies at a constant speed in a straight line from

(2, 1)

(8, 5)

, with units in kilometers. The trip takes 2 hours. (a) Write a parametric function for the drone's position

t

hours after it leaves. (b) Where is it after 30 minutes? (c) Give a different parametrization of the same segment that uses

0 \le t \le 1

(a) Over 2 hours,

x

changes by

8 - 2 = 6

km and

y

changes by

5 - 1 = 4

km, so the rates are

\frac{6 \text{ km}}{2 \text{ h}} = 3 \text{ km/h horizontally}, \qquad \frac{4 \text{ km}}{2 \text{ h}} = 2 \text{ km/h vertically}

x(t) = 2 + 3t, \qquad y(t) = 1 + 2t, \qquad 0 \le t \le 2

Check:

t = 2

gives

(2 + 6, 1 + 4) = (8, 5)

(b) 30 minutes is

t = 0.5

, so the position is

(2 + 3(0.5), \, 1 + 2(0.5)) = (3.5, 2)

Answer: After 30 minutes, the drone is at

(3.5, 2)

(c) Use the general formula:

x(t) = 2 + 6t

and

y(t) = 1 + 4t

for

0 \le t \le 1

. Now

x

changes 6 units and

y

changes 4 units per unit of

t

, so this version covers the segment in half as much

t

Answer:

(2 + 6t, \, 1 + 4t)

for

0 \le t \le 1

. Both versions trace the same segment with slope

\frac{2}{3}

, which shows that one path can have many parametrizations.

Common mistakes

Using degrees. The parameter is in radians. One revolution is $2\pi$ , not 360.

Forgetting the center. $(r\cos t, r\sin t)$ is centered at the origin. Add $h$ to $x$ and $k$ to $y$ .

Mixing up cosine and sine. $(\cos t, \sin t)$ starts at the right and goes counterclockwise. $(\sin t, \cos t)$ starts at the top and goes clockwise.

Wrong speed factor. If one revolution takes $T$ units of time, then $b = \frac{2\pi}{T}$ , not $T$ .

Wrong domain for a segment. With $x_1 + (x_2 - x_1)t$ , use $0 \le t \le 1$ . A bigger domain runs past the endpoint.

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