4.3 - Parametric Functions and Rates of Change

Overview

Topic 4.2 was about where a particle is. This topic is about which way it's moving and how fast its coordinates change. The key move is the same: look at

x(t)

and

y(t)

separately. The horizontal function tells you left or right, the vertical function tells you up or down, and dividing their average rates of change gives the slope of the path.

Direction of motion

t

increases, analyze each coordinate on its own:

If...	then the particle moves...
$x(t)$ is increasing	to the right
$x(t)$ is decreasing	to the left
$y(t)$ is increasing	up
$y(t)$ is decreasing	down

Put the two together to describe the motion. For example, if

x(t)

is decreasing while

y(t)

is increasing, the particle moves left and up. A max or min of

x(t)

inside the interval is where the particle switches between moving left and moving right, and a max or min of

y(t)

inside the interval is where it switches between moving up and moving down.

Example 1: Describing the direction

A particle's position is

(x(t), y(t))

, where

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

and

y(t) = -t^2 + 4t

for

0 \le t \le 4

(the same particle as Example 1 in Topic 4.2). Describe the direction of motion over the whole time interval.

Step 1: Analyze

x(t)

. It's an upward parabola in

t

with its vertex at

t = 1

, so

x(t)

decreases on

0 < t < 1

and increases on

1 < t < 4

Step 2: Analyze

y(t)

. It's a downward parabola in

t

with its vertex at

t = 2

, so

y(t)

increases on

0 < t < 2

and decreases on

2 < t < 4

4.3 - Parametric Functions and Rates of Change

Step 3: Combine.

Interval	$x(t)$	$y(t)$	Direction
$0 < t < 1$	decreasing	increasing	left and up
$1 < t < 2$	increasing	increasing	right and up
$2 < t < 4$	increasing	decreasing	right and down

Answer: The particle moves left and up for

0 < t < 1

, right and up for

1 < t < 2

, and right and down for

2 < t < 4

. At

t = 1

it switches from left to right (its leftmost point,

(-4, 3)

), and at

t = 2

it switches from up to down (its highest point,

(-3, 4)

Same point, different directions

A curve can pass through the same point more than once, and the particle doesn't have to move the same way each time. At a given point, the direction of motion can be different for different values of

t

. So a point alone doesn't tell you the direction; you also need the time.

Example 2: Two passes through the origin

The curve

f(t) = (t^2 - 4, \, t^3 - 4t)

from Topic 4.1 passes through

(0, 0)

t = -2

and at

t = 2

. Use the table to describe the direction of motion each time.

$t$	$x(t)$	$y(t)$
$-2.1$	$0.41$	$-0.861$
$-2$	$0$	$0$
$-1.9$	$-0.39$	$0.741$
$1.9$	$-0.39$	$-0.741$
$2$	$0$	$0$
$2.1$	$0.41$	$0.861$

Near

t = -2

: from

t = -2.1

t = -1.9

x

goes from

0.41

-0.39

(decreasing) and

y

goes from

-0.861

0.741

(increasing). The particle moves left and up.

Near

t = 2

: from

t = 1.9

t = 2.1

x

goes from

-0.39

0.41

(increasing) and

y

goes from

-0.741

0.861

(increasing). The particle moves right and up.

Answer: Both times the particle is at the origin, but at

t = -2

it's heading left and up (into the loop), and at

t = 2

it's heading right and up (out of the loop).

One curve, many parametrizations

The same curve can be parametrized (described by a parametric function) in many ways. Different parametrizations can trace the curve in different directions and at different speeds. For example, all three of these trace the part of the parabola

y = x^2

from

x = -2

x = 2

Parametric function	Domain	Starts at	Direction
$f(t) = (t, t^2)$	$-2 \le t \le 2$	$(-2, 4)$	left to right
$g(t) = (-t, t^2)$	$-2 \le t \le 2$	$(2, 4)$	right to left
$h(t) = (2t, 4t^2)$	$-1 \le t \le 1$	$(-2, 4)$	left to right, in half the time

Each one lies on

y = x^2

. For

g

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

. For

h

y = 4t^2 = (2t)^2 = x^2

So a question about direction is a question about the parametric function, not just the curve.

Average rate of change and slope

Over an interval

[t_1, t_2]

in the domain, you can compute the average rate of change (AROC) of each coordinate function on its own:

\text{AROC of } x = \frac{x(t_2) - x(t_1)}{t_2 - t_1}, \qquad \text{AROC of } y = \frac{y(t_2) - y(t_1)}{t_2 - t_1}

These are rates with respect to

t

, like feet per second of horizontal motion and feet per second of vertical motion.

Divide the second by the first, and the

t_2 - t_1

cancels:

\frac{\text{AROC of } y}{\text{AROC of } x} = \frac{\dfrac{y(t_2) - y(t_1)}{t_2 - t_1}}{\dfrac{x(t_2) - x(t_1)}{t_2 - t_1}} = \frac{y(t_2) - y(t_1)}{x(t_2) - x(t_1)}

That's the slope of the segment between the points on the curve at

t_1

and

t_2

: change in

y

over change in

x

. It works as long as the average rate of change of

x(t)

isn't

0

. If

x(t_2) = x(t_1)

, the two points line up vertically and the slope is undefined.

The slope is a rate of

y

with respect to

x

(for example, feet of height per foot of horizontal distance), not a rate with respect to time.

Example 3: The thrown ball

A ball's position

t

seconds after it's thrown is

(x(t), y(t)) = (10t, \, -16t^2 + 32t + 20)

, in feet (the ball from Topic 4.2). Find the average rates of change of

x

and

y

, and the slope of the path, between

t = 0

and

t = 1

and between

t = 1

and

t = 2.5

Step 1: Positions.

(x(0), y(0)) = (0, 20)

(x(1), y(1)) = (10, 36)

, and

(x(2.5), y(2.5)) = (25, 0)

Step 2: On

[0, 1]

\frac{x(1) - x(0)}{1 - 0} = \frac{10 - 0}{1} = 10 \text{ ft/s}, \qquad \frac{y(1) - y(0)}{1 - 0} = \frac{36 - 20}{1} = 16 \text{ ft/s}

\text{slope} = \frac{16}{10} = 1.6

Step 3: On

[1, 2.5]

\frac{x(2.5) - x(1)}{2.5 - 1} = \frac{25 - 10}{1.5} = 10 \text{ ft/s}, \qquad \frac{y(2.5) - y(1)}{2.5 - 1} = \frac{0 - 36}{1.5} = -24 \text{ ft/s}

\text{slope} = \frac{-24}{10} = -2.4

Answer: From

t = 0

t = 1

, the ball moves right at an average of 10 feet per second and up at an average of 16 feet per second, so the segment between those positions has slope

1.6

: it rises 1.6 feet for every foot of horizontal travel. From

t = 1

t = 2.5

, it still moves right at 10 feet per second but drops at an average of 24 feet per second, so the slope is

-2.4

: it falls 2.4 feet for every foot of horizontal travel.

Example 4: Working from a table

A particle moves in the plane. Selected values of

x(t)

and

y(t)

are shown.

$t$	$0$	$1$	$2$	$3$	$4$
$x(t)$	$2$	$5$	$6$	$5$	$1$
$y(t)$	$1$	$3$	$2$	$-1$	$-2$

(a) Find the slope of the segment between the particle's positions at

t = 0

and

t = 2

. (b) What happens if you try the same thing on

[1, 3]

? (c) Compare the particle's position at

t = 4

with its position at

t = 2

. Justify your answer.

(a) Average rates of change on

[0, 2]

\frac{x(2) - x(0)}{2 - 0} = \frac{6 - 2}{2} = 2, \qquad \frac{y(2) - y(0)}{2 - 0} = \frac{2 - 1}{2} = 0.5

\text{slope} = \frac{0.5}{2} = 0.25

Answer: The segment from

(2, 1)

(6, 2)

has slope

0.25

(b) On

[1, 3]

, the average rate of change of

x

\frac{5 - 5}{3 - 1} = 0

. You can't divide by

0

Answer: The slope is undefined. The positions at

t = 1

and

t = 3

are

(5, 3)

and

(5, -1)

, which line up vertically, so the segment between them is vertical.

(c) On

[2, 4]

, the average rate of change of

x

\frac{1 - 6}{4 - 2} = -2.5

and the average rate of change of

y

\frac{-2 - 2}{4 - 2} = -2

Answer: Both average rates of change are negative, so at

t = 4

the particle is to the left of and below where it was at

t = 2

. The table only gives a few times, so this describes the overall change, not the direction at every moment in between.

Common mistakes

Flipping the slope. The slope is the rate for $y$ divided by the rate for $x$ , not the other way around.

Calling a rate a slope. $\frac{y(t_2) - y(t_1)}{t_2 - t_1}$ is the vertical rate with respect to time. You need both rates to get the slope of the path.

Dividing by zero. If the average rate of change of $x$ is $0$ , the slope is undefined (a vertical segment).

Reading direction from the curve alone. One curve can be traced either way. Use $x(t)$ and $y(t)$ , or the arrows.

Assuming one direction per point. Where the curve passes through a point twice, the particle can move differently each time.

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