Unit 4 · Topic 4.3

Parametric Functions and Rates of Change

For a parametric function on [t_1, t_2] , the average rates of change of x(t) and y(t) say how fast each coordinate changed per unit of t .

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Part 1 · Hook

Why this matters

A skier heads down a slope while a tracker logs how far she has gone horizontally and how much height she has lost, every two seconds. From two rows of that log you can say how fast each coordinate changed, and how steep the stretch of hill between them was.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the average rate of change of $g(t) = t^2$ on $[1, 3]$ ?

Show the answer

$\frac{g(3) - g(1)}{3 - 1}$ $= \frac{9 - 1}{2}$ $= 4$ , the slope of the secant line through $(1, 1)$ and $(3, 9)$ .

Correct: $4$ :
$8$ :
$2$ :
$5$ :

2. For $x(t) = t + 2$ and $y(t) = t^2$ , where is the point at $t = -1$ ?

$(1, 1)$
$(1, -1)$
$(-1, 1)$
$(3, 1)$

Show the answer

$x(-1) = -1 + 2 = 1$ and $y(-1) = (-1)^2 = 1$ .

Correct: $(1, 1)$ :
$(1, -1)$ :
$(-1, 1)$ :
$(3, 1)$ :

Part 4 · See it

See it first

The curve x(t) = t squared plus 1, y(t) = 4t minus t squared for t from 0 to 4, with points at t = 0 (1, 0), t = 1 (2, 3), t = 2 (5, 4), t = 3 (10, 3) and t = 4 (17, 0). A segment joins the points for t = 0 and t = 2; dashed legs show a change in x of 4 and a change in y of 4, so the segment has slope 1. — From $t = 0$ to $t = 2$ on $(t^2 + 1, 4t - t^2)$ , $x$ changes by $4$ and $y$ changes by $4$ , so the segment joining the two points has slope $1$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Pick an interval of the parameter, $[t_1, t_2]$ find the two points $(x(t_1), y(t_1))$ and $(x(t_2), y(t_2))$
Divide the change in $x$ by $t_2 - t_1$ the average rate of change of $x$ on $[t_1, t_2]$ ; do the same for $y$
Divide the change in $y$ by the change in $x$ the slope of the segment joining the two points, how steep the path is there on average
The change in $x$ is $0$ but $y$ changesthe segment is vertical and its slope is undefined

Part 6 · Key ideas

Key ideas

The average rate of change of $x$ on $[t_1, t_2]$ is $\frac{x(t_2) - x(t_1)}{t_2 - t_1}$ ; the same for $y$ . Units: units of $x$ per unit of $t$ .
The slope of the segment joining the two points is $\frac{y(t_2) - y(t_1)}{x(t_2) - x(t_1)}$ : the average rate of change of $y$ with respect to $x$ .
That slope equals (rate of $y$ ) divided by (rate of $x$ ), as long as the change in $x$ is not $0$ .
Over equal intervals of $t$ , average rates that increase mean the rate of change is increasing.

Part 7 · Misconception

A common mistake

The wrong idea: The slope of a parametric path between two points is $\frac{y(t_2) - y(t_1)}{t_2 - t_1}$ .

What actually happens: That is the average rate of change of $y$ with respect to $t$ . The slope of the path compares $y$ with $x$ : divide the change in $y$ by the change in $x$ .

Part 8 · Check yourself

Check yourself

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

Graph

The curve $(t^2 + 1, 4t - t^2)$

The curve $x(t) = t^2 + 1$ , $y(t) = 4t - t^2$ for $0 \le t \le 4$ , with the points for $t = 0, 1, 2, 3, 4$ marked.

The curve x(t) = t squared plus 1, y(t) = 4t minus t squared for t from 0 to 4, moving right the whole time, rising to a top point and falling again. Points are marked at t = 0 (1, 0), t = 1 (2, 3), t = 2 (5, 4), t = 3 (10, 3) and t = 4 (17, 0). — LevlPrep original diagram.

1. What is the slope of the segment joining the points for $t = 0$ and $t = 2$ ?

$1$
$2$
$4$
$\frac{1}{2}$

Show the answer

The points are $(1, 0)$ and $(5, 4)$ . Slope $= \frac{4 - 0}{5 - 1}$ $= \frac{4}{4}$ $= 1$ .

Correct: $1$ : Change in $y$ is $4$ , change in $x$ is $4$ : slope $1$ .
$2$ : $2$ is the average rate of change of $y$ with respect to $t$ , $\frac{4}{2}$ .
$4$ : $4$ is the change in $y$ alone; divide by the change in $x$ .
$\frac{1}{2}$ : This divides the change in $t$ by the change in $x$ .

2. The points for $t = 1$ and $t = 3$ are $(2, 3)$ and $(10, 3)$ . Which statement is true?

The segment has slope $0$ , though the point moved right
The point did not move between $t = 1$ and $t = 3$
The slope is undefined, since the $y$ -values are equal
The average rate of change of $x$ on $[1, 3]$ is $0$

Show the answer

The change in $y$ is $3 - 3 = 0$ and the change in $x$ is $10 - 2 = 8$ , so the slope is $\frac{0}{8} = 0$ . The segment is horizontal, but the point moved right and went up and down in between.

Correct: The segment has slope $0$ , though the point moved right: Equal $y$ -values, different $x$ -values: a horizontal segment.
The point did not move between $t = 1$ and $t = 3$ : $x$ went from $2$ to $10$ , so the point moved.
The slope is undefined, since the $y$ -values are equal: Undefined slope comes from a change in $x$ of $0$ ; here the change in $y$ is $0$ .
The average rate of change of $x$ on $[1, 3]$ is $0$ : The rate of $x$ is $\frac{10 - 2}{2} = 4$ .

3. Find the slope of the segment joining the points for $t = 2$ and $t = 4$ .

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

Show the answer

The points are $(5, 4)$ and $(17, 0)$ . Slope $= \frac{0 - 4}{17 - 5}$ $= \frac{-4}{12}$ $= -\frac{1}{3}$ .

Answer: $\frac{-1}{3}$ ≈ -0.333

4. On which intervals is the average rate of change of $x(t)$ equal to $4$ ? Select all that apply.

$[1, 3]$
$[0, 4]$
$[0, 2]$
$[2, 4]$

Show the answer

$x(t) = t^2 + 1$ gives $x = 1, 2, 5, 10, 17$ at $t = 0$ to $4$ . On $[1, 3]$ : $\frac{10 - 2}{2} = 4$ . On $[0, 4]$ : $\frac{17 - 1}{4} = 4$ .

Correct: $[1, 3]$ : $\frac{10 - 2}{3 - 1} = 4$ .
Correct: $[0, 4]$ : $\frac{17 - 1}{4 - 0} = 4$ .
$[0, 2]$ : $\frac{5 - 1}{2} = 2$ .
$[2, 4]$ : $\frac{17 - 5}{2} = 6$ .

5. A curve is given by $x(t) = 3t + 1$ and $y(t) = 2 - t^2$ . Find the slope of the segment joining the points for $t = 0$ and $t = 2$ .

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

Show the answer

At $t = 0$ : $(1, 2)$ . At $t = 2$ : $(7, -2)$ . Slope $= \frac{-2 - 2}{7 - 1}$ $= \frac{-4}{6}$ $= -\frac{2}{3}$ .

Answer: $\frac{-2}{3}$ ≈ -0.667

6. Between $t = a$ and $t = b$ , a point ends up farther left and higher than it started. What is the sign of the slope of the segment joining its two positions?

Negative
Positive
Zero
Undefined

Show the answer

Farther left: change in $x$ is negative. Higher: change in $y$ is positive. Positive over negative is negative.

Correct: Negative: $\frac{+}{-}$ is negative.
Positive: Positive needs both changes with the same sign, such as right and up.
Zero: Zero needs no change in $y$ .
Undefined: Undefined needs no change in $x$ .

Part 9 · Summary

Summary

For a parametric function on

[t_1, t_2]

, the average rates of change of

x(t)

and

y(t)

say how fast each coordinate changed per unit of

t

. Their quotient, the change in

y

over the change in

x

, is the slope of the segment joining the two points, an average steepness of the path.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic

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

The curve (t2+1,4t−t2)(t^2 + 1, 4t - t^2)

Summary

What comes next

Connections

The curve $(t^2 + 1, 4t - t^2)$