Unit 4 Topic 4.3

Parametric Functions and Rates of Change

see every function move

3 min readfreeNot practiced

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

For an ordinary function you know how to measure change: the average rate of change on an interval, the slope of the secant line. A parametric function has two coordinates changing at once, so there are three rates worth asking about: how fast xx changes, how fast yy changes, and how steep the path is.

Two average rates, one for each coordinate

Take a parametric function (x(t),y(t))(x(t), y(t)) and an interval [t1,t2][t_1, t_2]. Each coordinate is a function of tt, so each has its own average rate of change:

rate of x=x(t2)−x(t1)t2−t1,rate of y=y(t2)−y(t1)t2−t1.\text{rate of } x = \frac{x(t_2) - x(t_1)}{t_2 - t_1}, \qquad \text{rate of } y = \frac{y(t_2) - y(t_1)}{t_2 - t_1}.

These are the slopes of secant lines on the graphs of x(t)x(t) and y(t)y(t) against tt. If xx is in meters and tt in seconds, the rate of xx is in meters per second. A positive rate of xx means the point ended up farther right; a negative rate of yy means it ended up lower.

The slope of the path

To measure how steep the path itself is between the two points, compare the change in yy with the change in xx:

slope of the segment=y(t2)−y(t1)x(t2)−x(t1).\text{slope of the segment} = \frac{y(t_2) - y(t_1)}{x(t_2) - x(t_1)}.

This is the average rate of change of yy with respect to xx on that piece of the path. Dividing the top and bottom by t2−t1t_2 - t_1 shows that it is also the rate of yy divided by the rate of xx.

Worked example. A skier's horizontal distance x(t)x(t) and height y(t)y(t), in meters, are logged every 2 seconds.

tt (seconds)00224466
x(t)x(t) (meters)006616163030
y(t)y(t) (meters)5050464638382626

On [2,6][2, 6]: the average rate of change of xx is 30−66−2=6\frac{30 - 6}{6 - 2} = 6 meters per second, and of yy is 26−466−2=−5\frac{26 - 46}{6 - 2} = -5 meters per second.

The slope of the segment from (6,46)(6, 46) to (30,26)(30, 26) is 26−4630−6\frac{26 - 46}{30 - 6}=−2024= \frac{-20}{24}=−56= -\frac{5}{6}. Check: −56\frac{-5}{6}, the rate of yy over the rate of xx. On average the skier drops 55 meters for every 66 meters across.

Reading the sign

Change in xxChange in yySlope of the segment
positive (right)positive (up)positive
negative (left)positive (up)negative
nonzero0000: a horizontal segment
00nonzeroundefined: a vertical segment

A slope of 00 does not mean the point stood still. On (t2+1,4t−t2)(t^2 + 1, 4t - t^2) from t=1t = 1 to t=3t = 3, the point goes from (2,3)(2, 3) to (10,3)(10, 3): the segment is horizontal, yet the point moved 88 units right and rose and fell along the way.

How the rates themselves change

Over equal intervals of tt, compare successive average rates. In the skier's log the rates of xx on [0,2][0, 2], [2,4][2, 4], [4,6][4, 6] are 33, 55 and 77 meters per second: they increase, so the rate of change of xx is increasing, and the skier covers more horizontal ground in each 2 seconds. The segment slopes −23-\frac{2}{3}, −45-\frac{4}{5}, −67-\frac{6}{7} decrease, so the path gets steeper downhill from one stretch to the next.

Shorter intervals give a closer picture. The slope of a segment over a small interval of tt estimates how steep the path is near those points, and the rates of xx and yy over a small interval estimate how fast each coordinate is changing there.

Spot a mistake on this page?