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Math 8 · 8.F.B.4

Rate of change

When one quantity moves another, how much does the second change for each step of the first?

You need
Rate and unit rateGrade 6
Unlocks
Slope from a graph or a tableMath 8The derivative at a pointCalculus 1
Where you are 4 in Slope
Algebra9
Linear relationships11
Slope4
Rate of change
Slope from a graph or a tableThe slope formula and the four kinds of slopeSlope triangles: why a line has one slope everywhere

Why this exists

A taxi meter, a burning candle, a hiker’s altimeter, a phone’s battery icon: each shows one quantity changing as another moves on. The distance traveled moves the fare; the time passing moves the candle’s height.

Knowing only how much the output changed is not enough to compare two of these. A candle that burns down 6 cm has burnt fast if that took 2 hours and slowly if it took 12. What lets you compare is the change in the output for each one unit of the input: centimeters per hour, dollars per kilometer, meters per minute. That number is the rate of change.

You have met rates already in rates and unit rates, and in a proportional relationship the constant of proportionality was the whole story. Most relationships do not start from zero, though. A taxi fare has a charge before the car has moved, and a candle starts tall. The rate of change is the part of the relationship that says how fast, kept apart from where it started.

Definition

Take two quantities where the first, the input, decides the second, the output. Pick two values of the input. Between them,

\text{rate of change} = \frac{\text{change in output}}{\text{change in input}}

Each change is the later value minus the earlier value, taken in the same order on top and underneath. The result says how much the output changes for each one-unit step of the input.

Its units are the output’s unit per the input’s unit: a fare in dollars against a distance in kilometers has a rate in dollars per kilometer. A rate written without its units has lost half its meaning.

The sign carries meaning too:

  • a positive rate: the output goes up as the input goes up;
  • a negative rate: the output goes down as the input goes up;
  • a zero rate: the output stays the same while the input moves.

A relationship has a constant rate of change when the rate comes out the same whichever two inputs you pick.

Intuition

A taxi charges 3 dollars to start and then 2 dollars for every kilometer.

distance (km)fare (dollars)
03
15
27
39
411
513
615

Each step of 1 km adds 2 dollars. The rate of change is 2 dollars per kilometer, and the steady differences in the table are that rate showing itself. Plotted, the pairs climb a staircase of identical steps, each 1 across and 2 up, and identical steps put every corner on one straight line (Figure 1).

0 1 2 3 4 5 6 0 2 4 6 8 10 12 14 16 distance (km) fare (dollars) 1 km $2 3 km $6
Figure 1Every kilometer adds the same 2 dollars, so the steps are identical and the dots lie on one line. A step three times as wide is three times as tall.

Take a longer step, from 3 km to 6 km. The fare goes from 9 to 15 dollars, and \displaystyle \frac{6}{3} = 2 dollars per kilometer again. The rate does not care how wide a step you measure, which is why it can be called the rate of this line.

The starting 3 dollars is inside every fare, so it cancels out of every difference. That is also why dividing one fare by its distance does not give the rate here: \displaystyle \frac{15}{6} = 2.5, because the 3 dollars has been spread over the ride. Only for a line through (0, 0), a proportional relationship, does output divided by input give the rate.

Results

Constant rate, three ways to see it. These go together:

  • the rate of change is the same between any two inputs;
  • in a table whose inputs go up in equal steps, the outputs change by equal steps;
  • the points of the graph lie on one straight line.

Units. A rate of change is measured in units of the output per unit of the input.

Sign. A positive rate makes the graph rise from left to right, a negative rate makes it fall, and a zero rate makes it flat.

Slope. On a straight-line graph the rate of change is called the slope of the line.

A changing rate. When the outputs of equally spaced inputs do not change by equal steps, there is no single rate of change. Each interval has its own, and the graph bends.

Why it is true

Suppose the output changes by the same amount r for every one unit of input, and by the matching share of r for part of a unit. From an input a to a larger input b the input moves b - a units, so the output changes by r(b - a), and

\frac{r(b - a)}{b - a} = r

whichever two inputs you pick. For the taxi, with fare 3 + 2d, the starting charge subtracts away: (3 + 2b) - (3 + 2a) = 2(b - a).

The graph is straight because every one-unit step is the same small right triangle, 1 across and r up, laid corner to corner, so each step leaves in exactly the direction the last one did. Points reached by always heading in the same direction lie on one straight line. The other way round, along a straight line every step of the same width is a copy of every other; that is the subject of slope triangles.

Worked examples

Example 1

A phone battery from a table

Saoirse watches her phone’s charge while she streams a film. Find the rate of change of the charge, then find when the phone will run out if the rate stays the same.

time (minutes)0306090
charge (percent of full)90786654
Solution

The input is time and the output is the charge. Each step of 30 minutes changes the charge by 78 - 90 = -12, then 66 - 78 = -12, then 54 - 66 = -12. Equal time steps give equal charge steps, so the rate is constant. Using the first step,

\begin{aligned} \text{rate of change} &= \frac{-12 \text{ percent}}{30 \text{ minutes}} \\ &= -0.4 \text{ percent per minute} \end{aligned}

The sign is negative because the charge falls as time goes on. Per hour the same rate is -0.4 \times 60 = -24 percent per hour: two units for one rate.

At 90 minutes, 54 percent is left. Losing 0.4 percent a minute, using it all takes

54 \div 0.4 = 135 \text{ minutes}

so the phone runs out 90 + 135 = 225 minutes after the start, which is 3 hours 45 minutes.

Check. From the start, 225 minutes at -0.4 percent per minute is a change of -90 percent, and 90 - 90 = 0.

Example 2

A taxi fare from a story

Odalys takes two rides with the same taxi company. A ride of 4 km costs 10.40 dollars and a ride of 10 km costs 20.00 dollars. The company charges a fixed amount to start plus a fixed amount per kilometer. Find the rate of change of the fare and the starting charge.

Solution

Between the two rides the distance changes by 10 - 4 = 6 km and the fare by 20.00 - 10.40 = 9.60 dollars, so

\begin{aligned} \text{rate of change} &= \frac{9.60 \text{ dollars}}{6 \text{ km}} \\ &= 1.60 \text{ dollars per km} \end{aligned}

The 4 km ride therefore includes 4 \times 1.60 = 6.40 dollars for distance, and the rest is the starting charge:

10.40 - 6.40 = 4.00 \text{ dollars}

Check. The 10 km ride costs 4.00 + 10 \times 1.60 = 20.00 dollars, as stated.

One fare divided by its own distance, \displaystyle \frac{20.00}{10} = 2.00 dollars per km, is not the rate: the starting charge is folded into it, so it changes from ride to ride. For the 4 km ride it is \displaystyle \frac{10.40}{4} = 2.60.

Example 3

A hike from a graph

Figure 2 shows Pieter’s elevation during a hike. Find the rate of change on each of the three pieces, with units, and say what each one means.

0 1 2 3 4 5 0 200 400 600 800 1000 1200 1400 1600 1800 2000 time (hours) elevation (m) climb rest descent
Figure 2A climb, a rest and a descent. Each straight piece has its own rate of change.
Solution

Each piece is straight, so any two points on it will do. The corners are (0, 1200), (2, 1800), (3, 1800) and (5, 1400), with time in hours and elevation in meters.

From 0 to 2 hours the rate is \displaystyle \frac{1800 - 1200}{2 - 0} = \frac{600}{2} = 300 meters per hour: he climbs 300 m each hour.

From 2 to 3 hours the rate is \displaystyle \frac{1800 - 1800}{3 - 2} = \frac{0}{1} = 0 meters per hour: time passes and his elevation does not change.

From 3 to 5 hours the rate is \displaystyle \frac{1400 - 1800}{5 - 3} = \frac{-400}{2} = -200 meters per hour: he comes down 200 m each hour.

The climb is steeper than the descent because 300 is more than 200. The whole hike at once gives \displaystyle \frac{1400 - 1200}{5 - 0} = 40 meters per hour, true of the trip overall but of no single stretch of it. When the rate changes, one number for the whole is an average rate of change.

Example 4

A candle measured at uneven times

Thandiwe lights a candle and measures its height whenever she remembers to.

time since lit (hours)height (cm)
021
119.8
317.4
416.2
712.6

Is the candle burning at a constant rate? If it is, find the rate and when it will burn out.

Solution

The height differences are -1.2, -2.4, -1.2 and -3.6, not all equal. That does not settle the question, because the time steps are not equal either: 1, 2, 1 and 3 hours. Divide each change in height by its own change in time:

\begin{aligned} \frac{-1.2}{1} &= -1.2 &\qquad \frac{-2.4}{2} &= -1.2 \\[4pt] \frac{-1.2}{1} &= -1.2 &\qquad \frac{-3.6}{3} &= -1.2 \end{aligned}

Every interval gives the same value, so the rate is constant: -1.2 cm per hour. Starting at 21 cm, the candle burns out after

21 \div 1.2 = 17.5 \text{ hours}

Check. At 7 hours the rate predicts 21 - 1.2 \times 7 = 12.6 cm, which matches the table.

Example 5

A plant whose rate changes

Qadira measures a bean plant once a week.

weekheight (cm)
02
13
25
38
412
517

Does the plant grow at a constant rate? Find the rate of change in the first week, in the last week, and over all five weeks.

Solution

The weeks go up in equal steps of 1, so steady growth would show as equal differences. The differences are 1, 2, 3, 4 and 5 cm, so the rate is not constant: the plant grows faster each week.

In the first week the rate is \displaystyle \frac{3 - 2}{1} = 1 cm per week, and in the last week it is \displaystyle \frac{17 - 12}{1} = 5 cm per week. Over all five weeks it is

\frac{17 - 2}{5 - 0} = \frac{15}{5} = 3 \text{ cm per week}

That 3 is an average, not the rate in any one week: a steady 3 cm per week would have made the plant 2 + 3 \times 2 = 8 cm tall at week 2, not 5. In Figure 3 the points curve upward, each step steeper than the last, and the middle ones sag below the dashed line from first to last.

0 1 2 3 4 5 0 2 4 6 8 10 12 14 16 18 week height (cm)
Figure 3The weekly gains grow, so the dots bend upward and do not lie on any line.

Boundaries

The two inputs must differ. A change in input of 0 cannot be divided by. On a graph that is a vertical line, which has no rate of change, and later no slope.

One rate for the whole relationship only when the rate is constant. For a changing rate, such as the plant’s, a rate belongs to the interval it was measured over.

The same order on top and underneath. Mixing the order flips the sign and turns a shrinking candle into a growing one.

Equal differences need equal input steps. Steady differences in a table show a constant rate only when the inputs go up in equal steps. With uneven steps, divide each interval separately, as in Example 4.

A constant rate holds over a stretch, not for ever. A candle stops at 0 cm and a battery at 0 and 100 percent. Using a rate beyond the measured rows assumes the situation carries on the same way.

Where this goes next

The rate of change of a line is its Slope. Slope from a graph counts the rise and the run on a grid, slope triangles show why a line has one slope everywhere, and the slope formula writes the fraction of this lesson in coordinates.

A starting value and a rate together make a linear function, like the taxi’s fare. Linear functions in context write such rules from a story, a table or a graph, and telling linear functions from nonlinear ones turns Example 5 into a test.

For a changing rate, the number over an interval is the average rate of change. Shrink the interval and that average closes in on the rate at a single instant, the derivative at a point, where calculus begins.

Exercises

Rates from tables

  1. A taxi meter shows fares of 2.50, 4.25, 6.00 and 7.75 dollars at distances of 0, 1, 2 and 3 km. Find the rate of change of the fare, with units, and say what the 2.50 dollars is.
  2. A candle is 30, 26.8, 23.6 and 20.4 cm tall at 0, 2, 4 and 6 hours after it is lit. Find the rate of change of its height.
  3. A switched-off phone shows a charge of 64 percent at 0, 15, 30 and 45 minutes. Find the rate of change of the charge and say what it means.
  4. A hiker’s elevation is 640, 700, 790 and 820 m at 0, 10, 25 and 30 minutes. Is the rate of change constant? If it is, find it.

Rates from stories

  1. A 5 km taxi ride costs 11.25 dollars and a 9 km ride with the same company costs 19.25 dollars. The company charges a starting amount plus a fixed amount per kilometer. Find the rate of change of the fare and the starting charge.
  2. A hiker walking steadily downhill is at an elevation of 1450 m, and 2.5 hours later at 1000 m. Find the rate of change of the elevation.
  3. A 25 cm candle burns down at a steady rate and is gone after 10 hours. Find the rate of change of its height, and its height 4 hours after it was lit.
  4. Wendeline’s phone reads 35 percent at 2:00 pm. It has been on a charger since before 1:45 pm, gaining a steady 1.3 percent of a full charge per minute. When will it be full? What did it read at 1:45 pm?

Rates from a graph

Figure 4 shows the heights of two candles, A and B, lit at the same moment.

0 1 2 3 4 5 6 7 8 9 10 0 3 6 9 12 15 18 21 24 time (hours) height (cm) A B
Figure 4Two candles lit at the same moment, for Exercises 9 to 11.
  1. Find the rate of change of candle A’s height, with units.
  2. Find the rate of change of candle B’s height. Which candle burns faster, and which lasts longer?
  3. The two lines cross. At what time are the candles the same height, and what is that height? Check your answer with both rates.

Constant or changing?

Decide whether each rate of change is constant. If it is, give it; if not, say how it changes.

  1. Inputs x = 0, 1, 2, 3, 4 give outputs y = 5, 8, 11, 14, 17.
  2. A seedling is 4, 4.5, 5.5, 7 and 9 cm tall at weeks 0, 1, 2, 3 and 4.
  3. A climber’s elevation is 300, 420, 480 and 660 m at 0, 2, 3 and 6 hours.
  4. A phone’s charge is 100, 96, 92 and 82 percent at 0, 5, 10 and 20 minutes.

Reading a rate

  1. Say whether each rate of change is positive, negative or zero, and give its units. (a) A candle’s height against time as it burns. (b) A taxi fare against the distance traveled. (c) A hiker’s elevation against time while she walks along a level ridge.
  2. A hiker’s elevation changes by -250 m in 50 minutes, at a steady rate. Find the rate of change and say in words what the hiker is doing.

Reasoning

  1. Explain why the starting charge of a taxi has no effect on the rate of change of its fare, but does change a fare divided by its distance.
  2. A table has inputs 0, 1, 3, 4 and outputs 10, 16, 22, 28. The outputs go up by 6 each time. Is the rate of change constant? Explain.
  3. Rufus-Lee says the plant in Example 5 “grows 3 cm a week” because it grew 15 cm in 5 weeks. In what sense is he right, and what does his number hide?
Answers
  1. 1.75 dollars per km, since every step is 4.25 - 2.50 = 1.75. The 2.50 dollars is the starting charge, the fare at 0 km.
  2. -1.6 cm per hour: each 2 hour step is -3.2 cm, and \displaystyle \frac{-3.2}{2} = -1.6.
  3. 0 percent per minute. The charge never changes: while switched off, this phone neither loses nor gains charge.
  4. Yes, 6 m per minute: \displaystyle \frac{60}{10} = \frac{90}{15} = \frac{30}{5} = 6, even though the time steps are uneven.
  5. 2 dollars per km, and a starting charge of 1.25 dollars. The rate is \displaystyle \frac{19.25 - 11.25}{9 - 5} = \frac{8}{4} = 2, and the start is 11.25 - 5 \times 2 = 1.25.
  6. -180 m per hour, since \displaystyle \frac{1000 - 1450}{2.5} = \frac{-450}{2.5} = -180.
  7. -2.5 cm per hour, and 15 cm after 4 hours. The rate is \displaystyle \frac{0 - 25}{10} = -2.5, and 25 - 2.5 \times 4 = 15.
  8. Full at 2:50 pm, and 15.5 percent at 1:45 pm. The charge must rise 100 - 35 = 65 percent, which takes 65 \div 1.3 = 50 minutes. Going back 15 minutes changes the charge by 1.3 \times (-15) = -19.5, and 35 - 19.5 = 15.5.
  9. -3 cm per hour. Line A runs from (0, 24) to (8, 0), and \displaystyle \frac{0 - 24}{8 - 0} = -3.
  10. -1.5 cm per hour, since \displaystyle \frac{0 - 15}{10 - 0} = -1.5. Candle A burns faster, 3 cm an hour against 1.5; candle B lasts longer, 10 hours against 8.
  11. At 6 hours, when both are 6 cm tall: 24 - 3 \times 6 = 6 and 15 - 1.5 \times 6 = 6. A starts 9 cm taller and loses 1.5 cm an hour more, so the gap closes after 9 \div 1.5 = 6 hours.
  12. Constant, 3. The inputs go up by 1 and the outputs by 3 every time.
  13. Not constant. The weekly growth is 0.5, 1, 1.5 and 2 cm, so it grows faster each week.
  14. Constant, 60 m per hour: \displaystyle \frac{120}{2} = \frac{60}{1} = \frac{180}{3} = 60.
  15. Not constant. The first two intervals give \displaystyle \frac{-4}{5} = -0.8 percent per minute and the last gives \displaystyle \frac{-10}{10} = -1 percent per minute, so the charge falls faster at the end.
  16. (a) Negative, centimeters per hour. (b) Positive, dollars per kilometer. (c) Zero, meters per hour (or per minute).
  17. -5 m per minute, since \displaystyle \frac{-250}{50} = -5: the hiker is going downhill, 5 m each minute.
  18. Every fare contains the same starting charge, so it cancels when one fare is subtracted from another, leaving only the charge per kilometer. A fare divided by its distance keeps the starting charge on top, spread over the ride: for the taxi of Figure 1, \displaystyle \frac{5}{1} = 5 and \displaystyle \frac{15}{6} = 2.5, while the rate is 2 throughout.
  19. No. The input steps are 1, 2 and 1, so the rates are \displaystyle \frac{6}{1} = 6, \displaystyle \frac{6}{2} = 3 and \displaystyle \frac{6}{1} = 6. Equal output steps show a constant rate only when the input steps are equal too.
  20. He is right on average: \displaystyle \frac{17 - 2}{5 - 0} = 3 cm per week over the five weeks. His number hides that the growth was not steady, 1 cm in the first week and 5 cm in the last.
kg1128 · updated 2026-10-09 · CC-BY-SA-4.0