Unit 1, lesson 6

Changing Quantities and Average Rate

How to measure the change in a quantity, compute its average rate of change over an interval with the right units, and read that rate as the slope of a secant line.

  • Theory part 11
  • Practice part 12
  • Lines, Rates, and Distance

A weather report says it was 12 degrees at 6 in the morning and 20 degrees at 10. That tells you a value at two moments, and also a change of 8 degrees over 4 hours, or 2 degrees per hour on average. The temperature may have stalled for an hour and then jumped, so the average says nothing about any single moment. Telling a value apart from a change, and a change apart from a rate, is the first step toward the derivative.

Key terms

Changing quantity

Definition A changing quantity is a quantity whose value depends on another quantity, most often time. It is modeled by a function ff, where the input xx is the quantity you control or measure along and the output f(x)f(x) is the value of the changing quantity at that input. Both carry units.

In plain words It is something that takes different values at different times or settings, written as a function so you can ask for its value at any allowed input.

Example Water flows into a tank, and after tt minutes the tank holds

V(t)=200+15t liters,0≤t≤40.\begin{gathered} V(t) = 200 + 15t \text{ liters,} \\[0.4em] 0 \le t \le 40. \end{gathered}

The input tt is time in minutes, and the output V(t)V(t) is a volume in liters. After 1010 minutes the tank holds V(10)=350V(10) = 350 liters.

Initial value

Definition The initial value of a changing quantity is its output at the starting input of the situation. When the input is time measured from the start, the initial value is f(0)f(0).

In plain words It is where the quantity begins, before any change has happened.

Example For the tank, the initial value is V(0)=200V(0) = 200 liters. The tank was not empty when the clock started. In the formula 200+15t200 + 15t the initial value is the constant term, and on the graph it is the height of the point on the vertical axis.

Change in input and change in output

Definition For a function ff and two inputs aa and bb in its domain, the change in input from aa to bb is

Δx=b−a,\Delta x = b - a,

and the corresponding change in output is

Δy=f(b)−f(a).\Delta y = f(b) - f(a).

Both are signed. A negative Δy\Delta y means the quantity decreased. A change is a difference of two outputs, which is not the same thing as either output.

In plain words Subtract the starting value from the ending value. The result tells you how much the quantity went up or down, not where it ended up.

Example For the tank from t=10t = 10 to t=30t = 30, the change in input is 30−10=2030 - 10 = 20 minutes. The outputs are V(10)=350V(10) = 350 and V(30)=650V(30) = 650, so the change in output is

V(30)−V(10)=650−350=300 liters.\begin{aligned} V(30) - V(10) &= 650 - 350 \\[0.4em] &= 300 \text{ liters.} \end{aligned}

The tank holds 650650 liters at the end, but it gained 300300 liters during those 2020 minutes.

Average rate of change

Definition Let ff be a real-valued function and let aa and bb be in its domain with a≠ba \ne b. The average rate of change of ff from aa to bb is

ΔyΔx=f(b)−f(a)b−a.\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}.

It describes the interval as a whole. It does not claim that the quantity changes at this rate at every moment of the interval.

In plain words Divide how much the output changed by how much the input changed. The result is the steady rate that would produce the same total change.

Example Compare the tank V(t)=200+15tV(t) = 200 + 15t with a second tank whose inflow speeds up, holding W(t)=200+3t2W(t) = 200 + 3t^2 liters after tt minutes.

IntervalAverage rate of VVAverage rate of WW
[0,2][0, 2]151566
[2,4][2, 4]15151818
[4,6][4, 6]15153030
[0,6][0, 6]15151818

The first tank has the same rate on every interval, which is what a constant-rate model means. The second tank averages 1818 liters per minute over [0,6][0, 6], yet its rate on the separate pieces is 66, then 1818, then 3030. Over the whole interval,

W(6)−W(0)6−0=308−2006=18.\frac{W(6) - W(0)}{6 - 0} = \frac{308 - 200}{6} = 18.

Units of rate

Definition The units of an average rate of change are the output units divided by the input units. When the output is the position s(t)s(t) of an object moving along a line, the change s(b)−s(a)s(b) - s(a) is the signed displacement and the average rate is the average velocity. Average speed is different. For a<ba < b it is the total distance traveled divided by the elapsed time b−ab - a. It is never negative and never less than the absolute value of the average velocity, and it is strictly larger when the object turns around during the interval.

In plain words Read the units as output per input, such as liters per minute or dollars per item. For motion, average velocity cares only about where you started and where you ended, while average speed counts every meter you covered.

Example A walker’s position along a straight path is s(t)s(t) meters east of a gate after tt minutes. She walks 300300 meters east in the first 33 minutes and then 100100 meters back west in the next 22, so s(0)=0s(0) = 0 and s(5)=200s(5) = 200. Her average velocity is

s(5)−s(0)5−0=2005=40 meters per minute.\begin{aligned} &\frac{s(5) - s(0)}{5 - 0} = \frac{200}{5} \\[0.4em] &\quad = 40 \text{ meters per minute.} \end{aligned}

She covered 300+100=400300 + 100 = 400 meters in total, so her average speed is 4005=80\frac{400}{5} = 80 meters per minute.

Secant slope

Definition For a≠ba \ne b in the domain of ff, the secant line is the line through the two graph points (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)). Its slope is exactly the average rate of change of ff from aa to bb,

msec=f(b)−f(a)b−a.m_{\text{sec}} = \frac{f(b) - f(a)}{b - a}.

In plain words Mark two points on the graph and join them with a straight line. Its steepness is the average rate between those inputs.

Example For W(t)=200+3t2W(t) = 200 + 3t^2 on [2,4][2, 4] the graph points are (2,212)(2, 212) and (4,248)(4, 248). The secant slope is

248−2124−2=18,\frac{248 - 212}{4 - 2} = 18,

the same 1818 liters per minute found in the table, and the secant line is

y=212+18(t−2).y = 212 + 18(t - 2).

The curve of WW bends, but this one straight line tells the same story about the interval [2,4][2, 4] as the physical tank does.

Check yourself

  1. Find the average rate of change of f(x)=x2−4xf(x) = x^2 - 4x from x=1x = 1 to x=5x = 5.
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The outputs are f(1)=−3f(1) = -3 and f(5)=5f(5) = 5, so the average rate is

5−(−3)5−1=84=2.\frac{5 - (-3)}{5 - 1} = \frac{8}{4} = 2.
  1. A town had 12,00012{,}000 people in 2010, 13,50013{,}500 in 2015 and 14,10014{,}100 in 2020. Find the average rate of change of the population over each five-year period and over the whole decade. Does a constant-rate model fit?
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From 2010 to 2015 the rate is 15005=300\frac{1500}{5} = 300 people per year. From 2015 to 2020 it is 6005=120\frac{600}{5} = 120 people per year. Over the decade it is 210010=210\frac{2100}{10} = 210 people per year. The rate changed from one period to the next, so a constant-rate model does not fit.

  1. A ball thrown straight up has height h(t)=20t−5t2h(t) = 20t - 5t^2 meters after tt seconds, and it reaches its highest point at t=2t = 2. Find its average velocity and its average speed over [0,4][0, 4].
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The heights are h(0)=0h(0) = 0 and h(4)=0h(4) = 0, so the average velocity is 0−04=0\frac{0 - 0}{4} = 0 meters per second. The ball still moved. It rose from 00 to h(2)=20h(2) = 20 meters and fell back, a total distance of 4040 meters, so its average speed is 404=10\frac{40}{4} = 10 meters per second.

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