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Average Value and Accumulated Quantities · Tutorial 1276 of 1495

Average Rate from a Total and an Interval

Find the average rate over an interval by comparing total change with elapsed time, and interpret what that average does—and does not—say about rates at particular instants.

Intermediate 8 min read

What You'll Learn

  • Calculate an average rate from a total change and an interval length.
  • Use the correct order of endpoints and interpret the sign of the result.
  • Track the units of total change, time, and average rate.
  • Find a total change from piecewise rates before calculating the interval average.
  • Distinguish an interval average from the rate at a particular instant.

From Total Change to Average Rate

A rate describes how one quantity changes in relation to another. For example, a tank’s water volume may change by a certain number of liters per minute, or an object’s position may change by a certain number of meters per second. When the rate varies during an interval, a single rate can still summarize the overall change across that interval: the average rate.

The basic calculation compares the total change in the quantity with the length of the interval. It is essential to use the change in the quantity—not merely its final value—and the elapsed interval—not merely its ending time. The sign and units of the result also carry information about what happened.

Definition: Let \(Q\) be a quantity measured at times \(t_1\) and \(t_2\), where \(t_1<t_2\). The average rate of change of \(Q\) over \([t_1,t_2]\) is the total change in \(Q\) divided by the interval length: $$ \text{average rate} = \frac{Q(t_2)-Q(t_1)}{t_2-t_1}. $$ The numerator is the change in the quantity, and the denominator is the elapsed time.

Writing \(\Delta Q=Q(t_2)-Q(t_1)\) and \(\Delta t=t_2-t_1\), the same definition is \(\Delta Q/\Delta t\). The order of subtraction matters: both changes are calculated from the later endpoint minus the earlier endpoint. Since \(t_2>t_1\), the time interval length is positive. The average rate is positive when the quantity’s net change is positive, negative when its net change is negative, and zero when the quantity ends where it started.

Why Dividing Change by Time Gives an Average

Imagine replacing a variable-rate process with a constant-rate process that has the same total change over the same length of time. If a constant rate is \(r\), then over an interval of length \(\Delta t\), it produces a change of \(r\Delta t\). To match the observed change \(\Delta Q\), this constant rate must satisfy \(r\Delta t=\Delta Q\). Because \(\Delta t>0\), dividing both sides by \(\Delta t\) gives the average rate.

$$ r\Delta t=\Delta Q \qquad\Longrightarrow\qquad r=\frac{\Delta Q}{\Delta t}. $$

This explains the interpretation of the quotient: it is the constant rate that would produce the same net change over the interval. It does not mean the actual rate was constant. Nor does the calculation, by itself, tell us the rate at any particular instant.

Worked Example: Average Change in a Tank’s Volume

A tank contains \(120\) liters of water at \(t=2\) minutes and \(204\) liters at \(t=8\) minutes. Find the average rate of change of the water volume over this interval.

First, calculate the change in volume by subtracting the earlier amount from the later amount:

$$ \Delta V=204-120=84\text{ liters}. $$

Next, calculate the elapsed time:

$$ \Delta t=8-2=6\text{ minutes}. $$

Divide the total change by the interval length:

$$ \frac{\Delta V}{\Delta t} = \frac{84\text{ liters}}{6\text{ minutes}} = 14\text{ liters per minute}. $$

The average rate is \(14\) liters per minute. As a check, a constant rate of \(14\) liters per minute for \(6\) minutes would give a change of \(14\cdot 6=84\) liters, which agrees with the observed change. This check confirms the calculation; it does not establish that the tank actually gained water at a constant rate.

Signs and Units

The sign tells whether the quantity’s net change was an increase or a decrease as time moved from \(t_1\) to \(t_2\). A negative average rate is not automatically an error. It can describe a quantity that decreased overall. The units come from dividing the units of the quantity by the units of time. For instance, liters divided by minutes gives liters per minute; meters divided by seconds gives meters per second.

A useful check is to ask whether the units fit the question. If a total change is measured in kilograms and an interval in hours, the average rate must be in kilograms per hour. If the units do not combine that way, revisit what was divided or which interval was used.

Worked Example: A Negative Average Rate of Position

An object has position \(x(2)=5\) meters at \(t=2\) seconds and position \(x(8)=-7\) meters at \(t=8\) seconds. Find its average rate of change in position over the interval.

The change in position is the later position minus the earlier position:

$$ \Delta x=x(8)-x(2)=(-7)-5=-12\text{ meters}. $$

The time interval has length

$$ \Delta t=8-2=6\text{ seconds}. $$

Therefore, the average rate of change in position is

$$ \frac{\Delta x}{\Delta t} = \frac{-12\text{ meters}}{6\text{ seconds}} = -2\text{ meters per second}. $$

The negative sign records a net change toward smaller position values. Check the result by multiplying the average rate by the interval length: \((-2\text{ meters per second})(6\text{ seconds})=-12\) meters, the displacement calculated from the endpoint positions. This is an average rate of change in position; it does not, by itself, give the object’s average speed or describe its path between the endpoints.

When Rates Vary Within the Interval

Sometimes the total change is not given directly. It may instead be found from information about rates over parts of the interval. For a constant rate on a segment, the change on that segment is the rate multiplied by the segment’s duration. For several segments, find each segment’s change, add those changes with their signs, and then divide the resulting total change by the full interval length.

This is a way to use the piecewise-rate information to obtain one interval average. It is important to add signed changes rather than simply average the rate values. Segments may have different durations, so each rate contributes for a different amount of time.

Worked Example: Averaging Across Two Rate Segments

A quantity increases at a constant rate of \(3\) units per minute for the first \(4\) minutes. It then decreases at a constant rate of \(1\) unit per minute for the next \(6\) minutes. Find its average rate of change over the full \(10\)-minute interval.

For the first segment, multiply the rate by its duration:

$$ \Delta Q_1=(3\text{ units per minute})(4\text{ minutes})=12\text{ units}. $$

For the second segment, the rate is negative because the quantity decreases:

$$ \Delta Q_2=(-1\text{ unit per minute})(6\text{ minutes})=-6\text{ units}. $$

Add the signed changes to find the net change over the full interval:

$$ \Delta Q=\Delta Q_1+\Delta Q_2=12+(-6)=6\text{ units}. $$

The full interval lasts \(4+6=10\) minutes. Thus the average rate is

$$ \frac{\Delta Q}{\Delta t} = \frac{6\text{ units}}{10\text{ minutes}} = 0.6\text{ units per minute}. $$

Check by applying this average rate to the whole duration: \((0.6\text{ units per minute})(10\text{ minutes})=6\) units, the same net change. The number \(0.6\) is not the rate on either segment: the actual rates provided were \(3\) and \(-1\) units per minute. It summarizes the net change across both segments.

Average Rate Is Not an Instantaneous Rate

An average rate summarizes an entire interval using the quantity’s change between its endpoints. An instantaneous rate describes how the quantity is changing at one specified time. These answer different questions. The average-rate calculation needs the endpoint values and the interval length; it does not require the rate to be the same at every moment.

For example, a tank might gain water slowly during one part of an interval and quickly during another. The average rate over the full interval still records the net gain divided by the total elapsed time. Knowing that average does not identify the inflow rate at the beginning, the end, or any chosen moment in between. Information about the rate at an instant requires information about the process at that time, not only its overall change.

Likewise, a positive average rate does not imply that the quantity increased throughout the interval. It says only that its final value exceeds its initial value. The quantity could have decreased during part of the interval and increased by a larger amount during another part. The endpoint difference determines the average rate.

A Common Mistake: Dividing by the Ending Time

A frequent error is to divide the total change by the ending time rather than by the elapsed time. Suppose a quantity increases by \(42\) units from \(t=3\) hours to \(t=9\) hours. Dividing by \(9\) gives \(42/9\), but the interval lasted \(9-3=6\) hours, not \(9\) hours.

$$ \text{Incorrect: }\frac{42}{9}=\frac{14}{3}\text{ units per hour}. \qquad \text{Correct: }\frac{42}{9-3}=\frac{42}{6}=7\text{ units per hour}. $$

Check the corrected answer: \(7\) units per hour over \(6\) hours corresponds to a total change of \(7\cdot6=42\) units. The incorrect calculation uses the wrong duration, so multiplying its result by the actual \(6\)-hour interval would give only \((14/3)\cdot6=28\) units, not the stated change.

Another related error is to report a positive rate whenever the quantity’s amount is positive. The rate depends on the change, not on whether the quantity itself is positive. A tank can contain a positive amount of water while its volume is decreasing, giving a negative average rate over that interval.

A Reliable Calculation Process

1
Identify the interval.
Record the earlier and later times, and make sure the later time is greater than the earlier time.
2
Find the net change.
Subtract the quantity at the earlier time from the quantity at the later time. If rates are given by segments, calculate and add the signed changes for those segments first.
3
Divide by elapsed time.
Use the total time elapsed from the earlier endpoint to the later endpoint, not just the later time’s label.
4
Interpret and check.
State the units and explain the sign. Multiply the average rate by the interval length to check that it reproduces the net change.

The average rate is useful whenever a total change must be compared across intervals of different lengths. It gives a common basis for describing overall change, while preserving a clear distinction between an interval summary and the process’s behavior at a particular instant.

Check Your Understanding

For each question, pay attention to the endpoints, the sign of the change, and the units.

  1. A quantity changes from \(18\) units at \(t=4\) minutes to \(42\) units at \(t=10\) minutes. What is its average rate of change, including units?
  2. A position changes from \(11\) meters to \(2\) meters over \(3\) seconds. What is the average rate of change in position, and what does its sign indicate?
  3. A quantity increases at \(2\) units per second for \(5\) seconds, then decreases at \(1\) unit per second for \(5\) seconds. What is its net change and its average rate over the full \(10\)-second interval?
  4. In your own words, why does an interval average rate not determine the rate at a particular instant?
  5. A quantity changes by \(30\) units between \(t=2\) hours and \(t=8\) hours. Explain why dividing by \(8\) does not give its average rate over this interval.

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