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Rates of Change and Tangents · Tutorial 322 of 1495

Relating Output Change To Average Rate

Use an average rate and an input interval to calculate the expected output change, then check the result using units and endpoint values.

Beginner 9 min read

What You'll Learn

  • Rearrange the average-rate relationship to solve for output change.
  • Distinguish an output change from an output’s ending value.
  • Use consistent endpoint order when finding an interval’s change.
  • Track units to confirm that a predicted change has the right units.
  • Check a prediction by comparing its implied rate with the stated rate.
  • Interpret negative rates as decreases over the chosen interval.

From a Rate to an Output Change

An average rate tells us how much an output changes, on average, for each unit of input change across an interval. It can also be used in the reverse direction: if we know the average rate and the input change, we can calculate the corresponding output change.

For instance, if a tank’s water volume decreases by an average of 2.5 liters per minute over 3 minutes, the total change in volume is the rate multiplied by the time change. The sign matters: a decrease is represented by a negative output change.

The calculation gives a change, not the ending amount by itself. To find an ending amount, we add the change to the starting amount. Keeping those two steps distinct helps prevent sign errors and makes it easier to check a prediction.

The Average-Rate Relationship

Let an output be described by a function \(f\), and consider input values \(a\) and \(b\), where \(a\ne b\). The average rate of change of \(f\) from \(a\) to \(b\) is the output change divided by the input change. The order of the endpoints must be the same in both differences.

Definition: The average rate of change of \(f\) from input \(a\) to input \(b\) is
$$ \frac{f(b)-f(a)}{b-a}, \qquad a\ne b. $$

Write this average rate as \(r\). Multiplying both sides of its defining equation by the input change \(b-a\) gives the output change. This rearrangement is useful whenever the average rate and the two input endpoints are known.

$$ r=\frac{f(b)-f(a)}{b-a} \qquad\Longrightarrow\qquad f(b)-f(a)=r(b-a). $$
Key relationship: Output change equals average rate multiplied by input change. In symbols, \(\Delta y=r\Delta x\), where \(\Delta x\) is the change in input and \(\Delta y\) is the change in output.

If the starting output \(f(a)\) is also known, add the output change to it to obtain the ending output:

$$ f(b)=f(a)+r(b-a). $$

These equations can describe an exact change when \(r\) is the actual average rate on the interval from \(a\) to \(b\). They can also be used for a prediction when \(r\) is an estimated or assumed average rate. In the latter case, the computed endpoint is a prediction based on that rate; it need not equal the eventual observed endpoint.

Why Multiplying by the Input Change Works

The definition of average rate says that the net output change is divided by the net input change. If the rate is known, multiplying by the input change reverses that division. This is an algebraic rearrangement, but it also matches the meaning of the units: output units per input unit, multiplied by input units, leaves output units.

1
Write the average-rate equation.
For endpoints \(a\) and \(b\), the rate is \(r=(f(b)-f(a))/(b-a)\).
2
Multiply by the input change.
Multiplying both sides by \(b-a\) gives \(r(b-a)=f(b)-f(a)\).
3
Add the starting output if needed.
From \(f(b)-f(a)=r(b-a)\), add \(f(a)\) to both sides to obtain \(f(b)=f(a)+r(b-a)\).

No assumption that the output changes at a constant rate at every moment is needed for this identity. It concerns the net change across the whole interval. If a rate is being used as a forecast rather than as the actual average over that interval, the calculation instead states what output change the forecast rate implies.

Worked Examples

Worked Example: Predicting a Decrease in Water Volume

A tank contains \(26\) liters of water at time \(t=8\) minutes. Over the interval from \(t=8\) to \(t=11\) minutes, its average rate of volume change is \(-2.5\) liters per minute. Find the volume change and predict the volume at \(t=11\).

First calculate the input change, using the later time minus the earlier time:

$$ \Delta t=11-8=3\text{ minutes}. $$

Multiply the average rate by this input change:

$$ \Delta V= \left(-2.5\frac{\text{liters}}{\text{minute}}\right) (3\text{ minutes}) =-7.5\text{ liters}. $$

The minutes cancel, leaving liters. The negative sign means the volume decreases by \(7.5\) liters. Add this change to the starting volume:

$$ V(11)=26\text{ liters}+(-7.5\text{ liters}) =18.5\text{ liters}. $$

Check the implied average rate using the two endpoint volumes:

$$ \frac{18.5-26}{11-8} = \frac{-7.5}{3} =-2.5\frac{\text{liters}}{\text{minute}}. $$

This matches the stated rate, so the change and predicted endpoint are consistent with the given information.

Worked Example: Predicting a Cost Increase

A cost model has an average rate of change of \(1.20\) dollars per item as the number of items increases from \(40\) to \(70\). The cost at \(40\) items is \(185\) dollars. Use the average rate to find the predicted cost increase and the cost at \(70\) items.

The input change is the difference between the item counts:

$$ \Delta q=70-40=30\text{ items}. $$

The output change is the rate multiplied by those \(30\) additional items:

$$ \Delta C= \left(1.20\frac{\text{dollars}}{\text{item}}\right) (30\text{ items}) =36\text{ dollars}. $$

The units reduce to dollars. Add this increase to the cost at \(40\) items:

$$ C(70)=185\text{ dollars}+36\text{ dollars} =221\text{ dollars}. $$

To check, calculate the average rate from the two endpoint costs:

$$ \frac{221-185}{70-40} = \frac{36}{30} =1.20\frac{\text{dollars}}{\text{item}}. $$

The endpoint values imply the stated average rate. Therefore, the predicted cost increase and ending cost are consistent.

Worked Example: Predicting a Temperature Change

At hour \(2\), a sample has a temperature of \(19.5\) degrees Celsius. Suppose its average rate of temperature change from hour \(2\) to hour \(6\) is \(-1.75\) degrees Celsius per hour. Find the predicted temperature change and temperature at hour \(6\).

The time interval has length:

$$ \Delta t=6-2=4\text{ hours}. $$

Multiply by the average rate:

$$ \Delta T= \left(-1.75\frac{\text{degrees Celsius}}{\text{hour}}\right) (4\text{ hours}) =-7\text{ degrees Celsius}. $$

The predicted change is a decrease of \(7\) degrees Celsius. Starting from \(19.5\) degrees Celsius, the predicted endpoint is:

$$ T(6)=19.5+(-7)=12.5\text{ degrees Celsius}. $$

Check by calculating the average rate implied by the endpoint values:

$$ \frac{12.5-19.5}{6-2} = \frac{-7}{4} =-1.75\frac{\text{degrees Celsius}}{\text{hour}}. $$

This agrees with the stated rate. If \(-1.75\) was an estimated rate, this check confirms the arithmetic and the rate implied by the prediction; an independently measured temperature at hour \(6\) would be needed to assess how accurate the prediction was.

Using Units and Endpoints as Checks

Units provide a quick way to detect whether the multiplication is set up correctly. An average rate has units of output per input. The input change has units of input. Their product must therefore have units of output:

$$ \frac{\text{output units}}{\text{input units}} \times \text{input units} = \text{output units}. $$

If the result still has units of input per output, or if an input unit has not canceled, review the calculation. Units cannot catch every arithmetic error, but they can reveal a mismatched quantity or an incorrect operation.

Endpoint values provide a second check. If you have calculated an output change \(\Delta y\), then the proposed ending output must be the starting output plus that signed change. If both endpoint outputs are available, their difference should equal the calculated change. Dividing that difference by the input change should recover the average rate.

For an exact average rate, recovering the given rate checks consistency of the calculations. For a forecast based on an estimated rate, it checks what the prediction implies, not whether the forecast matches reality. Keep that distinction clear when using a model.

Common Mistake: Losing the Sign

Suppose a quantity starts at \(26\) liters and changes at an average rate of \(-2.5\) liters per minute for \(3\) minutes. A common incorrect calculation is to ignore the negative sign and write:

$$ \Delta V=(2.5)(3)=7.5\text{ liters}, \qquad V_{\text{end}}=26+7.5=33.5\text{ liters}. $$

That calculation predicts an increase, even though the given rate is negative and describes a decrease. The correction is to preserve the sign of the rate throughout:

$$ \Delta V=(-2.5)(3)=-7.5\text{ liters}, \qquad V_{\text{end}}=26+(-7.5)=18.5\text{ liters}. $$

The endpoint check exposes the mistake: the incorrect endpoint would imply an average rate of \((33.5-26)/3=2.5\) liters per minute, the opposite sign from the stated rate.

Choosing the Interval Correctly

Use the same endpoint order for both changes. From \(a\) to \(b\), calculate the input change as \(b-a\) and the output change as \(f(b)-f(a)\). If you reverse the endpoints in one difference but not the other, the signs will not represent the average rate over the stated direction.

Also make sure the rate applies to the interval you are using. An average rate measured from one pair of endpoints does not automatically determine the output change on a different interval. To make a prediction for another interval, you need a rate assumed or estimated for that interval. A rate model may be useful, but its assumptions should be stated.

This relationship matters in many settings: it turns rates such as liters per minute, dollars per item, or degrees per hour into finite changes in the corresponding output. The essential checks are straightforward: identify the interval, preserve the signs, verify that the units reduce correctly, and relate the change to the starting endpoint.

Check Your Understanding

For each question, state the signed output change and, when enough information is provided, the predicted ending output.

  1. A quantity changes at an average rate of \(4\) meters per second over an interval of \(6\) seconds. What output change does this rate imply?
  2. A tank holds \(42\) liters at the beginning of a \(5\)-minute interval and drains at an average rate of \(1.6\) liters per minute. What volume change and ending volume are predicted?
  3. A model’s output decreases by \(15\) units as its input increases by \(3\) units. What is the average rate of change, including its sign and units?
  4. Why does multiplying an average rate by the input change produce output units?
  5. If a forecast rate is used to calculate an endpoint, what does checking the implied average rate establish, and what does it not establish?

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