Average Rates Describe an Interval
A model can describe an output that changes as its input changes. For example, a function might give temperature at each time, water depth at each time, or cost at each usage level. When the model is nonlinear, its output does not follow a straight-line relationship with its input. An average rate is still useful: it summarizes the overall change between two selected inputs.
The word average matters. An interval-wide average uses the model’s values at the two endpoints. It does not, on its own, tell us how the output behaves at every input between those endpoints.
The numerator measures the net output change over the interval; the denominator measures the corresponding input change. The rate’s units are output units divided by input units. If the output decreases as the input increases, the average rate is negative. The formula requires \(a\ne b\) and that the model be defined at both endpoints.
Why the Formula Gives an Interval Average
Suppose the output changes from \(f(a)\) to \(f(b)\) while the input changes from \(a\) to \(b\). The total output change is \(f(b)-f(a)\), and the total input change is \(b-a\). If one number \(r\) is to summarize the output change per input unit across the whole interval, it must satisfy \(r(b-a)=f(b)-f(a)\). Dividing by the nonzero input change gives the average-rate formula.
This also explains the geometric interpretation. The endpoint values give two points on the graph, \((a,f(a))\) and \((b,f(b))\). The average rate is the slope between those points, called the secant slope. The calculation uses the endpoints, not a claim that the graph has that same slope throughout the interval.
If a model is linear, its average rate is the same on every interval in its domain. A nonlinear model generally does not have this constant-rate property. Therefore, the average over one interval can be a useful summary without being a description of the model’s behavior at every input in that interval.
Worked Example: Average Cooling Rate
Worked Example: Average Cooling Rate
A simplified temperature model for an object is \(T(t)=18+\frac{72}{t+1}\), where \(t\) is time in hours and \(T(t)\) is temperature in degrees Celsius. Find the average rate of change from \(t=0\) to \(t=3\), and interpret it.
First evaluate the model at the endpoints:
The input changes by \(3-0=3\) hours, and the output changes by \(36-90=-54\) degrees Celsius. Thus,
Over the three-hour interval, the object’s temperature has an average decrease of \(18\) °C per hour. The negative sign records that the temperature is lower at the end than at the start. It does not say that the temperature drops by exactly \(18\) °C during every hour.
Check: Multiplying the average rate by the three-hour input change gives \((-18)(3)=-54\) °C, which matches \(T(3)-T(0)=36-90=-54\) °C.
Worked Example: Water Level in a Tank
Worked Example: Average Rise in Water Level
For a particular filling process, suppose the water depth is modeled by \(h(t)=1+\frac{1}{2}t^2\), where \(t\) is time in minutes and \(h(t)\) is depth in metres. Find the average rate of change from \(t=0\) to \(t=4\), and explain what the rate does and does not tell you.
Calculate the endpoint depths:
The depth change is \(9-1=8\) metres, while the time change is \(4-0=4\) minutes. Therefore,
The water depth rises by an average of \(2\) metres per minute across these four minutes. The model is nonlinear, so this interval average does not imply that the water rises at a constant \(2\) metres per minute throughout the interval.
Check: The calculated average rate times the elapsed time is \((2)(4)=8\) metres, equal to the endpoint depth change \(9-1=8\) metres.
The rate’s units describe how the output changes relative to the input; they do not automatically describe a physical flow such as volume per minute. Here the model gives depth as a function of time, so the result is metres per minute. A model of volume would be needed to calculate a volume-flow rate.
Worked Example: Average Cost Change
Worked Example: Average Cost per Usage Unit
Suppose a service’s cost at usage level \(q\) is modeled by \(C(q)=4+0.5q+0.02q^2\), where \(q\) is measured in usage units and \(C(q)\) is measured in dollars. Find the average rate of cost change from \(q=10\) to \(q=30\). Interpret the units carefully.
Evaluate the cost at \(q=10\):
Now evaluate the cost at \(q=30\):
The cost increases by \(37-11=26\) dollars as usage increases by \(30-10=20\) units. Hence,
Across this usage interval, cost increases by an average of \(1.30\) dollars per usage unit. This is an interval-wide average, not a claim that every additional usage unit has the same cost effect.
Check: The average rate multiplied by the \(20\)-unit change is \((1.3)(20)=26\) dollars, matching \(C(30)-C(10)=37-11=26\) dollars.
What an Average Does Not Tell You
An average rate preserves the total endpoint change, but it compresses the behavior over the interval into one number. In the cooling example, the value \(-18\) °C per hour correctly summarizes the change from \(90\) °C to \(36\) °C over three hours. It does not give the temperature at every time, nor does it show how the change is distributed across the interval.
For that reason, keep three questions separate:
- Which interval? The endpoints determine the input change being summarized.
- Which output change? Subtract the initial output from the final output, keeping the order consistent.
- What is the interpretation? State the average change per input unit and include the relevant units.
If a question asks for an interval-wide rate, use both endpoint values. If it asks what happens at one particular input, the interval average alone is not enough information to answer. A nonlinear model may change at different rates at different inputs, even though one average correctly describes its total change across a chosen interval.
A Common Mistake: Treating the Average as Constant
In the water-level example, the average rate from \(t=0\) to \(t=4\) is \(2\) metres per minute. A mistaken conclusion would be: “The level rises by \(2\) metres in every minute.” That conclusion incorrectly treats the interval average as a constant rate throughout the interval.
For instance, the model gives \(h(2)=1+\frac{1}{2}(4)=3\) metres. From \(t=0\) to \(t=2\), the average rate is
not \(2\) metres per minute. The value \(2\) belongs to the average over the full interval from \(0\) to \(4\); it cannot be assigned to each part of that interval. The correct interpretation is that the total eight-metre rise over four minutes corresponds to an average of \(2\) metres per minute.
Check Your Understanding
For each question, identify what the average rate summarizes and keep the input and output units in view.
- A model gives \(P(2)=15\) and \(P(6)=27\), where \(P\) is measured in litres and the input is time in minutes. What is the average rate from input \(2\) to input \(6\), including units?
- A temperature model has values \(T(1)=22\) °C and \(T(5)=10\) °C. Is its average rate over this interval positive or negative, and what does the sign mean?
- In your own words, why does an average rate of \(3\) output units per input unit over an interval not prove that the output changes by \(3\) units during every input unit?
- A nonlinear cost model gives \(C(4)=18\) dollars and \(C(10)=42\) dollars. Calculate the average cost change per usage unit and check it by multiplying by the input change.
- What information do you need from a function to calculate its average rate between two inputs?
