Two Different Questions About Change
A quantity can change across an interval, and it can also have a rate of change at one particular input. These questions are related, but they do not ask for the same thing.
For example, suppose \(Q(t)\) gives the amount of water in a tank at time \(t\). Comparing the amount at 1 hour with the amount at 4 hours asks how much the quantity changed over that entire time interval, on average. Asking how quickly the amount is changing at exactly 1 hour is a different question: it concerns the instantaneous rate represented by \(Q'(1)\).
The distinction matters because an interval rate summarizes change across a span, while a derivative describes local change at one input. The two rates have the same type of units, but they need not have the same numerical value.
Average Rate of Change
Let \(f\) describe a quantity in terms of an input \(x\). Choose two inputs \(a\) and \(b\), with \(a\ne b\). The change in the output is \(f(b)-f(a)\), and the change in the input is \(b-a\). Dividing these changes gives the average rate of change from \(a\) to \(b\).
The order of subtraction must match in the numerator and denominator: output at \(b\) minus output at \(a\), divided by input \(b\) minus input \(a\). Reversing both differences gives the same rate; reversing only one gives the wrong sign.
The units follow directly from the quotient. If \(f(x)\) is measured in liters and \(x\) in hours, the rate is measured in liters per hour. The average rate describes the net output change per input unit across the interval. It does not assert that the quantity changed at that rate at every point in the interval.
Instantaneous Rate of Change
The derivative \(f'(a)\) describes the instantaneous rate of change at the input \(a\), provided the derivative exists there. Its units are output units per input unit, just as for an average rate.
Each quotient inside this limit is an average rate: it compares the function values at \(a\) and \(a+h\). The derivative is the limiting value of these average rates as the second input approaches \(a\). It is not found by setting \(h=0\) in the quotient, since that would make its denominator zero.
Why the Two Rates Are Connected
The connection follows by writing an interval endpoint as \(a+h\). The average rate from \(a\) to \(a+h\) is \(\frac{f(a+h)-f(a)}{h}\). If \(f'(a)\) exists, the definition of the derivative says that this average rate approaches \(f'(a)\) as \(h\) approaches zero.
For inputs \(a\) and \(b\), calculate \(\frac{f(b)-f(a)}{b-a}\).
Write \(b=a+h\), so the interval rate becomes \(\frac{f(a+h)-f(a)}{h}\), with \(h\ne 0\).
If the limit exists as \(h\to 0\), it equals \(f'(a)\) by the definition of the derivative.
This is a complete reason for the relationship: an instantaneous rate is the limit of average rates over intervals whose second endpoint approaches the input of interest. The condition that the derivative exists is essential. If the limit does not exist, there is no derivative at that input, even though average rates on intervals with distinct endpoints can still be calculated.
Worked Example: Comparing an Interval Rate with a Derivative
Worked Example: Water in a Tank
A model for the amount of water in a tank is \(V(t)=100+12t-t^2\) liters, where \(t\) is time in hours. Find the average rate of change from \(t=1\) to \(t=4\), then find the instantaneous rate at \(t=2\).
Average rate: First evaluate the function at both endpoints:
The output change is \(132-111=21\) liters, and the time change is \(4-1=3\) hours. Therefore,
Instantaneous rate: Differentiate the model to obtain \(V'(t)=12-2t\). At \(t=2\),
The rates differ: the average rate over the three-hour interval is \(7\) liters per hour, whereas the instantaneous rate at hour 2 is \(8\) liters per hour.
Check: From the derivative definition, \(V(2)=100+24-4=120\), and \(V(2+h)=100+12(2+h)-(2+h)^2=120+8h-h^2\). Thus
This confirms the instantaneous rate at \(t=2\). The average-rate calculation also checks because \(7\) liters per hour for 3 hours corresponds to a total change of \(7(3)=21\) liters, matching \(V(4)-V(1)\).
Worked Example: An Average Rate Need Not Equal a Nearby Derivative
Worked Example: A Quantity Following a Cubic Model
Suppose \(Q(t)=t^3\) describes a quantity in units as a function of time \(t\) in days. Compare the average rate from day 1 to day 3 with the instantaneous rate at day 2.
The endpoint values are \(Q(1)=1^3=1\) unit and \(Q(3)=3^3=27\) units. The change in the input is \(3-1=2\) days, so
For the instantaneous rate, differentiate: \(Q'(t)=3t^2\). At day 2,
The two values are close, but not equal: the interval average is \(13\), while the derivative at day 2 is \(12\). The average rate uses the values at days 1 and 3; the derivative uses local change at day 2.
Check: \(Q(2+h)=(2+h)^3=8+12h+6h^2+h^3\), so
and its limit as \(h\to0\) is \(12\), as calculated by differentiation. The average rate also reconstructs the observed change: \(13\) units per day multiplied by \(2\) days is \(26\) units, and \(27-1=26\).
Worked Example: Both Rates Can Be Negative
Worked Example: A Decreasing Quantity
Let \(N(t)=200-3t^2\) be a quantity measured in units, with \(t\) in days. Find its average rate from day 2 to day 5 and its instantaneous rate at day 3.
At the endpoints,
The average rate is therefore
Differentiate to get \(N'(t)=-6t\). At day 3,
The negative average rate says that the quantity's net change across the interval is a decrease of 63 units over 3 days. The negative derivative says that at day 3 its instantaneous rate of change is \(-18\) units per day.
Check: \(N(3)=200-27=173\), and \(N(3+h)=200-3(9+6h+h^2)=173-18h-3h^2\). Hence
Also, \((-21\text{ units per day})(3\text{ days})=-63\text{ units}\), which matches \(N(5)-N(2)=125-188=-63\) units.
A Common Mistake: Using One Input Instead of an Interval
A frequent error is to treat an average rate as the derivative at one of the interval's endpoints. In the water-tank example, this would give the incorrect calculation
The left side is \(7\), not \(10\). The correction is to evaluate the two function values and divide their difference by the input difference:
The derivative \(V'(1)=10\) answers a different question: it is the instantaneous rate at hour 1. A further mistake is to assume that an average rate must equal the derivative at the midpoint of the interval. It does in some particular examples, but it is not part of the definition and is not true in general. The cubic example shows that the average from day 1 to day 3 is \(13\), while the derivative at the midpoint, day 2, is \(12\).
Choosing the Rate That Matches the Question
Look for the wording in a problem. If it asks for change “from \(a\) to \(b\),” “over the interval,” or “on average,” use two function values and calculate an average rate. If it asks for the rate “at \(a\),” use the derivative \(f'(a)\), assuming it exists.
In either case, identify the units before interpreting the result. Both rates have units of output per input. But the interpretation differs: an average rate summarizes the total change across an interval, while an instantaneous rate describes local change at one input. Keeping those meanings separate prevents a correct calculation from being used to answer the wrong question.
Check Your Understanding
For each question, distinguish an interval average from an instantaneous rate and keep track of the units.
- If \(f(2)=9\) and \(f(6)=25\), what is the average rate of change from \(x=2\) to \(x=6\)?
- In your own words, what does \(f'(a)\) describe that an average rate from \(a\) to \(b\) does not?
- For \(P(t)=40+2t^2\), where \(P\) is in units and \(t\) is in hours, calculate the average rate from \(t=1\) to \(t=3\).
- For the same function, calculate \(P'(2)\). Is it equal to the average rate in question 3?
- State the limit that connects average rates over intervals beginning at \(a\) to the derivative at \(a\), and state the condition required for that limit to represent an instantaneous rate.
