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Calculus AB · Unit 2 · 2.1
AP Calculus · Differentiation: Definition and Fundamental Properties
Average and Instantaneous Rates of Change
One needs an interval and gives a secant slope; the other needs a point and gives a tangent slope.
key idea$\text{Average}=\frac{f(b)-f(a)}{b-a};\qquad \text{Instantaneous}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$
What it says
Average rate of change
\[\frac{f(b)-f(a)}{b-a}\]
The total change in output divided by the total change in input, across an interval. Geometrically it is the slope of the secant line joining the two endpoints. It needs two points and says nothing about what happened between them.
Instantaneous rate of change
\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]
The value the average rates approach as the interval shrinks to nothing around $a$. Geometrically it is the slope of the tangent at that single point.
the distinction
- The difference quotient $\dfrac{f(a+h)-f(a)}{h}$ is an average rate over $[a,a+h]$.
- Average needs an interval; instantaneous needs a point.
- Both carry the same units: output units per input unit.
- An average rate can be $0$ while the instantaneous rate is never $0$ — and vice versa.
Reading the notation
Difference quotient$\dfrac{f(a+h)-f(a)}{h}$ — an average rate written with a gap $h$ rather than two named endpoints. The form the derivative definition uses.
$\Delta y/\Delta x$The same average rate in change-notation. $\Delta$ means ‘change in’.
Secant lineThe straight line through two points on the curve. Its slope is the average rate.
Tangent lineThe line touching the curve at one point, matching its direction there. Its slope is the instantaneous rate.
“Average velocity”Average rate of change of position. “Average speed” is different — it uses total distance, not net displacement.
Worked examples
EX 1
Both rates for one function
$f(x)=x^{2}$For $f(x)=x^{2}$, find the average rate of change on $[1,4]$ and the instantaneous rate at $x=1$.
1
Average rate
$\dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5$.
2
Set up the instantaneous rate
$\lim_{h\to0}\dfrac{(1+h)^{2}-1}{h}$.
3
Simplify
$\dfrac{1+2h+h^{2}-1}{h}=\dfrac{2h+h^{2}}{h}=2+h$ for $h\neq0$.
4
Take the limit
$\lim_{h\to0}(2+h)=2$.
5
Compare
Average $5$ over $[1,4]$; instantaneous $2$ at $x=1$. Different questions, different numbers — the curve steepens across the interval.
EX 2
Average zero, instantaneous not
$f(x)=x^{2}-4$ on $[-2,2]$Find the average rate of change of $f(x)=x^{2}-4$ on $[-2,2]$, and comment.
1
Evaluate the endpoints
$f(-2)=0$ and $f(2)=0$.
2
Compute the average
$\dfrac{0-0}{2-(-2)}=\dfrac{0}{4}=0$.
3
Interpret
The function ends where it started, so the net change is zero.
4
Contrast with the instantaneous rates
$f'(x)=2x$, which is $-4$ at $x=-2$ and $+4$ at $x=2$ — never zero except at the vertex.
5
The point
An average rate of $0$ means no net change, not no change. The function fell and then rose by the same amount.
EX 3
From a table
estimationA car's odometer reads $s(2)=110$ km and $s(5)=272$ km. Find the average velocity, and say what you cannot determine.
1
Apply the formula
$\dfrac{s(5)-s(2)}{5-2}=\dfrac{272-110}{3}$.
2
Compute
$=\dfrac{162}{3}=54$ km/h.
3
State the meaning
Over those three hours the car averaged $54$ km/h.
4
State the limitation
You cannot determine the speed at any particular moment. The car may have stopped, or exceeded $100$ km/h, and still produced this average.
5
What would fix it
Only readings taken arbitrarily close together — or a formula — give the instantaneous rate.
EX 4
Average velocity versus average speed
distinctionA ball is thrown up and returns to the thrower's hand after $4$ seconds. Find its average velocity, and explain why average speed differs.
1
Average velocity uses displacement
Start and end position are the same, so the net change is $0$.
2
Compute it
$\dfrac{0}{4}=0$ m/s.
3
Average speed uses distance
The ball travelled up and back down, so the total distance is twice the maximum height — a positive number.
4
Give the contrast
Average velocity $=0$; average speed $\gt 0$. The words are not interchangeable.
5
The rule
Velocity is a rate of change of position and carries a sign. Speed is its magnitude and never does.
Boundary cases
“Average rate at a point” is not a thing
An average rate always needs two inputs. If a question names a single point, it wants the instantaneous rate.
Zero average does not mean nothing happened
It means the net change was zero. Example 2 falls four units and climbs back — a great deal happened.
Two points cannot reveal the middle
An average rate is compatible with wildly different behaviour in between. Never infer an instantaneous value from a single average.
Velocity and speed are different quantities
Velocity carries a sign and uses displacement; speed is its absolute value and uses distance. Averages of the two rarely match.