Lesson 1.1 · Unit 1 · CED topic 1.1
Introducing calculus: can change occur at an instant?
A speedometer reads 62 miles per hour. But speed is distance divided by time, and at a single instant no time has passed and no distance has been covered. How can a car have a speed at an instant at all? That puzzle is the reason calculus exists. Algebra can only describe change over an interval. Calculus uses the idea of a limit to describe change at an instant, and everything in this course, the derivative, the integral, series, grows from that one idea.
Learning objective.
- CHA-1.A. Interpret the rate of change at an instant in terms of average rates of change over intervals containing that instant.
Essential knowledge.
- CHA-1.A.1. Calculus uses limits to understand and model dynamic change.
- CHA-1.A.2. Because an average rate of change divides the change in one variable by the change in another, the average rate of change is undefined at a point where the change in the independent variable would be zero.
- CHA-1.A.3. The limit concept allows us to define instantaneous rate of change in terms of average rates of change.
Average rate of change
For a function \(f\), the average rate of change over the interval from \(x=a\) to \(x=b\) is the change in the output divided by the change in the input:
\[ \frac{f(b)-f(a)}{b-a} \]
Geometrically it is the slope of the secant line through the two points \((a,f(a))\) and \((b,f(b))\) on the graph. Its units are the output units per input unit: if \(s(t)\) is distance in meters and \(t\) is time in seconds, the average rate of change is in meters per second. The same expression is often written with a step size \(h\): the average rate of change from \(x=a\) to \(x=a+h\) is \(\dfrac{f(a+h)-f(a)}{h}\). The two forms are equivalent (put \(b=a+h\)), and you will meet both throughout the course.
Notice what the formula needs: a denominator that is not zero. If the two input values are the same, \(b-a=0\), the quotient is \(\frac{0}{0}\) and the average rate of change is undefined. An average needs an interval of positive length to average over. A single point has no length, so on its face the formula says nothing about the rate of change at a point.
From average rates to an instantaneous rate
Here is the key move. We cannot divide by a zero-length interval, but we can use shorter and shorter intervals and watch what the averages do. If the averages settle toward a single number as the interval shrinks, that number is the best possible meaning of "the rate of change at the instant." The instantaneous rate of change at \(x=a\) is defined in terms of average rates of change: it is the number that the average rates over smaller and smaller intervals containing \(a\) approach.
In the figure, the point \(P\) is fixed and \(Q\) slides toward it. Each secant line through \(P\) and \(Q\) has a slope equal to an average rate of change. As \(Q\) approaches \(P\) the secant lines pivot toward one limiting line, the tangent line at \(P\), and the slopes approach the slope of that line. In the next lessons, the idea of "approaches" gets a precise name (a limit) and a precise notation. For now the point is the logic: the instantaneous rate of change is not a quotient with a zero denominator; it is the value that a family of ordinary quotients approaches.
A ball dropped from a bridge has fallen \(s(t)=5t^2\) meters after \(t\) seconds (using \(g=10\) m/s\(^2\)). Estimate its speed at the instant \(t=1\) by finding average velocities over shorter and shorter intervals starting at \(t=1\).
Over \([1,2]\). \(s(1)=5\) and \(s(2)=20\), so the average velocity is \(\dfrac{20-5}{2-1}=15\) m/s.
Over \([1,1.1]\). \(s(1.1)=5(1.21)=6.05\), so \(\dfrac{6.05-5}{0.1}=10.5\) m/s.
Over \([1,1.01]\). \(s(1.01)=5(1.0201)=5.1005\), so \(\dfrac{0.1005}{0.01}=10.05\) m/s.
Over \([1,1.001]\). \(s(1.001)=5.010005\), so \(\dfrac{0.010005}{0.001}=10.005\) m/s.
The averages 15, 10.5, 10.05, 10.005 are closing in on 10. The instantaneous velocity at \(t=1\) is 10 m/s. Algebra confirms it: over \([1,1+h]\) the average is \(\dfrac{5(1+h)^2-5}{h}=\dfrac{10h+5h^2}{h}=10+5h\), which approaches 10 as \(h\) shrinks.
The table gives the distance \(d(t)\), in feet, that a cyclist has traveled after \(t\) seconds.
| \(t\) (s) | 0 | 2 | 5 | 8 |
|---|---|---|---|---|
| \(d(t)\) (ft) | 0 | 30 | 90 | 192 |
Find the average rate of change over \([2,5]\) and over \([5,8]\) and say what each means.
\([2,5]\). \(\dfrac{90-30}{5-2}=\dfrac{60}{3}=20\) ft/s. \([5,8]\). \(\dfrac{192-90}{8-5}=\dfrac{102}{3}=34\) ft/s.
Over the first interval the cyclist averaged 20 feet per second, and over the second, 34 feet per second. The cyclist is speeding up. These are averages; the table alone cannot tell us the speed at \(t=5\) exactly, though it suggests a speed somewhere between 20 and 34 ft/s there.
Why the limit idea is needed
Every method in this course, finding slopes of curves, the speed of a particle at an instant, the area under a curve, the sum of an infinite list of numbers, runs into the same obstacle: the naive computation would divide by zero or add infinitely many things. Each time, calculus resolves the obstacle by replacing the impossible single computation with a sequence of possible ones and asking what they approach. That is why Unit 1 is about limits before any derivative is mentioned. When you reach the formal definition of the derivative in Unit 2, you will see that it is exactly the expression \(\dfrac{f(a+h)-f(a)}{h}\) with the instruction "let \(h\) approach \(0\)."
On the AP exam a graphing calculator can build a table of average rates for you. Enter \(Y_1=5X^2\), then compute \(\dfrac{Y_1(1+h)-Y_1(1)}{h}\) for \(h=0.1,\,0.01,\,0.001\). A table of values is a quick way to see what the averages are approaching, but a pattern in a table is evidence, not proof. Later lessons show how to find limits exactly.
- Average rate of change over \([a,b]\) is \(\dfrac{f(b)-f(a)}{b-a}\), the slope of the secant line; its units are output units per input unit.
- It is undefined when the two input values are equal, because the denominator is zero.
- The instantaneous rate of change at \(x=a\) is defined through the limit of average rates of change over shorter and shorter intervals containing \(a\).
- Tables and graphs give average rates; a value read from a table over a short interval is an estimate of the instantaneous rate, not its exact value.
- On free-response questions, "include units" is part of the answer: write them.
- Find the average rate of change of \(f(x)=x^2-3x\) over \([1,4]\).
Show answer
\(f(1)=-2\) and \(f(4)=4\), so the average rate is \(\dfrac{4-(-2)}{4-1}=\dfrac{6}{3}=2\).
- A student tries to find the average rate of change of \(g\) over the "interval" \([3,3]\). What goes wrong?
Show answer
The change in input is \(3-3=0\), so the quotient is \(\dfrac{g(3)-g(3)}{0}=\dfrac{0}{0}\), which is undefined. An average rate of change needs an interval of positive length.
- The temperature of a cup of tea is \(T(0)=95\) and \(T(10)=71\), where \(T\) is in degrees Fahrenheit and \(t\) is in minutes. Find the average rate of change of \(T\) over \([0,10]\) and include units.
Show answer
\(\dfrac{71-95}{10-0}=-2.4\) degrees Fahrenheit per minute. The negative sign means the tea is cooling.
- For \(f(x)=\sqrt{x}\), simplify \(\dfrac{\sqrt{4+h}-2}{h}\) by multiplying by the conjugate, then evaluate it for \(h=0.01\). What number do the averages seem to approach as \(h\) shrinks?
Show answer
\(\dfrac{\sqrt{4+h}-2}{h}\cdot\dfrac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\dfrac{h}{h(\sqrt{4+h}+2)}=\dfrac{1}{\sqrt{4+h}+2}\). For \(h=0.01\) this is about \(0.2498\). As \(h\) shrinks the values approach \(\dfrac{1}{2+2}=0.25\).
- The position of a runner is \(p(t)=2t^2\) meters at \(t\) seconds. Find the average velocity over \([3,3.01]\) and use it to estimate the velocity at \(t=3\).
Show answer
\(p(3)=18\) and \(p(3.01)=2(9.0601)=18.1202\), so the average velocity is \(\dfrac{0.1202}{0.01}=12.02\) m/s. The velocity at \(t=3\) is approximately 12 m/s (the exact value is 12).
- Explain in a sentence why a speedometer reading cannot be computed as distance divided by elapsed time at a single instant, and how calculus fixes this.
Show answer
At one instant the elapsed time is zero, so the quotient would be \(\frac{0}{0}\). Calculus instead takes average speeds over shorter and shorter time intervals around the instant and defines the instantaneous speed as the limit of those averages.