Introducing calculus: can change occur at an instant?

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Calculus is the mathematics of change.

Algebra can tell you how much something changed between two moments; calculus can tell you how fast it is changing at one moment.

The tool that makes this possible is the limit, and this whole unit is about limits: what they mean, how to find them from graphs, tables and formulas, and what they tell you about continuity and asymptotes.

Unit 1 is examined mostly in multiple choice and as a step inside free-response questions: an existence argument in a table question such as Q1, or the end behavior of a model written as a limit.

Nothing is printed for you in either section of the AP exam: no formula sheet, no table of limits, derivatives or integrals, and no list of theorems.

Every definition, rule and theorem in these notes is something you recall, understand and use, along with the algebra and trigonometry from precalculus that the questions rely on.

How the points work

Each free-response question is worth 9 points, split over 3 or 4 parts, and every question is compulsory.

Points are never printed on the exam paper. The [n] and Total [n] in these notes show what a scoring guideline would award.

One point per element, never half points. Points are numbered P1 to P9 across a question, and each label says what earns it: a setup point (a limit expression, the condition of a theorem) and then an answer point.

Some points need earlier ones: no setup, no answer point. A later part may use a wrong value from an earlier part and still earn its own points (an imported value).

A correct expression earns its answer point even if you do not simplify it, and a later arithmetic slip does not take that point back. Any simplification you attempt must be correct.

A bald answer, a number with no supporting work, usually earns nothing.

Decimal answers need three places after the decimal point, rounded or truncated, and you lose at most one point a question for rounding. Exact answers may be left unsimplified on the no-calculator part.

Average rate of change

Suppose a hot-air balloon lifts off and its height above the ground, in feet, $t$ minutes after liftoff is

h(t) = 8t^2 + 30t, \qquad 0 \le t \le 5.

Between $t = 1$ and $t = 4$ the height goes from $h(1) = 38$ feet to $h(4) = 248$ feet.

The average rate of change over that interval is the change in height divided by the change in time:

\frac{h(4) - h(1)}{4 - 1} = \frac{248 - 38}{3} = 70 \text{ feet per minute}.

On a graph, this is the slope of the secant line through the two points $(1, 38)$ and $(4, 248)$ .

Its units are always "units of the output per unit of the input", here feet per minute.

Can change happen at an instant?

Now ask a sharper question: how fast is the balloon rising at exactly $t = 2$ ?

Try the same recipe with both times equal to 2:

\frac{h(2) - h(2)}{2 - 2} = \frac{0}{0}.

That is not a number.

You need to know that 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.

A single instant has no length, so it has no average rate of change.

Yet the balloon clearly is rising at some speed at $t = 2$ : a passenger's phone could show it.

The way out is to take average rates over shorter and shorter intervals that contain $t = 2$ , and watch what number they settle toward.

With a step of $\Delta t$ minutes,

\frac{h(2 + \Delta t) - h(2)}{\Delta t} = \frac{8(2 + \Delta t)^2 + 30(2 + \Delta t) - 92}{\Delta t} = \frac{62\,\Delta t + 8(\Delta t)^2}{\Delta t} = 62 + 8\,\Delta t \quad (\Delta t \ne 0).

$\Delta t$ (minutes)

0.1

0.01

0.001

$-0.001$

$-0.01$

$-0.1$

average rate (feet per minute)

62.8

62.08

62.008

61.992

61.92

61.2

From both sides the average rates close in on 62.

The graph below shows what is happening: each average rate is the slope of a secant line from $(2, 92)$ , and as the second point slides toward $t = 2$ the secant lines turn toward one line that just touches the curve at $(2, 92)$ .

The graph of the balloon's height h(t) = 8t squared plus 30t, in feet, for t from 0 to about 4.4 minutes. Two dashed secant segments start at the point (2, 92) on the curve: one ends at (4, 248), with slope 78, and a shorter one ends at (3, 162), with slope 70. A solid line through (2, 92) with slope 62 just touches the curve there. As the second point moves toward t = 2, the secant slopes close in on 62.

The number the average rates approach, 62 feet per minute, is the instantaneous rate of change of $h$ at $t = 2$ .

You need to know that the limit concept allows us to define instantaneous rate of change in terms of average rates of change.

In unit 2 this limit gets its name, the derivative, and its notation, $h'(2)$ .

Why calculus needs limits

Notice what the limit did.

The expression $62 + 8\,\Delta t$ is only valid for $\Delta t \ne 0$ , so you cannot just set $\Delta t = 0$ in the original fraction.

The limit asks a different question: not "what is the value at $\Delta t = 0$ ?"

but "what value do the outputs get close to as $\Delta t$ gets close to 0?".

That question has an answer even when the value at the point does not exist.

This is the sense in which calculus uses limits to understand and model dynamic change: velocity, the rate a medication enters the blood, the rate a city's population grows, the slope of a curve, the area under a curve.

Every one of these is defined by a limit of simpler quantities you can already compute.

Key terms

Calculus and change: Calculus uses limits to understand and model dynamic change.

Average rate of change at a point: 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.

Instantaneous rate of change: The limit concept allows us to define instantaneous rate of change in terms of average rates of change.

Average rate of change of $f$ on $[a, b]$ : $\dfrac{f(b) - f(a)}{b - a}$ , the slope of the secant line through $(a, f(a))$ and $(b, f(b))$ , in units of the output per unit of the input.

Worked example 1: an average rate of change with units

No calculator.

A cyclist's distance from the start of a trail, in miles, $t$ hours after she sets off is $D(t) = 12t - t^2$ for $0 \le t \le 4$ .

Find the average rate of change of $D$ over $1 \le t \le 3$ .

Indicate units of measure.

1. $D(1) = 12 - 1 = 11$ and $D(3) = 36 - 9 = 27$ .

2. Average rate of change $= \dfrac{D(3) - D(1)}{3 - 1} = \dfrac{27 - 11}{2} = 8$ .

Answer: 8 miles per hour.

Over those two hours the cyclist's distance from the start grows by 8 miles each hour on average; she may have been faster at some moments and slower at others.

Worked example 2: estimating an instantaneous rate from average rates

Calculator allowed.

Let $f(x) = \sqrt{x}$ .

Find the average rate of change of $f$ over $[4, 4 + \Delta x]$ for $\Delta x = 0.1$ , $0.01$ , $0.001$ and $-0.1$ , $-0.01$ , $-0.001$ , and use the results to estimate the instantaneous rate of change of $f$ at $x = 4$ .

1. For each step, compute $\dfrac{\sqrt{4 + \Delta x} - 2}{\Delta x}$ on the calculator, storing the full values.

$\Delta x$

0.1

0.01

0.001

$-0.001$

$-0.01$

$-0.1$

average rate

0.248457

0.249844

0.249984

0.250016

0.250156

0.251582

2. From the right the values rise toward 0.25; from the left they fall toward 0.25.

Answer: the instantaneous rate of change of $f$ at $x = 4$ is about 0.25.

The table only estimates it; the algebra in the next sections (multiplying by a conjugate) shows the limit is exactly $\tfrac{1}{4}$ .

Worked example 3: from average rates to an instantaneous rate, with an interpretation

No calculator.

For the balloon, $h(t) = 8t^2 + 30t$ .

Part A

Explain why the expression $\dfrac{h(2) - h(2)}{2 - 2}$ cannot give the rate at which the balloon is rising at $t = 2$ .

Part B

Find the instantaneous rate of change of $h$ at $t = 2$ as a limit of average rates of change, and interpret it in the context of the problem.

Show the work that leads to your answer.

Part A: The change in time is $2 - 2 = 0$ , so the expression divides by zero.

An average rate of change is undefined when the change in the independent variable is zero.

P1 (reason)

Part B: For $\Delta t \ne 0$ , $\dfrac{h(2 + \Delta t) - h(2)}{\Delta t} = \dfrac{62\,\Delta t + 8(\Delta t)^2}{\Delta t} = 62 + 8\,\Delta t$ P2 (average rate simplified)

As $\Delta t \to 0$ , $62 + 8\,\Delta t \to 62$ .

At time $t = 2$ minutes the balloon's height is increasing at 62 feet per minute.

P3 (answer with interpretation)

Not earned: "the rate is 0 because $h(2) - h(2) = 0$ " (it reads the undefined quotient as zero); "62" with no average-rate work; "the balloon is 62 feet high" (an amount, not a rate).

Answer: 62 feet per minute at $t = 2$ .

Check with the table above: the average rates close in on 62 from both sides.

Exam tip

When you interpret a rate of change, say three things: the quantity that is changing, the value with its units (units of the quantity per unit of time), and the instant or interval.

"At $t = 2$ minutes the balloon's height is increasing at 62 feet per minute" earns the point; "it goes up 62" does not.

Common mistakes

Dividing by the wrong change. The average rate of change is (change in output) divided by (change in input), in that order, with both changes taken in the same direction: $\frac{f(b) - f(a)}{b - a}$ . Mixing $f(b) - f(a)$ with $a - b$ flips the sign.

Plugging in the instant. Setting the step to zero in the difference quotient gives $\frac{0}{0}$ , which is not 0 and not "undefined, so no rate". It is a signal to take a limit.

Confusing an amount with a rate. $h(2) = 92$ feet is where the balloon is; 62 feet per minute is how fast it is rising. Their units differ, and an interpretation with the wrong units loses its point.

Using only one side. An estimate from steps on one side only can mislead when the function behaves differently on the other side. Check both signs of the step.

Check your understanding

1. What is the average rate of change of $f(x) = x^3 - 2x$ over the interval $[1, 3]$ ?

(A) 10
(B) 11
(C) 22
(D) 25 Total [1]

2. Which of the following expressions gives the instantaneous rate of change of a function $g$ at $x = 5$ ?

(A) $\dfrac{g(5) - g(0)}{5 - 0}$
(B) $\displaystyle\lim_{h \to 0} \big(g(5 + h) - g(5)\big)$
(C) $\displaystyle\lim_{h \to 0} \frac{g(5 + h) - g(5)}{h}$
(D) $\displaystyle\lim_{h \to 0} \frac{g(5 + h)}{h}$ Total [1]

3. A freight train moves along a straight track. Its position $s(t)$ , in meters from a signal post, at selected times $t$ seconds is given in the table below.

$t$ (seconds)

2.99

3.01

$s(t)$ (meters)

38.7504

39.2504

Part A

Find the average velocity of the train over $2.99 \le t \le 3$ and over $3 \le t \le 3.01$ .

Show the work that leads to your answers.

Part B

What does the table suggest about the velocity of the train at $t = 3$ ?

Give a reason for your answer.

Total [3]

4. A student is asked for the rate of change of $f(x) = x^2 + 1$ at $x = 2$ and writes $\dfrac{f(2) - f(2)}{2 - 2} = 0$ .

Part A

Is the student's answer correct?

Give a reason for your answer.

Part B

Find the instantaneous rate of change of $f$ at $x = 2$ as a limit of average rates of change.

Show the work that leads to your answer.

Total [2]

5. Calculator allowed. The number of bacteria in a culture $t$ hours after an experiment starts is modeled by $P(t) = 200 \cdot 2^t$ .

Part A

Find the average rate of change of $P$ over $1 \le t \le 3$ .

Indicate units of measure.

Part B

Find the average rates of change of $P$ over $2 \le t \le 2.1$ and over $2 \le t \le 2.01$ .

Show the setup for your calculations.

Based on these values, is the instantaneous rate of change of $P$ at $t = 2$ greater than or less than the average rate of change over $2 \le t \le 2.01$ ?

Give a reason for your answer.

Total [4]

6. Of the following, which is the best description of the instantaneous rate of change of a function $f$ at $x = a$ ?

(A) The value $f(a)$ .
(B) The average rate of change of $f$ over the interval $[0, a]$ .
(C) The value that the average rates of change of $f$ over intervals containing $a$ approach as the intervals shrink to $a$ .
(D) The average rate of change of $f$ over $[a - 1, a + 1]$ . Total [1]

7. The total number of visitors who have entered a national park since January 1, in thousands, is $V(t)$ , where $t$ is measured in months after January 1. Selected values of $V(t)$ are given in the table below. What is the average rate of change of $V$ over $2 \le t \le 7$ ?

$t$ (months)

$V(t)$ (thousands of visitors)

(A) $\frac{50}{7}$ thousand visitors per month

(B) 10 thousand visitors per month

(D) 50 thousand visitors per month Total [1]

Answers

1. (B) $f(3) = 27 - 6 = 21$ and $f(1) = 1 - 2 = -1$ , so $\frac{21 - (-1)}{3 - 1} = \frac{22}{2} = 11$ . (A) adds the two outputs, $\frac{21 + (-1)}{2}$ , a sign slip; (C) forgets to divide by the change in $x$ ; (D) is $3x^2 - 2$ at $x = 3$ , a rate at the endpoint, not an average over the interval.

2. (C) The instantaneous rate is the limit of the average rates $\frac{g(5 + h) - g(5)}{h}$ as $h \to 0$ . (A) is an average rate over $[0, 5]$ , not a rate at an instant; (B) drops the division by $h$ , so its limit is just 0 for a continuous $g$ ; (D) forgets to subtract $g(5)$ .

3. Part A and Part B (3 points)

P1 (Part A) Left interval: $\dfrac{s(3) - s(2.99)}{3 - 2.99} = \dfrac{39 - 38.7504}{0.01} = 24.96$ meters per second.
P2 (Part A) Right interval: $\dfrac{s(3.01) - s(3)}{3.01 - 3} = \dfrac{39.2504 - 39}{0.01} = 25.04$ meters per second.
P3 (Part B) Answer with reason: the velocity at $t = 3$ is about 25 meters per second, because the average velocities over the shortest intervals on either side of $t = 3$ are 24.96 and 25.04, which close in on 25.
Not earned: P1 or P2 for a quotient with the time difference missing; P3 for "25" with no reference to the two average velocities.

4. Part A and Part B (2 points)

P1 (Part A) Answer with reason: no. The change in $x$ is $2 - 2 = 0$ , so the quotient is $\frac{0}{0}$ , which is undefined; it is not 0. An instantaneous rate is a limit of average rates.
P2 (Part B) Answer: $\dfrac{f(2 + h) - f(2)}{h} = \dfrac{4h + h^2}{h} = 4 + h$ for $h \ne 0$ , which approaches 4 as $h \to 0$ . The rate of change at $x = 2$ is 4.

5. Part A and Part B (4 points)

P1 (Part A) Average rate: $\dfrac{P(3) - P(1)}{3 - 1} = \dfrac{1600 - 400}{2} = 600$ .
P2 (Part A) Units: bacteria per hour.
P3 (Part B) Setup and values: $\dfrac{P(2.1) - P(2)}{0.1} = 574.188$ and $\dfrac{P(2.01) - P(2)}{0.01} = 556.444$ bacteria per hour.
P4 (Part B) Answer with reason: less than the average rate over $2 \le t \le 2.01$ (556.444), because the average rates over $[2, 2 + \Delta t]$ decrease as the interval shrinks toward $t = 2$ (from 574.188 to 556.444), so the value they approach is below 556.444. (The exact value is $800\ln 2 = 554.518$ .)
Not earned: P3 for values with no quotient written; P4 for "greater, because the bacteria are increasing" (the size of $P$ is not the question).

6. (C) This is what the limit of average rates means. (A) is the output, an amount, not a rate; (B) is an average over a fixed interval that does not shrink; (D) is one average rate, an estimate at best.

7. (B) $\dfrac{V(7) - V(2)}{7 - 2} = \dfrac{74 - 24}{5} = 10$ thousand visitors per month. (A) divides by the end time 7, not by the change in time $7 - 2$ ; (C) uses $V(7)$ alone, $\frac{74}{5}$ , forgetting to subtract $V(2)$ ; (D) is the change in $V$ , an amount, never divided by the change in time.

Introducing calculus: can change occur at an instant?

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