Unit 1 · Topic 1.1 Beta

Can Change Happen at an Instant?

An average rate of change, (f(b) − f(a))/(b − a) , is a slope over an interval, with units of output per input.

Practice 2: Connecting RepresentationsPractice 3: Justification

Part 1 · Hook

Why this matters

A speedometer shows 64 miles per hour at this very instant. But a speed is distance divided by time, and in an instant no time passes and no distance is covered. Zero divided by zero is not a number. So what does the speedometer mean? Answering that question is where calculus begins.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the slope of the line through $(1, 4)$ and $(3, 10)$ ?

$3$
$\frac{1}{3}$
$6$
$2$

Show the answer

$\frac{10 - 4}{3 - 1}$ $= \frac{6}{2}$ $= 3$ .

Correct: $3$ :
$\frac{1}{3}$ :
$6$ :
$2$ :

2. If $s(t) = 16t^2$ , what is $s(1.5)$ ?

$36$
$24$
$48$
$12$

Show the answer

$16(1.5)^2 = 16(2.25) = 36$ .

Correct: $36$ :
$24$ :
$48$ :
$12$ :

Part 4 · See it

See it first

A stone falls s(t) = 16t squared feet. Lines join the point (1, 16) on the curve to points at t = 2, 1.5 and 1.2, with slopes 48, 40 and 35.2; a dashed line through (1, 16) has slope 32. A table lists the average speeds from t = 1 over [1, 2], [1, 1.5], [1, 1.2], [1, 1.1], [1, 1.01] and [1, 1.001]: 48, 40, 35.2, 33.6, 32.16 and 32.016 feet per second, closing in on 32. — A falling stone, $s(t) = 16t^2$ : the average speed from $t = 1$ over shorter and shorter intervals closes in on 32 feet per second. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Over an interval [a, b], divide the change in a quantity by the change in timethe average rate of change $\frac{f(b) - f(a)}{b - a}$ , the slope of the line through two points of the graph
At a single instant the interval has length 0the quotient becomes zero divided by zero, which has no value
Shrink the interval around the instant insteadthe average rates close in on one number
That number is what the average rates approachit is the instantaneous rate of change: the speed the speedometer shows

Part 6 · Key ideas

Key ideas

The average rate of change of $f$ over $[a, b]$ is $\frac{f(b) - f(a)}{b - a}$ : the slope of the line through $(a, f(a))$ and $(b, f(b))$ . Its units are output units per input unit.
A rate of change at one instant cannot be computed as a quotient, because the interval would have length 0.
The instantaneous rate of change at $t = a$ is the number the average rates over shorter and shorter intervals around $a$ close in on.
From a table, estimate the rate at an instant with the shortest interval that contains it; from both sides, average the two closest rates.
A negative rate means the quantity is decreasing; its size tells how fast.

Part 7 · Misconception

A common mistake

The wrong idea: The average rate over $[0, 5]$ tells how fast the quantity is changing at $t = 5$ .

What actually happens: An average rate describes a whole interval. The rate at one instant can be very different; estimate it from the shortest interval around that instant, and interpret averages with the words "on average over the interval".

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A cooling cup of coffee

Temperature $T(t)$ of a cup of coffee, in degrees Fahrenheit, $t$ minutes after it is poured
$t$ (minutes)	0	2	5	9	10
$T(t)$ (°F)	180	168	153	137	134

1. Find the average rate of change of $T$ over $0 \le t \le 5$ . Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\frac{T(5) - T(0)}{5 - 0}$ $= \frac{153 - 180}{5}$ $= -\frac{27}{5}$ $= -5.4$ °F per minute.

Answer: $\frac{-27}{5}$ ≈ -5.400 °F per minute

2. Which sentence correctly interprets the value $\frac{T(5) - T(0)}{5 - 0}$ $= -5.4$ ?

Over the first 5 minutes, the temperature fell by an average of 5.4 °F per minute.
At exactly 5 minutes, the temperature is falling at 5.4 °F per minute.
After 5 minutes, the coffee is 5.4 °F cooler than when it was poured.
The temperature fell 5.4 °F in each of the first 5 minutes, minute by minute.

Show the answer

An average rate describes a whole interval: from $t = 0$ to $t = 5$ the coffee lost 27 °F, which is 5.4 °F per minute on average. It does not say how fast it cooled at any one moment.

Correct: Over the first 5 minutes, the temperature fell by an average of 5.4 °F per minute.: It names the interval, the quantity, the units and "on average".
At exactly 5 minutes, the temperature is falling at 5.4 °F per minute.: That describes one instant. An average over an interval does not tell the rate at its end.
After 5 minutes, the coffee is 5.4 °F cooler than when it was poured.: The total drop is 27 °F; 5.4 is the drop per minute.
The temperature fell 5.4 °F in each of the first 5 minutes, minute by minute.: An average does not say the cooling was steady; the table shows it slowed down.

3. Using the table, which is the best estimate of the rate at which the temperature is changing at $t = 9.5$ minutes?

−3 °F per minute, from the interval [9, 10]
−3.8 °F per minute, from the interval [5, 10]
−4 °F per minute, from the interval [5, 9]
137 °F, the temperature at t = 9

Show the answer

The narrowest interval in the table that contains $t = 9.5$ is $[9, 10]$ : $\frac{134 - 137}{10 - 9}$ $= -3$ °F per minute. Shorter intervals give averages closer to the rate at an instant.

Correct: −3 °F per minute, from the interval [9, 10]: $[9, 10]$ is the shortest interval around 9.5.
−3.8 °F per minute, from the interval [5, 10]: $[5, 10]$ contains 9.5 but is five times longer, so its average blends in the faster cooling from earlier.
−4 °F per minute, from the interval [5, 9]: $[5, 9]$ does not contain $t = 9.5$ .
137 °F, the temperature at t = 9: That is a temperature, not a rate of change; a rate has units of °F per minute.

4. What do the average rates of change over $[0, 2]$ , $[2, 5]$ , $[5, 9]$ and $[9, 10]$ show about the coffee?

It cools more and more slowly: the averages are −6, −5, −4 and −3 °F per minute.
It cools faster and faster: the averages are −3, −4, −5 and −6 °F per minute.
It cools at a steady rate, because the temperature falls on each interval.
It warms up at the end, because −3 is greater than −6.

Show the answer

The averages are $\frac{168 - 180}{2} = -6$ , $\frac{153 - 168}{3} = -5$ , $\frac{137 - 153}{4} = -4$ and $\frac{134 - 137}{1} = -3$ . They are all negative (cooling) and shrink in size, so the cooling slows down.

Correct: It cools more and more slowly: the averages are −6, −5, −4 and −3 °F per minute.: The averages go −6, −5, −4, −3: still cooling, but more slowly.
It cools faster and faster: the averages are −3, −4, −5 and −6 °F per minute.: That lists them in reverse time order.
It cools at a steady rate, because the temperature falls on each interval.: Falling on every interval does not mean falling at the same rate; the averages differ.
It warms up at the end, because −3 is greater than −6.: A rate of −3 still means the temperature is going down, just more slowly.

5. A stone dropped from a bridge falls $s(t) = 16t^2$ feet in $t$ seconds. Write its average speed over the interval from $t = 2$ to $t = 2 + h$ , for $h > 0$ , as a simplified expression in $h$ .

Type an expression for $\text{average speed} =$ .

Show the answer

$\frac{s(2 + h) - s(2)}{h}$ $= \frac{16(4 + 4h + h^2) - 64}{h}$ $= \frac{64h + 16h^2}{h}$ $= 64$ $+ 16h$ feet per second, for $h > 0$ .

Answer: $\text{average speed} =$ $64 + 16h$

6. For the stone above, the average speed over $[2, 2 + h]$ is $64 + 16h$ feet per second. As $h$ shrinks toward 0, what value do these average speeds close in on, and what does it represent?

64 feet per second, the speed at the instant t = 2
80 feet per second, the speed at the instant t = 2
0 feet per second, because the interval disappears
64 feet, the distance fallen by t = 2

Show the answer

As $h$ gets closer to 0, $16h$ gets closer to 0, so $64 + 16h$ closes in on 64. That number is the instantaneous rate of change: the stone's speed at $t = 2$ , 64 feet per second.

Correct: 64 feet per second, the speed at the instant t = 2: $64 + 16h$ heads to 64 as $h$ heads to 0.
80 feet per second, the speed at the instant t = 2: 80 is the average over $[2, 3]$ ( $h = 1$ ), not what the averages approach.
0 feet per second, because the interval disappears: The averages never get near 0; they get near 64.
64 feet, the distance fallen by t = 2: $s(2) = 64$ feet happens to have the same number, but a speed has units of feet per second.

7. A train passes mile marker 40 at 1:00 p.m. and mile marker 100 at 2:00 p.m. on a straight track. Which statement must be true?

Its average speed from 1:00 to 2:00 was 60 miles per hour.
At 1:30 p.m. it was traveling at 60 miles per hour.
It traveled at 60 miles per hour for the entire hour.
At 2:00 p.m. it was traveling at 100 miles per hour.

Show the answer

The data give an average over the interval: $\frac{100 - 40}{1} = 60$ miles per hour. They say nothing certain about the speed at any single instant, such as 1:30.

Correct: Its average speed from 1:00 to 2:00 was 60 miles per hour.: $\frac{60 \text{ miles}}{1 \text{ hour}}$ .
At 1:30 p.m. it was traveling at 60 miles per hour.: The train could have sped up and slowed down; the average says nothing sure about one moment.
It traveled at 60 miles per hour for the entire hour.: An average does not mean a steady speed.
At 2:00 p.m. it was traveling at 100 miles per hour.: The mile marker 100 is a position, not a speed.

Part 9 · Summary

Summary

An average rate of change,

\frac{f(b) - f(a)}{b - a}

, is a slope over an interval, with units of output per input. A rate at one instant cannot be a quotient, since nothing changes in zero time; instead, the averages over shrinking intervals close in on one number, the instantaneous rate of change. The next topic gives that closing in a name and a notation.

Part 10 · Up next

What comes next

Part 11 · Connections