The Derivative in Context

When a function measures something, such as gallons of water, a temperature, or a number of tickets, its derivative measures how fast that quantity is changing.

The derivative as a rate of change

f(t)

is a quantity at time

t

, then

f'(a)

is the instantaneous rate of change of that quantity at

t = a

. The units of

f'

are the units of

f

per unit of

t

. The second derivative

f''(a)

is the rate of change of

f'

, so its units are those of

f'

per unit of

t

again.

	Meaning	Units, for $V$ in gallons and $t$ in minutes
$V(12)$	the amount at $t = 12$	gallons
$V'(12)$	the rate the amount changes at $t = 12$	gallons per minute
$V''(12)$	the rate that rate changes at $t = 12$	gallons per minute per minute

A rate from a tangent line

Example 1

Water drains from a tank. The graph shows the volume

V(t)

of water in the tank, in gallons, at time

t

minutes for

0 \le t \le 24

, and the line tangent to the graph at

t = 12

The volume of water in the tank, and its tangent line at t = 12 through (4, 120) and (20, 24).

(a) Find

V'(12)

. Using correct units, interpret its meaning. (b) It's also known that

V''(12) = 0.5

. Using correct units, interpret its meaning.

(a) The tangent line passes through

(4, 120)

and

(20, 24)

, so

V'(12) = \frac{24 - 120}{20 - 4} = -6.

At time

t = 12

minutes, the volume of water in the tank is decreasing at a rate of

6

gallons per minute.

(b) At time

t = 12

minutes, the rate of change of the volume is increasing at

0.5

gallon per minute per minute. Since

V'(12)

is negative and increasing, the water is draining more slowly at that moment.

Inputs other than time

Four different numbers

Example 3

The amount of a medication in a patient's bloodstream is

M(t) = 20te^{-t/2}

milligrams,

t

hours after a dose. Find each value and interpret it with units: (a)

M(3)

, (b) the average rate of change of

M

over

1 \le t \le 3

, (c)

M'(3)

, and (d)

M''(3)

The product rule gives

\begin{aligned} M'(t) &= 20e^{-t/2} - 10te^{-t/2} \\ &= 10(2 - t)e^{-t/2}, \\[6pt] M''(t) &= -10e^{-t/2} - 5(2 - t)e^{-t/2} \\ &= 5(t - 4)e^{-t/2}. \end{aligned}

(a)

M(3) = 60e^{-3/2} \approx 13.388

. Three hours after the dose, the bloodstream holds about

13.388

milligrams of the medication.

(b) The average rate of change is

\begin{aligned} \frac{M(3) - M(1)}{3 - 1} &= 30e^{-3/2} - 10e^{-1/2} \\[4pt] &\approx 0.629. \end{aligned}

From

t = 1

t = 3

hours, the amount increased at an average rate of

0.629

milligram per hour.

(c)

M'(3) = -10e^{-3/2} \approx -2.231

. At time

t = 3

hours, the amount is decreasing at a rate of

2.231

milligrams per hour.

(d)

M''(3) = -5e^{-3/2} \approx -1.116

. At time

t = 3

hours, the rate of change of the amount is decreasing at

1.116

milligrams per hour per hour. With

M'(3) < 0

as well, the amount is falling faster and faster at that moment.

The secant from t = 1 to t = 3 (blue) has slope 0.629, the average rate. The tangent at t = 3 (amber) has slope −2.231, the rate at that instant.

Clock times, and the moment a value is reached

Example 4

An online petition has

S(t) = 300\sqrt{t + 4}

signatures

t

hours after 9 A.M. Find the rate at which the number of signatures is changing (a) at 2 P.M., and (b) when the petition doubles its 9 A.M. count. (c) Find the time when signatures are arriving at a rate of

30

per hour.

The chain rule gives

S'(t) = \frac{150}{\sqrt{t + 4}}

signatures per hour.

(a) 2 P.M. is

5

hours after 9 A.M., so

t = 5

and

S'(5) = \frac{150}{\sqrt{9}} = 50

. At 2 P.M., the number of signatures is increasing at a rate of

50

signatures per hour.

(b) At 9 A.M.,

t = 0

and

S(0) = 300\sqrt{4} = 600

, so the condition is

S(t) = 1200

. Find the time from

S

300\sqrt{t + 4} = 1200

gives

\sqrt{t + 4} = 4

, so

t = 12

. Then

S'(12) = \frac{150}{4} = 37.5

. When the petition reaches twice its 9 A.M. count, at 9 P.M., it's gaining

37.5

signatures per hour.

(c) This time the condition is on the rate, so solve

S'(t) = 30

\sqrt{t + 4} = 5

, so

t = 21

. That's 6 A.M. the next day.

Rates from a table

Example 5

The temperature

H(t)

of water in a heater, in degrees Fahrenheit, is recorded at selected times

t

minutes.

H

is differentiable.

$t$ (minutes)	$H(t)$ (°F)
$0$	$52$
$5$	$70$
$8$	$79$
$14$	$93$
$20$	$101$

(a) Use the data to estimate

H'(11)

. Using correct units, interpret your answer. (b) Find the average rate of change of

H

over

0 \le t \le 20

, and interpret it.

(a) The closest times on either side of

11

are

8

and

14

, so

\begin{aligned} H'(11) &\approx \frac{H(14) - H(8)}{14 - 8} \\[4pt] &= \frac{93 - 79}{6} = \frac{7}{3}. \end{aligned}

At time

t = 11

minutes, the temperature of the water is increasing at about

\frac{7}{3} \approx 2.333

degrees Fahrenheit per minute.

(b)

\frac{H(20) - H(0)}{20 - 0} = \frac{101 - 52}{20} = 2.45

. Over the first

20

minutes, the temperature rose at an average rate of

2.45

degrees Fahrenheit per minute.