4.3 - Rates of Change in Applied Contexts Other Than Motion

Determining Increasing vs. Decreasing

The sign of the derivative tells you whether a function is increasing or decreasing.

\text{If } f'(x) > 0 \implies f(x) \text{ is INCREASING}

\text{If } f'(x) < 0 \implies f(x) \text{ is DECREASING}

To figure out if something is increasing or decreasing, check the sign of that thing’s derivative. The word “something” is doing a lot of work here, because this applies to any quantity:

Is height increasing? Check if $h'(t) > 0$

Is velocity decreasing? Check if $v'(t) < 0$ , which means $a(t) < 0$

Is temperature increasing? Check if $T'(t) > 0$

Is population decreasing? Check if $P'(t) < 0$

Examples

Example 1: height function

h(t) = -4.9t^2 + 20t + 5

. Is height increasing at

t = 1

h'(t) = -9.8t + 20

h'(1) = -9.8 + 20 = 10.2

Since

h'(1) = 10.2 > 0

, height is increasing at

t = 1

second.

Example 2: population

P(t) = 1000e^{0.05t}

. Is population decreasing at

t = 10

P'(t) = 1000e^{0.05t} \cdot 0.05 = 50e^{0.05t}

P'(10) = 50e^{0.5} \approx 82.4

Since

P'(10) > 0

, population is increasing, not decreasing, at

t = 10

years.

Example 3: cost function

C(x) = 0.02x^3 - 3x^2 + 150x + 2000

. Is cost decreasing when

x = 80

C'(x) = 0.06x^2 - 6x + 150

C'(80) = 0.06(6400) - 6(80) + 150 = 384 - 480 + 150 = 54

Since

C'(80) = 54 > 0

, cost is increasing, not decreasing, at 80 items.

Using a Calculator to Find Derivatives

On TI-83/84 calculators, use the numerical derivative feature to evaluate derivatives at specific points.

Steps

Press MATH, select option 8 (nDeriv)

Enter nDeriv(function, variable, value)

Press ENTER

Example

Find the derivative of

f(x) = \sin(x^2) + 3x

x = 2


MATH 8
nDeriv(sin(x²) + 3x, x, 2)

Make sure your calculator is in RADIAN mode (press MODE to check).

Result: about

0.385

Interpretation: since

f'(2) \approx 0.385 > 0

, the function is increasing at

x = 2

, at a rate of about 0.385 units per unit.

Another example

Evaluate the derivative of

g(x) = e^{-x}\cos(2x)

x = 1


nDeriv(e^(-x)cos(2x), x, 1)

Result: about

-0.516

Interpretation: since

g'(1) \approx -0.516 < 0

, the function is decreasing at

x = 1

, at a rate of about 0.516 units per unit.

Functions That Are Already Rates

Pay close attention to whether a function is a rate or represents a quantity whose rate you need to find.

Case 1: the function represents a quantity

Given:

P(t)

represents the rabbit population after

t

years.

What is

P'(t)

? It’s the rate of change of the rabbit population, in rabbits per year, at time

t

Units: rabbits/year

Case 2: the function already represents a rate

Given:

R(t)

represents the rate at which the rabbit population is increasing (in rabbits per year) at time

t

years.

What is

R'(t)

? It’s the rate of change of the rate at which the rabbit population is increasing.

Units: rabbits/year/year, or rabbits/year² (“rabbits per year squared”)

Think of it like acceleration. If

v(t)

is velocity (already a rate), then

v'(t) = a(t)

is acceleration (a rate of a rate), measured in m/s².

Comparison

Context	Function	Derivative	Units
Temperature is $T$ degrees	$T(t)$	$T'(t)$ = rate temp changes	°F/hour
Rate of temp change is $R$	$R(t)$	$R'(t)$ = rate that rate changes	°F/hour²
Position is $s$ meters	$s(t)$	$s'(t) = v(t)$ = velocity	m/s
Velocity is $v$ m/s	$v(t)$	$v'(t) = a(t)$ = acceleration	m/s²

Key insight: look for the word “rate” in the problem description. If the function already describes a rate, its derivative has squared units.

Estimating Rates of Change from Tables

With discrete data in a table, we estimate derivatives using average rates of change.

Year ( $t$ )	Population ( $P$ )
0	85
10	120
20	150
30	165
40	172

Estimating at a point between data values

Estimate

P'(15)

Since 15 sits between 10 and 20, use those two points:

P'(15) \approx \frac{P(20) - P(10)}{20 - 10} = \frac{150 - 120}{10} = 3 \text{ people/year}

At year 15, the population is increasing at roughly 3 people per year.

Estimating at a point that’s on the table

Estimate

P'(20)

Since 20 is on the table, you have three options.

Option 1: centered difference (surrounding points)

P'(20) \approx \frac{P(30) - P(10)}{30 - 10} = \frac{165 - 120}{20} = 2.25 \text{ people/year}

Option 2: left-side points

P'(20) \approx \frac{P(20) - P(10)}{20 - 10} = \frac{30}{10} = 3 \text{ people/year}

Option 3: right-side points

P'(20) \approx \frac{P(30) - P(20)}{30 - 20} = \frac{15}{10} = 1.5 \text{ people/year}

Which one? Any of them is acceptable, but the centered difference (Option 1) is usually the most accurate. Always show your calculation on AP exams when they say “show your work” or “show your computations.”

Practice Problem

Time ( $t$ , hours)	Temperature ( $T$ , °F)
0	68
2	75
4	79
6	80
8	78

Part A: estimate $T'(3)$

Since 3 is between 2 and 4:

T'(3) \approx \frac{T(4) - T(2)}{4 - 2} = \frac{79 - 75}{2} = 2 \text{ °F/hour}

At 3 hours, temperature is increasing at about 2°F per hour.

Part B: estimate $T'(6)$

Since 6 is on the table, use the centered difference:

T'(6) \approx \frac{T(8) - T(4)}{8 - 4} = \frac{78 - 79}{4} = -0.25 \text{ °F/hour}

At 6 hours, temperature is decreasing at about 0.25°F per hour.

Part C: is temperature increasing or decreasing at $t = 7$ ?

Estimate

T'(7)

using the points at 6 and 8:

T'(7) \approx \frac{T(8) - T(6)}{8 - 6} = \frac{78 - 80}{2} = -1 \text{ °F/hour}

Since

T'(7) < 0

, temperature is decreasing at

t = 7

hours.

Writing Proper Justifications

When answering whether something is increasing or decreasing, give a complete justification.

Format: “[Quantity] is [increasing/decreasing] because [derivative] is [positive/negative] at [point].”

Good justifications

“Height is increasing at $t = 2$ because $h'(2) = 5 > 0$ .”

“Population is decreasing at $t = 10$ because $P'(10) = -3.2 < 0$ .”

“Velocity is increasing at $t = 4$ because $v'(4) = a(4) = 2 > 0$ .”

“Cost is decreasing when $x = 50$ because $C'(50) = -0.8 < 0$ .”

Incomplete justifications

“Height is increasing.” Missing the why.

“Because $h'(2) = 5$ .” Missing the conclusion about what’s increasing.

“Positive derivative.” Missing what quantity, and at what point.

Key Takeaways

Increasing/decreasing test: check the sign of the derivative (positive is increasing, negative is decreasing)

Applies to ANY quantity: whatever is increasing or decreasing, check THAT thing’s derivative

Calculator: MATH 8 on the TI-84 for numerical derivatives, and always verify radian mode

Already a rate: if the function already represents a rate, its derivative has squared units

Table estimates between points: use the surrounding data points for the average rate of change

Table estimates at points: three options (centered, left, right), and centered is usually most accurate

Show your work: on AP exams, always show the calculation behind your answer

Complete justifications: state what’s changing, the direction, and why (the derivative’s sign)

Back Next