Derivatives in context | Contextual uses | Achievable AP Calculus AB

What you’ll learn:

Interpreting derivatives as rates of change in word problems
Understanding the units of a derivative in context

Given a function $f (x)$ representing a real-world quantity, its derivative $f^{'} (x)$ represents how quantity or output $f$ changes with respect to $x$ - whether it’s increasing, decreasing, or not changing. Interpreting this rate helps us understand how a system responds to small changes. The units of $f^{'} (x)$ are always

$\frac{Units of output f ( x )}{Units of input x}$

The sign of $f^{'} (x)$ indicates whether $f (x)$ is increasing ( $+$ ), decreasing ( $-$ ), or unchanging ( $0$ ).

Examples

1. A tank is being filled with water, and its volume (in gallons) at time $t$ (in minutes) is given by $V (t)$ .
a) What sign would you expect for $V^{'} (t)$ ?
b) What does $V (5) = 15$ mean? Include units of measure.
c) What does $V^{'} (5) = 3$ mean? Include units of measure.

Solutions

a) $V^{'} (t)$ represents how the volume changes over time. Since the tank is being filled with water, the volume will increase, meaning the rate of change of volume will be positive.

b) $V (5) = 15$ means that after $5$ minutes, there are $15$ gallons of water in the tank.

c) $V^{'} (5) = 3$ means that at $t = 5$ minutes, the water volume is increasing at a rate of $3$ gallons per minute.

AP tip:

If a free-response question does not ask for units of measure, it isn’t necessary to write. Points are sometimes even docked if your answer happens to be incorrect.

2. The temperature (in degrees Celsius) of a cooling liquid is given by the function $T (t)$ , where $t$ is measured in minutes. The following table provides temperature readings:

$t$ $T (t)$

0 95

5 82

10 70

15 60

20 52

25 45

30 40

a) Approximate $T^{'} (15)$ . Include units.
b) A scientist models the cooling process using the function $G (t) = 95 e^{- 0.05 t}$ . Find $G^{'} (15)$ and interpret its meaning in the context of cooling.

$t$	$T (t)$
0	95
5	82
10	70
15	60
20	52
25	45
30	40

Solutions

(spoiler)

a) Approximate $T^{'} (15)$ . Include units.

$T^{'} (15)$ represents the rate of change of the temperature at $t = 15$ minutes and can be approximated with a difference quotient of the surrounding values. Since $t = 15$ is also in the table, there are 3 ways to do this:

$\frac{T ( 20 ) - T ( 10 )}{20 - 10}$
$\frac{T ( 20 ) - T ( 15 )}{20 - 15}$
$\frac{T ( 15 ) - T ( 10 )}{15 - 10}$

Any would be fine to approximate $T^{'} (15)$ as long as you show your work. Using the first difference quotient,

$\frac{T ( 20 ) - T ( 10 )}{20 - 10}$

$= \frac{52 - 70}{10}$

$= - 1.8° C/min$

Interpretation: At $t = 15$ minutes, the temperature is decreasing at a rate of approximately $1.8°$ C/min.

b) Given $G (t) = 95 e^{- 0.05 t}$ , find $G^{'} (15)$ .

Taking the derivative,

$\frac{d G}{d t} = 95 e^{- 0.05 t} \cdot (- 0.05) = - 4.75 e^{- 0.05 t}$

$G^{'} (15) = - 4.75 e^{- 0.05 (15)} = - 2.244$

Interpretation: At $t = 15$ minutes, the liquid is cooling (its temperature is decreasing) at a rate of $2.244°$ C/min.

AP tip:

Unless otherwise specified, final answers on the calculator-allowed FRQ portion must be accurate to 3 decimal places.

3. In economics, derivative can measure how the cost and revenue change as production increases. The marginal cost is the derivative of the cost function and gives the change in cost when one additional unit of product is made.

Suppose a company’s cost function is given by:

$C (x) = 500 + 10 x + 0.5 x^{2}$

Find the marginal cost when 20 units are produced.

Solution

(spoiler)

Taking the derivative of the cost function,

$C^{'} (x) = 10 + 1 x$

When $x = 20$ units, $C^{'} (20) = 30$ .

Because the derivative is a rate of change, the units are the same as the units for the output $C (x)$ (dollars) divided by the units for the input $x$ (unit of product).

Interpretation: When $x = 20$ units produced, producing one more unit costs approximately $$30$ per unit.

4. The number of customers $N$ who subscribe to a video streaming service depends on the monthly price $p$ (in dollars), so $N = f (p)$ .

a) What are the units of $\frac{d N}{d p}$ ?
b) What does $\frac{d N}{d p}$ represent?
c) Would you expect this derivative to be positive or negative?
d) If $\frac{d N}{d p} = - 2000$ at $p = 10$ , estimate the effect of increasing the price by $1.

Solutions

(spoiler)

a) The units of the derivative are units of the output $N$ divided by the units of the input $p$ . This would be the number of customers subscribed per dollar change in price.

b) The derivative represents how the number of customers changes as the price of the streaming service changes.

c) As the price increases, the number of customers likely decreases, which means the derivative is most likely negative.

d) The rate of change is negative at $p = 10$ , which means that increasing the price by another dollar would result in a decrease in the number of customers $N$ .

What you’ll learn:

Interpreting derivatives as rates of change in word problems
Understanding the units of a derivative in context

$\frac{Units of output f ( x )}{Units of input x}$

The sign of $f^{'} (x)$ indicates whether $f (x)$ is increasing ( $+$ ), decreasing ( $-$ ), or unchanging ( $0$ ).

Examples

1. A tank is being filled with water, and its volume (in gallons) at time $t$ (in minutes) is given by $V (t)$ .
a) What sign would you expect for $V^{'} (t)$ ?
b) What does $V (5) = 15$ mean? Include units of measure.
c) What does $V^{'} (5) = 3$ mean? Include units of measure.

Solutions

a) $V^{'} (t)$ represents how the volume changes over time. Since the tank is being filled with water, the volume will increase, meaning the rate of change of volume will be positive.

b) $V (5) = 15$ means that after $5$ minutes, there are $15$ gallons of water in the tank.

c) $V^{'} (5) = 3$ means that at $t = 5$ minutes, the water volume is increasing at a rate of $3$ gallons per minute.

AP tip:

If a free-response question does not ask for units of measure, it isn’t necessary to write. Points are sometimes even docked if your answer happens to be incorrect.

2. The temperature (in degrees Celsius) of a cooling liquid is given by the function $T (t)$ , where $t$ is measured in minutes. The following table provides temperature readings:

$t$ $T (t)$

0 95

5 82

10 70

15 60

20 52

25 45

30 40

a) Approximate $T^{'} (15)$ . Include units.
b) A scientist models the cooling process using the function $G (t) = 95 e^{- 0.05 t}$ . Find $G^{'} (15)$ and interpret its meaning in the context of cooling.

$t$	$T (t)$
0	95
5	82
10	70
15	60
20	52
25	45
30	40

Solutions

(spoiler)

a) Approximate $T^{'} (15)$ . Include units.

$\frac{T ( 20 ) - T ( 10 )}{20 - 10}$
$\frac{T ( 20 ) - T ( 15 )}{20 - 15}$
$\frac{T ( 15 ) - T ( 10 )}{15 - 10}$

Any would be fine to approximate $T^{'} (15)$ as long as you show your work. Using the first difference quotient,

$\frac{T ( 20 ) - T ( 10 )}{20 - 10}$

$= \frac{52 - 70}{10}$

$= - 1.8° C/min$

Interpretation: At $t = 15$ minutes, the temperature is decreasing at a rate of approximately $1.8°$ C/min.

b) Given $G (t) = 95 e^{- 0.05 t}$ , find $G^{'} (15)$ .

Taking the derivative,

$\frac{d G}{d t} = 95 e^{- 0.05 t} \cdot (- 0.05) = - 4.75 e^{- 0.05 t}$

$G^{'} (15) = - 4.75 e^{- 0.05 (15)} = - 2.244$

Interpretation: At $t = 15$ minutes, the liquid is cooling (its temperature is decreasing) at a rate of $2.244°$ C/min.

AP tip:

Unless otherwise specified, final answers on the calculator-allowed FRQ portion must be accurate to 3 decimal places.

3. In economics, derivative can measure how the cost and revenue change as production increases. The marginal cost is the derivative of the cost function and gives the change in cost when one additional unit of product is made.

Suppose a company’s cost function is given by:

$C (x) = 500 + 10 x + 0.5 x^{2}$

Find the marginal cost when 20 units are produced.

Solution

(spoiler)

Taking the derivative of the cost function,

$C^{'} (x) = 10 + 1 x$

When $x = 20$ units, $C^{'} (20) = 30$ .

Because the derivative is a rate of change, the units are the same as the units for the output $C (x)$ (dollars) divided by the units for the input $x$ (unit of product).

Interpretation: When $x = 20$ units produced, producing one more unit costs approximately $$30$ per unit.

4. The number of customers $N$ who subscribe to a video streaming service depends on the monthly price $p$ (in dollars), so $N = f (p)$ .

a) What are the units of $\frac{d N}{d p}$ ?
b) What does $\frac{d N}{d p}$ represent?
c) Would you expect this derivative to be positive or negative?
d) If $\frac{d N}{d p} = - 2000$ at $p = 10$ , estimate the effect of increasing the price by $1.

Solutions

(spoiler)

a) The units of the derivative are units of the output $N$ divided by the units of the input $p$ . This would be the number of customers subscribed per dollar change in price.

b) The derivative represents how the number of customers changes as the price of the streaming service changes.

c) As the price increases, the number of customers likely decreases, which means the derivative is most likely negative.

d) The rate of change is negative at $p = 10$ , which means that increasing the price by another dollar would result in a decrease in the number of customers $N$ .