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Introduction
1. Limits
2. Derivative basics
3. Advanced differentiation
4. Contextual uses
4.1 Derivatives in context
4.2 Straight-line motion
4.3 Related rates
4.4 Linear approximations
4.5 L'Hopital's rule
5. Analytical uses
6. Integration
7. Differential equations
8. Applications of integrals
9. Testing details tag
Wrapping up
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4.1 Derivatives in context
Achievable AP Calculus AB
4. Contextual uses
Our AP Calculus AB course is currently in development and is a work-in-progress.

Derivatives in context

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What you’ll learn:

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

Suppose a function f(x) represents a real-world quantity. Its derivative f′(x) tells you how the output f(x) changes as the input x changes. In other words, f′(x) is a rate of change.

The units of f′(x) are always

Units of input xUnits of output f(x)​

The sign of f′(x) tells you the direction of change:

  • f′(x)>0: f(x) is increasing
  • f′(x)<0: f(x) is decreasing
  • f′(x)=0: f(x) is not changing (at that instant)

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 doesV(5)=15mean? Include units of measure.
    c) What doesV′(5)=3 mean? Include units of measure.

Solutions

a) V′(t) is the rate at which the volume changes with respect to time. Since the tank is being filled, the volume is increasing, so V′(t) should be positive.

b) V(5)=15 means that at t=5 minutes, the tank contains 15 gallons of water.

c) V′(5)=3 means that at t=5 minutes, the 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.

  1. 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 functionG(t)=95e−0.05t. Find G′(15) and interpret its meaning in the context of cooling.

Solutions

(spoiler)

a) Approximate T′(15). Include units.

T′(15) is the instantaneous rate of change of temperature at t=15 minutes. We can approximate it using a difference quotient based on nearby table values. Since t=15 is in the table, there are three common choices:

20−10T(20)−T(10)​

20−15T(20)−T(15)​

15−10T(15)−T(10)​

Any of these can be used as long as you show your work. Using the first (a symmetric difference around t=15):

20−10T(20)−T(10)​

=1052−70​

=−1.8°C/min

Interpretation: At t=15 minutes, the temperature is decreasing at a rate of about 1.8° C per minute.

b) Given G(t)=95e−0.05t, find G′(15).

Differentiate G(t) with respect to t:

dtdG​=95e−0.05t⋅(−0.05)=−4.75e−0.05t

Now evaluate at t=15:

G′(15)=−4.75e−0.05(15)=−2.244

Interpretation: At t=15 minutes, the model predicts the liquid’s temperature is decreasing at a rate of 2.244° C per minute.

AP tip:

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

  1. 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+10x+0.5x2

Find the marginal cost when 20 units are produced.

Solution

(spoiler)

Taking the derivative of the cost function,

C′(x)=10+1x

When x=20 units,

C′(20)=30

Because a derivative is a rate of change, its units are the units of the output C(x) (dollars) divided by the units of the input x (units of product).

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

  1. 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 dpdN​?
b) What does dpdN​ represent?
c) Would you expect this derivative to be positive or negative?
d) If dpdN​=−2000 at p=10, estimate the effect of increasing the price by $1.

Solutions

(spoiler)

a) The units of the derivative are the units of the output N divided by the units of the input p. That is, customers per dollar.

b) dpdN​ represents the rate at which the number of subscribers changes as the monthly price changes.

c) As price increases, the number of customers typically decreases, so you’d expect the derivative to be negative.

d) If dpdN​=−2000 at p=10, then increasing the price by $1 (from $10 to $11) would change the number of subscribers by about

−2000 customers

So the model predicts about 2000 fewer customers for a $1 increase in price (near p=10).

Interpreting derivatives as rates of change

  • Derivative f′(x): rate of change of f(x) with respect to x
  • Units: units of xunits of f(x)​
  • Sign interpretation:
    • f′(x)>0: increasing
    • f′(x)<0: decreasing
    • f′(x)=0: no change

Example 1: Tank filling problem

  • V′(t) positive: volume increasing as tank fills
  • V(5)=15: 15 gallons at 5 minutes
  • V′(5)=3: increasing at 3 gallons per minute at t=5

Example 2: Cooling liquid

  • T′(15): approximate using difference quotient; units are °C/min
    • Negative value: temperature decreasing
  • G′(15) for G(t)=95e−0.05t: −2.244°C/min at t=15
    • Interpretation: temperature decreasing at this rate

Example 3: Marginal cost in economics

  • Marginal cost C′(x): rate of change of cost per unit produced
  • For C(x)=500+10x+0.5x2, C′(x)=10+x
  • At x=20, C′(20)=30: cost increases by $30 per additional unit at 20 units

Example 4: Subscribers vs. price

  • Units of dpdN​: customers per dollar
  • dpdN​: rate subscribers change as price changes
  • Expected sign: negative (higher price, fewer customers)
  • If dpdN​=−2000 at p=10: raising price by $1 loses about 2000 customers

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Derivatives in context

What you’ll learn:

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

Suppose a function f(x) represents a real-world quantity. Its derivative f′(x) tells you how the output f(x) changes as the input x changes. In other words, f′(x) is a rate of change.

The units of f′(x) are always

Units of input xUnits of output f(x)​

The sign of f′(x) tells you the direction of change:

  • f′(x)>0: f(x) is increasing
  • f′(x)<0: f(x) is decreasing
  • f′(x)=0: f(x) is not changing (at that instant)

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 doesV(5)=15mean? Include units of measure.
    c) What doesV′(5)=3 mean? Include units of measure.

Solutions

a) V′(t) is the rate at which the volume changes with respect to time. Since the tank is being filled, the volume is increasing, so V′(t) should be positive.

b) V(5)=15 means that at t=5 minutes, the tank contains 15 gallons of water.

c) V′(5)=3 means that at t=5 minutes, the 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.

  1. 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 functionG(t)=95e−0.05t. Find G′(15) and interpret its meaning in the context of cooling.

Solutions

(spoiler)

a) Approximate T′(15). Include units.

T′(15) is the instantaneous rate of change of temperature at t=15 minutes. We can approximate it using a difference quotient based on nearby table values. Since t=15 is in the table, there are three common choices:

20−10T(20)−T(10)​

20−15T(20)−T(15)​

15−10T(15)−T(10)​

Any of these can be used as long as you show your work. Using the first (a symmetric difference around t=15):

20−10T(20)−T(10)​

=1052−70​

=−1.8°C/min

Interpretation: At t=15 minutes, the temperature is decreasing at a rate of about 1.8° C per minute.

b) Given G(t)=95e−0.05t, find G′(15).

Differentiate G(t) with respect to t:

dtdG​=95e−0.05t⋅(−0.05)=−4.75e−0.05t

Now evaluate at t=15:

G′(15)=−4.75e−0.05(15)=−2.244

Interpretation: At t=15 minutes, the model predicts the liquid’s temperature is decreasing at a rate of 2.244° C per minute.

AP tip:

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

  1. 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+10x+0.5x2

Find the marginal cost when 20 units are produced.

Solution

(spoiler)

Taking the derivative of the cost function,

C′(x)=10+1x

When x=20 units,

C′(20)=30

Because a derivative is a rate of change, its units are the units of the output C(x) (dollars) divided by the units of the input x (units of product).

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

  1. 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 dpdN​?
b) What does dpdN​ represent?
c) Would you expect this derivative to be positive or negative?
d) If dpdN​=−2000 at p=10, estimate the effect of increasing the price by $1.

Solutions

(spoiler)

a) The units of the derivative are the units of the output N divided by the units of the input p. That is, customers per dollar.

b) dpdN​ represents the rate at which the number of subscribers changes as the monthly price changes.

c) As price increases, the number of customers typically decreases, so you’d expect the derivative to be negative.

d) If dpdN​=−2000 at p=10, then increasing the price by $1 (from $10 to $11) would change the number of subscribers by about

−2000 customers

So the model predicts about 2000 fewer customers for a $1 increase in price (near p=10).

Key points

Interpreting derivatives as rates of change

  • Derivative f′(x): rate of change of f(x) with respect to x
  • Units: units of xunits of f(x)​
  • Sign interpretation:
    • f′(x)>0: increasing
    • f′(x)<0: decreasing
    • f′(x)=0: no change

Example 1: Tank filling problem

  • V′(t) positive: volume increasing as tank fills
  • V(5)=15: 15 gallons at 5 minutes
  • V′(5)=3: increasing at 3 gallons per minute at t=5

Example 2: Cooling liquid

  • T′(15): approximate using difference quotient; units are °C/min
    • Negative value: temperature decreasing
  • G′(15) for G(t)=95e−0.05t: −2.244°C/min at t=15
    • Interpretation: temperature decreasing at this rate

Example 3: Marginal cost in economics

  • Marginal cost C′(x): rate of change of cost per unit produced
  • For C(x)=500+10x+0.5x2, C′(x)=10+x
  • At x=20, C′(20)=30: cost increases by $30 per additional unit at 20 units

Example 4: Subscribers vs. price

  • Units of dpdN​: customers per dollar
  • dpdN​: rate subscribers change as price changes
  • Expected sign: negative (higher price, fewer customers)
  • If dpdN​=−2000 at p=10: raising price by $1 loses about 2000 customers