AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/4/4-3)
Unit 4 · Topic 4.3
4.3 Rates of Change in Applied Contexts Other Than Motion
Derivatives describe rates in any setting: populations, costs, temperatures, water in a tank, medicine in the bloodstream. The math is the same as for motion. What changes is the meaning, which comes from the quantities and units in the problem.
Key terms
- rate of change in context
- units
- marginal cost
- applied rate
Same math, new meanings
Any time one quantity depends on another, its derivative is a rate. The structure of the problem doesn't change: find the derivative, evaluate it, and explain what it means with units.
| Setting | Function | Derivative means |
|---|---|---|
| Biology | P(t), bacteria count | Growth rate, in bacteria per hour |
| Economics | C(x), cost of x items | Marginal cost, in dollars per item |
| Medicine | A(t), mg of drug in blood | Rate the drug amount changes, in mg per hour |
| Physics | T(t), temperature | Rate of heating or cooling, in degrees per minute |
| Geometry | A(r), area of a circle | Rate area grows per unit of radius |
Marginal cost and revenue
In economics, the derivative of cost with respect to the number of items made is called marginal cost. C′(100) approximates the extra cost of making one more item, the 101st, once 100 have been made. It's an approximation because C′ is an instantaneous rate, but it's usually very close.
Marginal revenue R′(x) and marginal profit P′(x) work the same way.
Rates of rates
The second derivative in context tells you how a rate is changing. If P(t) is a population, P″(t) > 0 means the population's growth rate is increasing. That's a statement about the rate, not about the population going up. Units are the function's units per input unit squared, like people per year per year.
Amounts vs. rates in a table
Context problems often give a table of values of some quantity, like the number of gallons in a tank at several times. Differences between table entries give average rates; you estimate instantaneous rates with difference quotients (2.3).
If instead the table lists a rate, like gallons per hour, then the table values themselves are derivatives. A positive entry means the amount is increasing at that time, even if the entries are getting smaller. Ask yourself whether each number is an amount or a rate before using it.
Working with models
Exam models often use eˣ, ln x or trig functions, like P(t) = 500e^(0.04t) or T(t) = 60 + 10 sin(πt/12). Differentiate with the chain rule and evaluate. On calculator-active questions, you can use the calculator's numerical derivative instead, and give three decimal places.
Read the question for what it asks: a value of the function (an amount), a value of the derivative (a rate), or a sign of the derivative (increasing or decreasing). Each needs a different kind of answer.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Growth rate of a population model
A fish population is modeled by P(t) = 500e^(0.04t), where t is in years. Find P′(10) and interpret it.
Show the solutionHide the solution
- Step 1: Chain rule: P′(t) = 500·0.04·e^(0.04t) = 20e^(0.04t).
- Step 2: P′(10) = 20e^(0.4) ≈ 29.836.
- Step 3: Units: fish per year. Positive, so the population is growing.
Answer: P′(10) ≈ 29.836: ten years in, the fish population is increasing at about 29.8 fish per year.
- Example 2
Marginal cost
The cost of producing x phone cases is C(x) = 2000 + 30x − 0.01x² dollars, for 0 ≤ x ≤ 1000. Find C′(100) and explain what it tells you.
Show the solutionHide the solution
- Step 1: C′(x) = 30 − 0.02x.
- Step 2: C′(100) = 30 − 2 = 28.
- Step 3: Units: dollars per case.
- Step 4: Check: the actual cost of the 101st case is C(101) − C(100) = 27.99 dollars, very close to 28.
Answer: C′(100) = 28 dollars per case: when 100 cases have been made, the cost of making one more is about 28 dollars.
- Example 3
Trap: forgetting the chain rule factor
The temperature in a town is modeled by T(t) = 60 + 10 sin(πt/12) degrees Fahrenheit, t hours after midnight. Is the temperature increasing or decreasing at t = 14 (2 p.m.), and how fast?
Show the solutionHide the solution
- Step 1: Chain rule: T′(t) = 10 cos(πt/12)·(π/12) = (10π/12) cos(πt/12).
- Step 2: At t = 14: cos(14π/12) = cos(7π/6) = −√3/2.
- Step 3: T′(14) = (10π/12)(−√3/2) = −5√3π/12 ≈ −2.267.
- Step 4: Forgetting the inner factor π/12 would give 10 cos(7π/6) ≈ −8.660, almost four times too big.
Answer: T′(14) ≈ −2.267, so at 2 p.m. the temperature is decreasing at about 2.267°F per hour.
Common mistakes
- Giving the value of the function when the question asks for a rate, or the other way around.
- Writing the wrong units, such as dollars instead of dollars per item.
- Interpreting P″ > 0 as “the population is increasing.” It means the growth rate is increasing.
On the exam
- Applied rate questions appear in free response regularly: rates of water flow, people entering a building, temperatures and so on. Units and context words are part of the scoring.
- If a question asks whether an amount is increasing or decreasing at a time, give the sign of its derivative as your reason.
Connected topics
Videos
Check yourself
4 questions on 4.3 Rates of Change in Applied Contexts Other Than Motion. Pick an answer to see if you got it, and why.
The cost, in dollars, of producing x units of a product is C(x) = 0.01x³ − 0.6x² + 15x + 200. Use the marginal cost at x = 30 to estimate the cost of producing the 31st unit.
Water flows into a storage tank at a rate of R(t) = 30 + 10 sin(t/2) gallons per hour and leaks out at a rate of L(t) = 4t gallons per hour, where t is measured in hours, 0 ≤ t ≤ 12. At time t = 8, is the amount of water in the tank increasing or decreasing, and at what rate?
The temperature of a cup of tea, in degrees Fahrenheit, is modeled by H(t) = 70 + 110e^(−0.08t), where t is measured in minutes. At what time t is the temperature of the tea decreasing at a rate of 5 degrees Fahrenheit per minute?
At a school of 1,200 students, the number of students who have heard a rumor t days after it starts is modeled by N(t) = 1200/(1 + 49e^(−0.6t)). At what rate is the number of students who have heard the rumor increasing at t = 5 days?
0 of 4 answered