AP Calculus AB · Free-response skill

A rate model in context

The move. From a differential equation in an invented context, give a rate with units and say what it means, describe the slope field in words, and solve with an initial condition.

How to do it

  1. A value of dydt\frac{dy}{dt} is a rate. Give it with units, the output unit per input unit, such as liters per minute.
  2. To say what a rate means, name the quantity, whether it is rising or falling, how fast with units, and at what moment or for what value of the quantity.
  3. To reason about a slope field in words, put a few values of the variable into the right side. Say where the slope is positive, negative or zero, and whether it gets steeper or flatter as the variable changes. A slope that shrinks toward 0 means the quantity levels off.
  4. To solve, separate, integrate both sides with one constant, use the starting value to find the constant, then solve for the quantity. When a log of ∣M−y∣|M - y| appears, decide the sign from the context.
  5. For a "when" question, set your solution equal to the target value and solve with a log. Practice the log steps by hand, since a question like this may come in the part with no calculator, and keep an exact form, such as 3ln⁡23\ln 2, before rounding.

Practice prompt Try it

A fictional town has 1,200 households. On day 0 the town announces a new recycling program. Let S(t)S(t) be the number of households that have heard about the program t days after the announcement. A planner models the spread of the news with dSdt=0.25(1200−S)\frac{dS}{dt} = 0.25(1200 - S), where S(0)=0S(0) = 0.

Use the model to answer each part.

  1. Work out the value of dSdt\frac{dS}{dt} when S=400S = 400. Then tell what that number says about the households in the town, with units.
  2. Without solving the equation, describe how the slopes in a slope field for this equation change as S goes from 0 up toward 1,200, and what that says about how S grows.
  3. Separate the variables to find S(t)S(t), using S(0)=0S(0) = 0. Show your work.
  4. Use your answer to part c to find how many days pass before 900 households have heard about the program. Round to the nearest tenth of a day.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) When S=400S = 400, dSdt=0.25(1200−400)=0.25(800)=200\frac{dS}{dt} = 0.25(1200 - 400) = 0.25(800) = 200. At the moment when 400 households have heard about the program, the number of households who have heard is increasing at 200 households per day.

  • Point 1 Substitutes the given value of S into the right side of the equation.
  • Point 2 Names the quantity, the direction, the units and the moment the rate describes.

(b) For 0≤S<12000 \le S < 1200, the slope 0.25(1200−S)0.25(1200 - S) is positive, and it gets smaller as S gets closer to 1,200: it is 300 at S = 0, 150 at S = 600 and 0 at S = 1,200. So the segments get flatter as S rises. S keeps increasing, but more and more slowly, and its graph levels off toward 1,200.

  • Point 3 Uses values of the right side to show the slope is positive and shrinking toward 0.
  • Point 4 Turns the slope pattern into a statement about how S grows over time.

(c) Separating gives dS1200−S=0.25 dt\frac{dS}{1200 - S} = 0.25\,dt. Integrating, −ln⁡∣1200−S∣=0.25t+C-\ln|1200 - S| = 0.25t + C. At t=0t = 0, S=0S = 0, so C=−ln⁡1200C = -\ln 1200. Since S<1200S < 1200, ln⁡(1200−S)=ln⁡1200−0.25t\ln(1200 - S) = \ln 1200 - 0.25t, so 1200−S=1200e−0.25t1200 - S = 1200e^{-0.25t} and S(t)=1200−1200e−0.25tS(t) = 1200 - 1200e^{-0.25t}.

  • Point 5 Moves every S to the dS side and integrates both sides with one constant.
  • Point 6 Finds the constant from the starting value, then solves for S.

(d) Set 1200−1200e−0.25t=9001200 - 1200e^{-0.25t} = 900. Then e−0.25t=3001200=14e^{-0.25t} = \frac{300}{1200} = \frac{1}{4}, so −0.25t=ln⁡14-0.25t = \ln\frac{1}{4} and t=4ln⁡4≈5.5t = 4\ln 4 \approx 5.5 days.

  • Point 7 Sets the solution equal to the target, isolates the exponential and solves with a log.