AP Calculus AB · Lesson 3

Exponential growth and decay

Know Know it

If dydt=ky\frac{dy}{dt} = ky, separating gives dyy=k dt\frac{dy}{y} = k\,dt, so ln⁡∣y∣=kt+C\ln|y| = kt + C and y=y0ekty = y_0e^{kt}, where y0y_0 is the amount at t=0t = 0. This is the only family of solutions, so "rate proportional to the amount" always means an exponential model.

If k>0k > 0 the quantity grows; if k<0k < 0 it decays toward 0 without reaching it. Equal time steps multiply y by the same factor ekΔte^{k \Delta t}.

The doubling time comes from ekt=2e^{kt} = 2, so t=ln⁡2kt = \frac{\ln 2}{k}. For decay with dydt=−ry\frac{dy}{dt} = -ry, the half-life is ln⁡2r\frac{\ln 2}{r}. Neither depends on the starting amount.

Reading the model y=y0ekty = y_0e^{kt}
PartWhat it tells you
y0y_0The amount when t=0t = 0
k>0k > 0Growth; each unit of time multiplies y by eke^k
k<0k < 0Decay toward 0
ln⁡2∣k∣\frac{\ln 2}{|k|}The doubling time (growth) or the half-life (decay)

Calculator On the exam

Each section of the exam has a part with a graphing calculator and a part without. You need a graphing calculator for the second multiple-choice part (13 questions) and the first free-response part (2 questions), and the testing app has one built in for those parts. The first multiple-choice part (29 questions) and the second free-response part (4 questions) allow no calculator.

So practice exponential models both ways. Without a calculator, keep exact forms such as ln⁡25\frac{\ln 2}{5} or 3e−13e^{-1}. With one, use it for decimal values of ekte^{kt} and logs, and keep full precision until the last step.

Trap k is not the growth factor

For y=50e0.2ty = 50e^{0.2t}, each unit of time multiplies y by e0.2≈1.221e^{0.2} \approx 1.221, a gain of about 22.1 percent, not 20 percent. The constant k is the rate of change per unit of y, so dydt=0.2y\frac{dy}{dt} = 0.2y; the factor per unit of time is eke^k.

Worked example Easy

A bacteria culture satisfies dPdt=0.4P\frac{dP}{dt} = 0.4P, and P(0)=300P(0) = 300. Which expression gives P(t)P(t)?

  1. 0.4e300t0.4e^{300t}
  2. 300e0.4t300e^{0.4t}Answer
  3. 300+0.4t300 + 0.4t
  4. 300(0.4)t300(0.4)^t

How to solve it

  1. dPdt=kP\frac{dP}{dt} = kP has the solution P=P0ektP = P_0e^{kt}.
  2. Here k=0.4k = 0.4 and P0=300P_0 = 300, so P(t)=300e0.4tP(t) = 300e^{0.4t}.

Why each choice is right or wrong

  • A. Incorrect. 0.4e300t0.4e^{300t} swaps the roles of the two numbers: 300 is the starting amount and 0.4 belongs in the exponent.
  • B. Correct. A rate proportional to P gives P=P(0)ektP = P(0)e^{kt} with k=0.4k = 0.4 and P(0)=300P(0) = 300.
  • C. Incorrect. 300+0.4t300 + 0.4t grows by a fixed 0.4 each unit of time, but the rate 0.4P0.4P grows as P grows.
  • D. Incorrect. 300(0.4)t300(0.4)^t multiplies by 0.4 each unit of time, which is decay, while k=0.4>0k = 0.4 > 0 means growth.

Worked example Medium

A population satisfies dPdt=kP\frac{dP}{dt} = kP and doubles every 5 hours. What is the value of k, per hour?

  1. 5ln⁡2\frac{5}{\ln 2}
  2. 25\frac{2}{5}
  3. ln⁡25\frac{\ln 2}{5}Answer
  4. 5ln⁡25\ln 2

How to solve it

  1. P=P0ektP = P_0e^{kt}, and after 5 hours P=2P0P = 2P_0.
  2. So e5k=2e^{5k} = 2, 5k=ln⁡25k = \ln 2 and k=ln⁡25k = \frac{\ln 2}{5}.

Why each choice is right or wrong

  • A. Incorrect. 5ln⁡2≈7.21\frac{5}{\ln 2} \approx 7.21 flips the fraction; solving 5k=ln⁡25k = \ln 2 divides ln⁡2\ln 2 by 5.
  • B. Incorrect. 25\frac{2}{5} divides the doubling factor by the time, but the factor must go inside a log: e5k=2e^{5k} = 2.
  • C. Correct. Doubling in 5 hours means e5k=2e^{5k} = 2, so 5k=ln⁡25k = \ln 2 and k=ln⁡25≈0.139k = \frac{\ln 2}{5} \approx 0.139.
  • D. Incorrect. 5ln⁡2≈3.475\ln 2 \approx 3.47 multiplies by 5 instead of dividing, which would double the population in a fraction of an hour.

Explain Explain it

Task verbs: determine, interpret

With two data points, divide them so y0y_0 cancels: y(t2)y(t1)=ek(t2−t1)\frac{y(t_2)}{y(t_1)} = e^{k(t_2 - t_1)}. Take the log to get k. A ratio shared by equal time steps is the clue that the data fit an exponential model.

To interpret k, say it with units: k=0.03k = 0.03 per year means the quantity is changing at 3 percent of its current size per year at every moment. To interpret a value of dydt\frac{dy}{dt}, give the quantity, the direction, the rate with units and the moment.

Worked example Hard

A substance decays so that its mass A satisfies dAdt=kA\frac{dA}{dt} = kA. There are 80 milligrams at t=0t = 0 and 20 milligrams at t=6t = 6 hours. How much is left at t=9t = 9 hours?

  1. 10 milligramsAnswer
  2. 2.5 milligrams
  3. about 28.3 milligrams
  4. 0 milligrams

How to solve it

  1. A(6)A(0)=2080=14=e6k\frac{A(6)}{A(0)} = \frac{20}{80} = \frac{1}{4} = e^{6k}, so k=ln⁡(1/4)6≈−0.231k = \frac{\ln(1/4)}{6} \approx -0.231 per hour.
  2. Equivalently, the half-life is 3 hours: 80 to 40 to 20 in 6 hours.
  3. Nine hours is three half-lives: 80 to 40 to 20 to 10 milligrams.

Why each choice is right or wrong

  • A. Correct. The mass falls to a quarter every 6 hours, so A=80(14)t/6A = 80\left(\frac{1}{4}\right)^{t/6} and A(9)=80(14)1.5=808=10A(9) = 80\left(\frac{1}{4}\right)^{1.5} = \frac{80}{8} = 10 milligrams.
  • B. Incorrect. 2.5 milligrams applies 9 more hours of decay to the 20 milligrams at t=6t = 6, which is the amount at t=15t = 15.
  • C. Incorrect. About 28.3 milligrams, 80(12)1.580\left(\frac{1}{2}\right)^{1.5}, assumes a 6-hour half-life, but the mass fell to a quarter, not a half, in 6 hours.
  • D. Incorrect. 0 milligrams assumes a steady drop of 10 milligrams per hour, which would empty it by t=8t = 8, but exponential decay never reaches 0.

Interpret Your turn

A savings account balance B, in dollars, satisfies dBdt=0.05B\frac{dB}{dt} = 0.05B, where t is in years. Say what the 0.05 tells you about the balance, and work out how many years it takes the balance to double.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says that at every moment the balance is growing at a rate equal to 0.05 times (5 percent of) its current value, per year
  • Finds the doubling time ln⁡20.05≈13.9\frac{\ln 2}{0.05} \approx 13.9 years

Model answer

The 0.05 means the balance grows at a rate equal to 5 percent of its current value per year: whatever B is at a moment, it is increasing at 0.05B0.05B dollars per year then. Solving gives B=B0e0.05tB = B_0e^{0.05t}. The balance doubles when e0.05t=2e^{0.05t} = 2, so t=ln⁡20.05≈13.86t = \frac{\ln 2}{0.05} \approx 13.86 years, about 13.9 years, whatever the starting balance.

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