Lesson 34: Exponential Growth & Decay

(a) Opening Question

A biologist starts a bacterial culture with 500 cells. Two hours later she counts 800 cells. She assumes the population grows so that the rate of growth is proportional to the current population — that is, dy/dt = ky for some constant k.

Before learning any formulas, think about three things:

1. Why does "rate proportional to the amount present" rule out a straight-line (linear) model like y = 500 + 150t?
2. If you knew the value of k, how would you predict the population after 5 hours?
3. The phrase "proportional to the current population" is the entire physical assumption. What kind of function has a derivative proportional to itself?

You already know the answer to #3 from Unit 3: the only functions whose derivative is a constant multiple of themselves are exponentials, y = Ceᵏᵗ. That single fact is the engine of this entire lesson. Hold onto the bacteria numbers — we will solve this exact problem in Worked Example 3, finding k, the doubling time, and the population at t = 5.

(b) Core Concepts

The exponential model

Many quantities change at a rate proportional to how much is present: money earning continuous interest, a radioactive sample decaying, a population with unlimited resources. The mathematical statement of "rate of change is proportional to the amount" is the exponential differential equation:

dy/dt = ky

Here k is the constant of proportionality (also called the growth/decay constant). This is a separable differential equation — exactly the type from Lesson 33 — so we can solve it from scratch.

Deriving the general solution

Start with dy/dt = ky and separate the variables (assume y > 0):

(1/y) dy = k dt

Integrate both sides:

∫ (1/y) dy = ∫ k dt
ln|y| = kt + C

Exponentiate to undo the logarithm:

|y| = e^(kt + C) = e^C · e^(kt)

Since e^C is just a positive constant, write it as a single constant and absorb the sign; call it y₀. The general solution is:

y = y₀ eᵏᵗ

What is y₀? Set t = 0: y(0) = y₀e⁰ = y₀. So y₀ is the initial amount — the value of y when t = 0. The model has exactly two parameters: the starting amount y₀ and the rate constant k.

Verify by substitution. If y = y₀eᵏᵗ, then dy/dt = y₀ · k eᵏᵗ = k(y₀eᵏᵗ) = ky. ✓ The function satisfies the differential equation, and y(0) = y₀. On the AP exam you are often asked to verify a proposed solution this way.

Growth versus decay — the sign of k

• If k > 0, then eᵏᵗ → ∞ as t → ∞: this is exponential growth.
• If k < 0, then eᵏᵗ → 0 as t → ∞: this is exponential decay.

The sign of k is the single most important qualitative fact about the model. A decay constant is negative — watch for this.

Finding k from data

You almost never are given k. Instead you are given two data points and must solve for it. If you know y₀ and one other value y(t₁) = y₁:

y₁ = y₀ e^(k t₁)   ⟹   e^(k t₁) = y₁/y₀   ⟹   k = (1/t₁) ln(y₁/y₀)

The logarithm is how you isolate k. A logarithm is required whenever you solve for k or solve for a time t.

Half-life and doubling time

Two special times have names.

Half-life (decay, k < 0): the time t_½ for the amount to fall to half its value. Set y = y₀/2:

y₀/2 = y₀ e^(k t_½)   ⟹   1/2 = e^(k t_½)   ⟹   k t_½ = ln(1/2) = −ln 2

t_½ = −(ln 2)/k        (equivalently  k = −(ln 2)/t_½)

Since k < 0, the half-life t_½ = −(ln 2)/k comes out positive, as it must.

Doubling time (growth, k > 0): the time T_d for the amount to double. Set y = 2y₀:

2 = e^(k T_d)   ⟹   k T_d = ln 2   ⟹   T_d = (ln 2)/k        (and  k = (ln 2)/T_d)

Notice the half-life and doubling time depend only on k, not on y₀ — a property unique to exponential models. The amount halves (or doubles) over every interval of the same length, no matter where you start.

Worked mini-example (decay)

A radioactive sample has half-life 8 days, starting at 100 mg. Find k, then the amount after 20 days.

From the half-life: k = −(ln 2)/8 ≈ −0.0866 /day. The model is y = 100 e^(−0.0866 t). After 20 days:

y(20) = 100 e^(−0.0866 · 20) ≈ 17.68 mg

Equivalently, 20 days is 20/8 = 2.5 half-lives, so y = 100 · (½)^2.5 ≈ 17.68 mg — same answer. Always interpret with units (mg, days).

Newton's Law of Cooling — a shifted model

A cooling object does not decay toward zero — it decays toward the ambient (room) temperature T. The rate of cooling is proportional to the difference between the object's temperature and the surroundings:

dy/dt = k(y − T)

This is still separable. Let u = y − T, so du/dt = dy/dt (because T is constant), giving du/dt = ku — our familiar equation in u. Its solution is u = u₀eᵏᵗ, and substituting back u = y − T, u₀ = y₀ − T:

y = T + (y₀ − T) eᵏᵗ

Here y₀ is the initial temperature and T is the constant surrounding temperature. Because a cooling object loses heat, k < 0, so eᵏᵗ → 0 and y → T: the object's temperature approaches the room temperature as t → ∞. The (y₀ − T) factor is the initial gap between the object and its surroundings. The shift T is the most common source of error — see Common Mistakes.

Beyond AB (not tested): When growth is limited by carrying capacity, BC students study the logistic model dy/dt = ky(1 − y/L). It is not on the AP Calculus AB exam, and nothing in this lesson's practice or FRQ uses it. All graded content here is the exponential (and Newton's-cooling) model.

(c) Worked Examples

Example 1 — Growth: find k from data, then predict [NO CALC for setup; CALC for evaluation]

Problem. A town's population was 1000 in 1990 and 1500 in 2000. Assume dy/dt = ky. Let t = years since 1990. Find k, the doubling time, and the predicted population in 2010.

Strategy. y₀ = 1000. Use the 2000 data point (t = 10, y = 1500) to solve for k with a logarithm, then evaluate.

Solution.

1500 = 1000 e^(10k)  ⟹  e^(10k) = 1.5  ⟹  k = (ln 1.5)/10 ≈ 0.04055 /year

Doubling time: T_d = (ln 2)/k = (ln 2)/((ln 1.5)/10) = 10 ln2/ln1.5 ≈ 17.10 years.

Prediction for 2010 (t = 20):

y(20) = 1000 e^(20 · 0.04055) = 1000 e^(ln 1.5 · 2) = 1000 (1.5)² = 2250 people

Justification/interpretation. Because k = (ln 1.5)/10 > 0, the model predicts growth; the population is predicted to be 2250 people in 2010, and it doubles roughly every 17.1 years.

Example 2 — Radioactive decay + half-life [CALC]

Problem. Strontium-90 has a half-life of 28 years. A sample starts at 200 g. Find the decay constant k and the amount remaining after 50 years.

Strategy. Get k from the half-life, then evaluate y(50).

Solution.

k = −(ln 2)/28 ≈ −0.02476 /year
y(50) = 200 e^(−0.02476 · 50) ≈ 58.01 g

TI-84: 200e^(-0.0247650) → 58.01. Interpretation: about 58.0 g of Strontium-90 remain after 50 years. The negative k confirms decay.

Example 3 — The opening problem: doubling time [CALC]

Problem. A culture starts at 500 cells and grows to 800 in 2 hours with dy/dt = ky. Find k, the doubling time, and the population at t = 5 hours.

Solution.

800 = 500 e^(2k)  ⟹  k = (1/2) ln(800/500) = (1/2) ln(1.6) ≈ 0.2350 /hour
T_d = (ln 2)/0.2350 ≈ 2.95 hours
y(5) = 500 e^(0.2350 · 5) ≈ 1619 cells

Interpretation: the population doubles about every 2.95 hours and reaches roughly 1619 cells at t = 5 hours.

Example 4 — Newton's Law of Cooling [CALC]

Problem. Coffee at 90 °C is left in a 20 °C room. After 5 minutes it is 70 °C. Find k, the temperature at t = 15 min, and the time to cool to 40 °C.

Strategy. Use y = T + (y₀ − T)eᵏᵗ with T = 20, y₀ = 90, so y = 20 + 70 eᵏᵗ. Find k from the t = 5 data, subtracting the shift first.

Solution.

70 = 20 + 70 e^(5k)  ⟹  e^(5k) = 50/70  ⟹  k = (1/5) ln(5/7) ≈ −0.06729 /min

Temperature at t = 15:

y(15) = 20 + 70 e^(−0.06729 · 15) ≈ 20 + 25.51 ≈ 45.51 °C

Time to reach 40 °C:

40 = 20 + 70 eᵏᵗ  ⟹  e^(kt) = 20/70  ⟹  t = ln(2/7)/k ≈ 18.62 minutes

Interpretation: the coffee is about 45.5 °C after 15 minutes and reaches 40 °C after about 18.6 minutes. Note that it cools toward 20 °C, never below it.

(d) Common Mistakes

Using a linear model instead of exponential. Students see "1000 then 1500" and write y = 1000 + 50t. Why wrong: "rate proportional to the amount" forces dy/dt = ky, whose solution is exponential, y = y₀eᵏᵗ, not linear. Fix: the phrase "rate proportional to amount present" always means exponential.

Wrong sign on k for decay. Writing k = +(ln 2)/t_½ for a decaying quantity. Why wrong: a decaying amount needs eᵏᵗ → 0, which requires k < 0. Fix: for decay, k = −(ln 2)/t_½; the formula carries the minus sign.

Confusing half-life with the decay constant. Plugging the half-life in for k (or vice versa). Why wrong: t_½ is a time (in days, years…); k is a rate (per day, per year). They are reciprocally related by t_½ = −(ln 2)/k, not equal. Fix: convert with the formula, and check units.

Mishandling the shift in Newton's cooling. Writing 70 = 90 e^(5k) instead of subtracting the room temperature. Why wrong: in y = T + (y₀ − T)eᵏᵗ, the exponential multiplies the gap (y₀ − T), not the raw temperature. Fix: isolate the exponential as (y − T)/(y₀ − T) = eᵏᵗ before taking the log.

Forgetting units (and interpretation). Reporting "≈ 58" with no grams, or "k = −0.025" with no "per year." Why wrong: AP scoring requires units and a sentence of interpretation. Fix: always attach units and answer the question in words.

(e) Practice Problems

(Justify.) A quantity satisfies dy/dt = ky with k < 0. Explain what happens to y as t → ∞, and state whether this models growth or decay. Justify using the form of the solution.

(Justify/interpret.) Soup at 100 °C is placed in a 20 °C room and cools so that after 10 minutes it is 60 °C. Write the cooling model y = T + (y₀ − T)eᵏᵗ with numbers filled in, set up (do not solve numerically) the equation for k, and state the value of k in exact form. Interpret the sign of k.

A town of 8000 grows continuously at 2% per year. To the nearest year, when does it reach 10,000? (Short answer — show the equation and solve.)

(f) AP Exam Focus — Free Response

FRQ (9 points total). [CALC permitted on this question]

A turkey is removed from an oven at temperature 165 °F and placed in a room held at a constant 70 °F. The turkey cools so that its temperature y, in °F, satisfies Newton's Law of Cooling

dy/dt = k(y − 70)

where t is the time in minutes after the turkey is removed and k is a constant. Twenty minutes after it is removed, the turkey's temperature is 140 °F.

(a) (2 points) Verify that y = 70 + (y₀ − 70)eᵏᵗ is a solution of the differential equation dy/dt = k(y − 70), and identify the value of y₀ for this turkey.

(b) (3 points) Use the given data to find the value of k. Show the equation you solve.

(c) (2 points) Find the temperature of the turkey 45 minutes after it is removed from the oven. Include units.

(d) (2 points) It is safe to carve the turkey once it has cooled to 100 °F. According to the model, how many minutes after removal does this occur? Include units, and interpret the long-run behavior of the turkey's temperature.

Model Solution

(a) Differentiate y = 70 + (y₀ − 70)eᵏᵗ:

dy/dt = (y₀ − 70) · k eᵏᵗ = k[(y₀ − 70)eᵏᵗ] = k[y − 70]

since (y₀ − 70)eᵏᵗ = y − 70. Thus dy/dt = k(y − 70), so the function satisfies the differential equation. At t = 0, y = 70 + (y₀ − 70) = y₀, the initial temperature, so y₀ = 165 (the oven temperature). The model is y = 70 + 95 eᵏᵗ.

(b) Use t = 20, y = 140:

140 = 70 + 95 e^(20k)
70 = 95 e^(20k)
e^(20k) = 70/95 = 14/19
k = (1/20) ln(14/19) ≈ −0.01527  (per minute)

(c) At t = 45:

y(45) = 70 + 95 e^(−0.01527 · 45) ≈ 70 + 47.79 ≈ 117.8 °F

The turkey is about 117.8 °F 45 minutes after removal.

(d) Set y = 100:

100 = 70 + 95 eᵏᵗ
eᵏᵗ = 30/95
t = ln(30/95)/k = ln(30/95)/((1/20)ln(14/19)) ≈ 75.5 minutes

The turkey reaches 100 °F about 75.5 minutes after removal. Long-run behavior: since k < 0, eᵏᵗ → 0 as t → ∞, so y → 70 °F — the turkey's temperature approaches the room temperature of 70 °F but never goes below it.

Scoring Commentary — where students lose points

• (a) 1 pt for the differentiation showing dy/dt = k(y − 70); 1 pt for identifying y₀ = 165. Students lose the first point if they only assert the answer without showing (y₀ − 70)eᵏᵗ = y − 70.
• (b) 1 pt for the equation 140 = 70 + 95e^(20k) (correctly subtracting the 70 shift); 1 pt for isolating e^(20k) = 14/19; 1 pt for k ≈ −0.01527. The classic error is 140 = 165 e^(20k), forgetting the shift — this loses all 3 points for part (b).
• (c) 1 pt for the substitution t = 45; 1 pt for ≈ 117.8 °F with units. No units → lose the answer point.
• (d) 1 pt for t ≈ 75.5 min with the equation shown; 1 pt for the interpretation that y → 70 °F as t → ∞ because k < 0. A bare number with no interpretation earns only 1 of 2.

60 = 20 + 80 e^(10k)  ⟹  e^(10k) = 40/80 = 1/2  ⟹  k = (1/10) ln(1/2) = −(ln 2)/10