Lesson 1: Exponential Growth and Decay
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Derive and solve the exponential growth/decay equation dP/dt = kP
- Compute half-life and doubling time from the growth/decay constant k
- Apply the model to radioactive decay problems
- Apply the model to continuously compounded interest
- Determine k from experimental data (two data points)
The Fundamental Model
The simplest and most important first-order ODE application is the exponential model. It arises whenever the rate of change of a quantity is proportional to the quantity itself.
Exponential Growth/Decay Equation: dP/dt = kP, where P(t) is the quantity at time t and k is the growth (k > 0) or decay (k < 0) constant. The general solution is P(t) = P0ekt, where P0 = P(0) is the initial amount.
This equation is separable. Separating variables: dP/P = k dt. Integrating both sides: ln|P| = kt + C, so P = Aekt. Setting t = 0 gives A = P0.
Half-Life and Doubling Time
Half-Life (t1/2): The time for a decaying quantity to reach half its initial value. From P0ekt1/2 = P0/2, we get t1/2 = ln(2)/|k|.
Doubling Time (td): The time for a growing quantity to double. From P0ektd = 2P0, we get td = ln(2)/k.
Notice the beautiful symmetry: both formulas involve ln(2) ≈ 0.693. This gives us a quick estimation rule -- if a quantity grows at r% per unit time (continuously), its doubling time is approximately 69.3/r.
Radioactive Decay
Radioactive substances decay at a rate proportional to the amount present. The decay constant k is negative, and each isotope has a characteristic half-life.
Example 1: Carbon-14 Dating
Carbon-14 has a half-life of 5730 years. A bone fragment contains 40% of its original C-14. How old is it?
Step 1: Find k from the half-life.
t1/2 = ln(2)/|k| ⇒ 5730 = ln(2)/|k| ⇒ k = -ln(2)/5730 ≈ -0.000121 per year.
Step 2: Use P(t) = P0ekt with P(t)/P0 = 0.40.
0.40 = e-0.000121t ⇒ ln(0.40) = -0.000121t ⇒ t = -ln(0.40)/0.000121 ≈ 7573 years.
Example 2: Iodine-131 Decay
Iodine-131 (used in thyroid treatment) has a half-life of 8 days. If a patient receives 100 mCi, how much remains after 20 days?
Step 1: k = -ln(2)/8 ≈ -0.08664 per day.
Step 2: P(20) = 100 e-0.08664(20) = 100 e-1.7328 ≈ 100(0.1768) ≈ 17.68 mCi.
Continuously Compounded Interest
If a principal A0 is invested at annual rate r compounded continuously, the balance satisfies dA/dt = rA with solution A(t) = A0ert.
Example 3: Investment Growth
You invest $5,000 at 6% annual interest compounded continuously. How long until the investment doubles?
Solution: We need A(t) = 2A0.
5000 e0.06t = 10000 ⇒ e0.06t = 2 ⇒ 0.06t = ln 2 ⇒ t = ln(2)/0.06 ≈ 11.55 years.
This matches the doubling time formula: td = ln(2)/0.06 ≈ 11.55.
Determining k from Data
Often we are given two measurements and must find k. The strategy: use P(t1) = P0ekt1 and P(t2) = P0ekt2, then divide to eliminate P0.
Example 4: Bacterial Culture
A bacterial culture has 800 cells at t = 0 and 2200 cells after 3 hours. Find k and predict the population at t = 5 hours.
Step 1: P(3) = 800 e3k = 2200.
e3k = 2200/800 = 2.75 ⇒ 3k = ln(2.75) ⇒ k = ln(2.75)/3 ≈ 0.3370 per hour.
Step 2: P(5) = 800 e0.3370(5) = 800 e1.685 ≈ 800(5.393) ≈ 4314 cells.
Doubling time: td = ln(2)/0.3370 ≈ 2.06 hours.
When Exponential Models Break Down
The exponential model assumes unlimited resources and no competition. In reality:
- Populations encounter limited food, space, and other resources.
- Growth rates slow as populations become large.
- The exponential model is best for short time horizons or small populations far from capacity limits.
This motivates the logistic model in the next lesson, which adds a carrying capacity to account for resource limitations.
Check Your Understanding
1. A radioactive isotope has k = -0.05 per hour. What is its half-life?
2. A population doubles every 4 years. What is the continuous growth rate k?
3. You deposit $10,000 at 4% compounded continuously. What is the balance after 10 years?
4. A sample of 500 mg of a substance decays to 350 mg in 10 days. Find k and the half-life.
5. Why does the exponential growth model predict infinite population as t → ∞? Is this realistic?
Key Takeaways
- dP/dt = kP has solution P(t) = P0ekt; k > 0 is growth, k < 0 is decay.
- Half-life = ln(2)/|k| and doubling time = ln(2)/k -- both depend only on k, not on P0.
- To find k from two data points, use the ratio P(t2)/P(t1) to eliminate P0.
- Applications include radioactive decay, carbon dating, bacterial growth, and continuously compounded interest.
- Exponential growth is unrealistic for large populations -- the logistic model (next lesson) corrects this.
Ready for More?
Next Lesson
In Lesson 2, you will study the logistic equation, which adds a carrying capacity to model realistic population growth.
Start Lesson 2