6.8 Exponential Growth and Decay

Exponential growth and decay models describe how quantities change at a rate proportional to their current value. This idea connects directly to differential equations, making it one of the most important applications of integration in Calculus I. Whether you're modeling population growth, radioactive decay, or cooling temperatures, the underlying math is the same.

Exponential Growth Models

Applications of exponential growth

Exponential growth happens when a quantity increases at a rate proportional to its current size. The bigger it gets, the faster it grows.

The continuous exponential growth model comes from solving the differential equation $\frac{dy}{dt} = ky$, where $k > 0$. Separating variables and integrating both sides gives:

$y(t) = y_0 e^{kt}$

• $y_0$: initial value (at $t = 0$)
• $k$: continuous growth rate
• $t$: time

This is the form you'll use most in Calculus I. There's also a discrete version, $y = a(1 + r)^t$, where $r$ is the growth rate per time interval, but the continuous model is what falls out of the integration naturally.

Population dynamics are a classic example. Bacteria starting at 100 cells and doubling every hour follow $P(t) = 100 \cdot 2^t$. You can rewrite this in continuous form as $P(t) = 100 \cdot e^{(\ln 2)t}$, so the continuous growth rate is $k = \ln 2 \approx 0.693$.

Compound interest also uses exponential growth. The standard formula is:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

• $P$: initial principal
• $r$: annual interest rate (decimal)
• $n$: compounding periods per year
• $t$: years

As $n \to \infty$, this approaches continuous compounding: $A = Pe^{rt}$. This limit is a direct application of the definition of $e$.

Doubling time in growth models

Doubling time is how long it takes a quantity to double. To find it, set $y(t_d) = 2y_0$ in the continuous model:

1. Start with $2y_0 = y_0 e^{k t_d}$
2. Divide both sides by $y_0$: $2 = e^{k t_d}$
3. Take the natural log: $\ln 2 = k t_d$
4. Solve: $t_d = \frac{\ln 2}{k}$

A key property: doubling time is constant. It doesn't matter whether you have 100 or 10,000 of something. If $k$ stays the same, the time to double is always $\frac{\ln 2}{k}$.

Exponential Decay Models

Uses of exponential decay

Exponential decay works the same way as growth, but the rate constant is negative. The quantity shrinks at a rate proportional to its current size.

From the differential equation $\frac{dy}{dt} = -\lambda y$ (where $\lambda > 0$), integration gives:

$y(t) = y_0 e^{-\lambda t}$

• $y_0$: initial amount
• $\lambda$: decay constant (positive number)

There's also a discrete version, $y = a(1 - r)^t$, but again, the continuous form is what you get from solving the differential equation.

Radioactive decay is the textbook example. If you start with 500 grams of a substance with decay constant $\lambda = 0.05 \text{ yr}^{-1}$, then after 10 years you have:

$A(10) = 500 e^{-0.05(10)} = 500 e^{-0.5} \approx 303 \text{ grams}$

Newton's Law of Cooling models temperature change as exponential decay toward an ambient temperature $T_a$:

$T(t) = T_a + (T_0 - T_a)e^{-kt}$

• $T_0$: initial object temperature
• $T_a$: surrounding (ambient) temperature
• $k$: cooling constant (positive)

The quantity that decays here isn't the temperature itself but the difference between the object's temperature and the ambient temperature. That difference shrinks exponentially toward zero, so $T(t)$ approaches $T_a$ over time.

Half-life in decay processes

Half-life is the time for a quantity to drop to half its value. The derivation mirrors doubling time exactly:

1. Set $\frac{1}{2}y_0 = y_0 e^{-\lambda t_{1/2}}$
2. Divide by $y_0$: $\frac{1}{2} = e^{-\lambda t_{1/2}}$
3. Take the natural log: $-\ln 2 = -\lambda t_{1/2}$
4. Solve: $t_{1/2} = \frac{\ln 2}{\lambda}$

Notice this formula looks identical to the doubling time formula, just with $\lambda$ in place of $k$. Half-life is also constant: it takes the same amount of time to go from 1000 to 500 as from 500 to 250.

Half-life is especially useful in radiometric dating. Carbon-14 has a half-life of about 5,730 years. If a sample has 25% of its original Carbon-14, that's two half-lives, so the sample is roughly 11,460 years old.

Mathematical Foundations

All exponential growth and decay models trace back to one differential equation:

$\frac{dy}{dt} = ky$

When $k > 0$, you get growth. When $k < 0$, you get decay. Solving it requires separation of variables:

1. Rewrite as $\frac{dy}{y} = k \, dt$
2. Integrate both sides: $\ln|y| = kt + C$
3. Exponentiate: $y = e^{kt + C} = e^C \cdot e^{kt}$
4. Since $y(0) = y_0$, you get $e^C = y_0$
5. Final solution: $y(t) = y_0 e^{kt}$

This is why the topic lives in "Applications of Integration." The exponential function is the unique function that is its own derivative (up to a constant multiple), and that property is exactly what makes it the solution here.

One practical note: if you're given data and need to find $k$, take the natural log of both sides to turn the exponential relationship into a linear one. Plotting $\ln(y)$ vs. $t$ should give a straight line with slope $k$. This technique, called linearization, makes it much easier to extract the rate constant from real-world data.