2.8 Exponential Growth and Decay

Exponential Growth and Decay

Exponential growth and decay describe how quantities change at a rate proportional to their current value. This idea connects directly to differential equations from earlier in the course: the equation $\frac{dy}{dt} = ky$ has the solution $y = A_0 e^{kt}$, and that solution is the foundation for everything in this section. You'll use it to model population growth, radioactive decay, compound interest, and cooling processes.

Exponential Growth and Decay

Exponential growth in real-world scenarios

The exponential growth equation is:

$A(t) = A_0 e^{kt}$

where:

• $A(t)$ is the quantity at time $t$
• $A_0$ is the initial value at $t = 0$
• $k$ is the growth rate (a positive constant)
• $t$ is the time elapsed

This equation comes from solving the differential equation $\frac{dA}{dt} = kA$, which says the rate of change of $A$ is proportional to $A$ itself. That proportionality is what makes growth "exponential" rather than linear.

Population dynamics is a classic application. If a bacteria colony starts at 500 cells and grows at a rate of $k = 0.04$ per minute, you can plug into $A(t) = 500e^{0.04t}$ to estimate the colony size at any future time.

Compound interest uses the same structure. When interest is compounded continuously, the formula $A(t) = A_0 e^{rt}$ applies, where $r$ is the annual interest rate. For example, $1,000 invested at 5% compounded continuously grows to $A(t) = 1000e^{0.05t}$. After 10 years, that's $1000e^{0.5} \approx \$1,648.72$.

Doubling time in exponential growth

Doubling time ($t_d$) is how long it takes a quantity to double. To find it, set $A(t_d) = 2A_0$ and solve:

1. Start with $2A_0 = A_0 e^{kt_d}$
2. Divide both sides by $A_0$: $2 = e^{kt_d}$
3. Take the natural log: $\ln(2) = kt_d$
4. Solve for $t_d$:

$t_d = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$

Notice that doubling time depends only on $k$, not on the initial amount. A population growing at 2% per year ($k = 0.02$) has a doubling time of $\frac{0.693}{0.02} \approx 34.7$ years. Bump that rate to 3%, and the doubling time drops to about 23.1 years. Small changes in $k$ have a big impact.

Exponential decay applications

Exponential decay uses the same model but with a negative exponent:

$A(t) = A_0 e^{-kt}$

Here $k$ is still a positive constant, and the negative sign ensures the quantity decreases over time.

Radioactive decay is the textbook example. Carbon-14 has a decay constant of approximately $k = 0.000121$ per year. If a sample starts with 100 grams, the amount remaining after $t$ years is $A(t) = 100e^{-0.000121t}$. This is the basis of carbon dating.

Newton's Law of Cooling models how an object's temperature approaches the ambient (surrounding) temperature:

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

• $T_a$ is the ambient temperature
• $T_0$ is the initial temperature of the object
• $k$ is the cooling rate constant

For example, if coffee at 90°C is placed in a 20°C room with $k = 0.1$ per minute, its temperature after 5 minutes is $T(5) = 20 + 70e^{-0.5} \approx 62.5°C$. The quantity $(T_0 - T_a)$ decays exponentially, so the object never quite reaches ambient temperature, only approaches it asymptotically.

Half-life in exponential decay

Half-life ($t_{1/2}$) is the time for a quantity to fall to half its value. The derivation mirrors doubling time exactly:

1. Set $\frac{1}{2}A_0 = A_0 e^{-kt_{1/2}}$
2. Divide by $A_0$: $\frac{1}{2} = e^{-kt_{1/2}}$
3. Take the natural log: $\ln\left(\frac{1}{2}\right) = -kt_{1/2}$
4. Since $\ln\left(\frac{1}{2}\right) = -\ln(2)$, you get:

$t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$

This is the same formula as doubling time, which makes sense: doubling time and half-life are two sides of the same relationship between $k$ and $\ln(2)$.

A few reference points that show up in problems:

• Carbon-14: $t_{1/2} \approx 5,730$ years (used in archaeological dating)
• Iodine-131: $t_{1/2} \approx 8$ days (used in medical treatments)
• Caffeine in the body: $t_{1/2} \approx 5$ hours

Shorter half-life means faster decay and a larger $k$. If a problem gives you the half-life, you can find $k$ by rearranging: $k = \frac{\ln(2)}{t_{1/2}}$.

Mathematical modeling and analysis

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. The key property is that the rate of change is proportional to the current value.

Real-world growth doesn't continue exponentially forever. The logistic model accounts for this by introducing a carrying capacity $L$, the maximum sustainable value:

$\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)$

In this model, growth starts out nearly exponential when $y$ is small relative to $L$, but slows and levels off as $y$ approaches $L$. You may or may not cover logistic growth in detail depending on your course, but it's worth knowing that pure exponential growth is an idealization that works best over short time scales or when resources are unlimited.