Growth and decay for MATH 140

The Most Important Equation You've Never Heard Of

In 1965, Intel co-founder Gordon Moore made a prediction: the number of transistors on a computer chip would double approximately every two years.

This prediction — "Moore's Law" — has held remarkably true for over 50 years. It's why the smartphone in your pocket has more computing power than all of NASA had when they landed on the moon.

But here's what's fascinating: Moore didn't need to know anything about semiconductor physics to make this prediction. He just needed one equation:

$$A(t) = A_0 \cdot b^{t}$$

This is the exponential growth model, and it describes everything from computer chips to bacteria colonies to your retirement savings. Master this equation, and you can predict the future.

The Exponential Growth Model

When a quantity grows by a constant percentage over equal time intervals, it follows exponential growth:

$$A(t) = A_0 \cdot b^{t}$$

Where:

• $A(t)$ = the amount at time $t$
• $A_0$ = the initial amount (at $t = 0$)
• $b$ = the growth factor (how much it multiplies per time unit)
• $t$ = time

The key insight: Each time period, the quantity multiplies by the same factor, rather than adding the same amount.

Growth Factor vs. Growth Rate

These two are related but different:

Growth rate ($r$): The percentage increase per time period
Growth factor ($b$): What you multiply by: $b = 1 + r$

So the model can also be written as:

$$A(t) = A_0 \cdot (1 + r)^{t}$$

Example: A population of 1,000 bacteria grows by 15% per hour.

• $A_0 = 1000$
• $r = 0.15$, so $b = 1.15$
• $A(t) = 1000 \cdot (1.15)^t$

After 10 hours: $A(10) = 1000 \cdot (1.15)^{10} \approx 4,046$ bacteria

The Exponential Decay Model

When a quantity decreases by a constant percentage, we get exponential decay:

$$A(t) = A_0 \cdot b^{t} \quad \text{where } 0 < b < 1$$

Or equivalently:

$$A(t) = A_0 \cdot (1 - r)^{t}$$

Where $r$ is the decay rate (the percentage lost per time period).

Example: A car worth 30,000 dollars depreciates by 12% per year.

• $A_0 = 30000$
• $r = 0.12$, so $b = 0.88$
• $A(t) = 30000 \cdot (0.88)^t$

After 5 years: $A(5) = 30000 \cdot (0.88)^5 \approx 15{,}837$ dollars

The Natural Exponential Model

In many real-world applications, especially in science, we use base $e$:

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

Where:

• $k > 0$ means growth
• $k < 0$ means decay
• $|k|$ is called the continuous growth/decay rate

Why base $e$?

1. It arises naturally from continuous compounding
2. The calculus is cleaner: $\frac{d}{dt}[e^{kt}] = k \cdot e^{kt}$
3. Many natural processes (radioactive decay, population growth, cooling) naturally follow this form

The two forms are equivalent — you can always convert between them: $$A_0 \cdot b^t = A_0 \cdot e^{kt} \quad \text{where } k = \ln(b)$$

Half-Life: The Signature of Decay

Half-life is the time it takes for a quantity to reduce to half its value.

This is the go-to concept for radioactive decay, but it applies to anything that decays exponentially.

Finding Half-Life from the Decay Constant

If $A(t) = A_0 \cdot e^{kt}$ (where $k < 0$), the half-life is:

Using Half-Life in Problems

If you know the half-life, you can write the decay model as:

Example: Carbon-14 has a half-life of 5,730 years.

If a fossil originally had 100 units of C-14:

• After 5,730 years: 50 units remain
• After 11,460 years: 25 units remain
• After 17,190 years: 12.5 units remain

Using the formula: $A(t) = 100 \cdot \left(\frac{1}{2}\right)^{t/5730}$

Doubling Time: The Signature of Growth

Doubling time is the time it takes for a quantity to double.

Finding Doubling Time from the Growth Constant

If $A(t) = A_0 \cdot e^{kt}$ (where $k > 0$), the doubling time is:

The Rule of 70

A quick approximation: if something grows at $r$% per period, it doubles in approximately:

$$t_{\text{double}} \approx \frac{70}{r}$$

Examples:

• 7% annual growth → doubles in ~10 years
• 10% annual growth → doubles in ~7 years
• 2% annual growth → doubles in ~35 years

This is incredibly useful for quick mental math about investments, populations, or any growing quantity.

Application: Compound Interest

When money earns interest, the interest itself earns interest. This is compound interest — exponential growth applied to money.

Compounding $n$ Times Per Year

Where:

• $P$ = principal (initial investment)
• $r$ = annual interest rate (as a decimal)
• $n$ = number of times interest compounds per year
• $t$ = time in years

Example: 5,000 dollars invested at 6% annual interest, compounded monthly, for 10 years:

Continuous Compounding

What if we compound infinitely often? As $n \to \infty$:

$$A(t) = Pe^{rt}$$

This is continuous compounding — the theoretical maximum growth rate.

Same example with continuous compounding:

$$A = 5000 \cdot e^{0.06 \cdot 10} = 5000 \cdot e^{0.6} \approx 9{,}111$$

Only about 14 dollars more than monthly compounding — the difference is small!

Application: Population Growth

Populations often grow exponentially when resources are abundant:

$$P(t) = P_0 \cdot e^{kt}$$

Example: A city has 500,000 people and is growing at 2.5% per year continuously.

• $P_0 = 500{,}000$
• $k = 0.025$
• $P(t) = 500{,}000 \cdot e^{0.025t}$

When will the population reach 1 million?

$1{,}000{,}000 = 500{,}000 \cdot e^{0.025t}$

$2 = e^{0.025t}$

$\ln(2) = 0.025t$

$t = \frac{\ln(2)}{0.025} \approx 27.7$ years

Application: Radioactive Decay

Radioactive substances decay at a rate proportional to their current amount:

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

The decay constant $k$ is related to half-life by: $k = \frac{\ln(2)}{t_{1/2}}$

Carbon Dating

Living organisms contain Carbon-14, which decays with a half-life of 5,730 years. When an organism dies, it stops absorbing new C-14, so the amount decreases.

Example: A fossil contains 30% of its original C-14. How old is it?

$$0.30 = e^{-kt}$$

where $k = \frac{\ln(2)}{5730} \approx 0.000121$

$$\ln(0.30) = -0.000121 \cdot t$$

Application: Newton's Law of Cooling

When an object is placed in an environment with a different temperature, it cools (or warms) exponentially toward the ambient temperature:

Where:

• $T(t)$ = temperature at time $t$
• $T_{\text{ambient}}$ = surrounding temperature
• $T_0$ = initial temperature of the object
• $k$ = cooling constant (depends on the object and environment)

Example: Coffee at 180°F is placed in a 70°F room. After 10 minutes, it's 120°F. What will it be after 20 minutes?

First, find $k$: $$120 = 70 + (180 - 70)e^{-10k}$$ $$50 = 110e^{-10k}$$ $$e^{-10k} = \frac{50}{110} = \frac{5}{11}$$ $$k = \frac{-\ln(5/11)}{10} \approx 0.0788$$

Now find $T(20)$:

Application: Medicine and Drug Concentration

When you take medication, your body eliminates it exponentially:

$$C(t) = C_0 \cdot e^{-kt}$$

Example: A painkiller has a half-life of 4 hours in the bloodstream. If you take a 500mg dose:

$k = \frac{\ln(2)}{4} \approx 0.173$

After 8 hours: $C(8) = 500 \cdot e^{-0.173 \cdot 8} = 500 \cdot e^{-1.386} \approx 125$ mg

This is why medications have dosing schedules — they're designed to keep the drug concentration within an effective range!

Solving Growth and Decay Problems

Type 1: Find the Amount

Given: initial amount, rate/factor, time
Find: final amount

Just plug into the formula: $$A(t) = A_0 \cdot b^t \quad \text{or} \quad A(t) = A_0 \cdot e^{kt}$$

Type 2: Find the Time

Given: initial amount, final amount, rate/factor
Find: time

Use logarithms:

Starting with $A = A_0 \cdot b^t$:

1. Divide: $\frac{A}{A_0} = b^t$
2. Take log: $\log_b\left(\frac{A}{A_0}\right) = t$
3. Or use natural log: $t = \frac{\ln(A/A_0)}{\ln(b)}$

Type 3: Find the Rate

Given: initial amount, final amount, time
Find: rate

Use logarithms:

Starting with $A = A_0 \cdot e^{kt}$:

1. Divide: $\frac{A}{A_0} = e^{kt}$
2. Take ln: $\ln\left(\frac{A}{A_0}\right) = kt$
3. Solve: $k = \frac{1}{t}\ln\left(\frac{A}{A_0}\right)$

Practice Problems

Common Mistakes and Misunderstandings

❌ Mistake: Confusing growth rate with growth factor

Wrong: "It grows 5% per year, so $A = A_0 \cdot (0.05)^t$"

Why it's wrong: The 5% is the rate of increase, not the multiplier. If something grows by 5%, you have 100% + 5% = 105% of what you started with.

Correct: $A = A_0 \cdot (1.05)^t$. The growth factor is $1 + r = 1 + 0.05 = 1.05$.

❌ Mistake: Using the wrong sign for decay

Wrong: "It decays at 3% per year, so $A = A_0 \cdot (1.03)^t$"

Why it's wrong: That's growth, not decay! Decay means you're losing 3% each period.

Correct: $A = A_0 \cdot (0.97)^t$ or $A = A_0 \cdot (1 - 0.03)^t$. For decay, the factor is less than 1.

❌ Mistake: Forgetting that half-life means HALF

Wrong: After 2 half-lives, "none remains"

Why it's wrong: After each half-life, half remains. After 2 half-lives, you have $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ remaining, not zero.

Correct: After $n$ half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^n$. The quantity approaches zero but never reaches it.

❌ Mistake: Not matching time units

Wrong: "Interest is 6% per year, invested for 18 months, so $A = P(1.06)^{18}$"

Why it's wrong: The rate is per year, but you used 18 (months) as the exponent. The units must match!

Correct: Either convert time to years: $A = P(1.06)^{1.5}$, or convert rate to monthly: $A = P(1.005)^{18}$.

❌ Mistake: Confusing $A_0$ with $A(t)$

Wrong: Setting up "I have 50,000 dollars now and want to know what I started with" as $50000 \cdot e^{rt} = ?$

Why it's wrong: You're solving for the initial amount, not the final amount. The 50,000 dollars is $A(t)$, not $A_0$.

Correct: $50000 = A_0 \cdot e^{rt}$, so $A_0 = \frac{50000}{e^{rt}}$

Summary: Growth vs. Decay at a Glance

The Big Picture

Exponential growth and decay are everywhere:

• Your money grows exponentially with compound interest
• Diseases spread exponentially in early stages
• Technology improves exponentially (Moore's Law)
• Radioactive waste decays exponentially over millennia
• Your morning coffee cools exponentially toward room temperature
• Medicine leaves your bloodstream exponentially

The same mathematics describes all of these phenomena. Once you master the exponential model, you have a powerful tool for understanding — and predicting — how quantities change over time.