Exponential Models

Part 1 of 7 — Exponential Growth and Decay

The Differential Equation

rac{dy}{dt} = ky implies y = y_0 e^{kt}

• $k > 0$: exponential growth
• $k < 0$: exponential decay

Worked Example

A bacteria culture starts with 500 and grows to 1500 in 2 hours.

$1500 = 500e^{2k}$

$3 = e^{2k}$

k = rac{ln 3}{2} approx 0.549

Population at time $t$: $P(t) = 500e^{0.549t}$

Exponential Growth 🎯

Key Takeaways — Part 1

1. $\frac{dy}{dt} = ky$ always has solution $y = y_0 e^{kt}$
2. Use two data points to find $k$

Exponential Models

Part 2 of 7 — Newton's Law of Cooling

The Model

rac{dT}{dt} = k(T - T_s)

where $T_s$ is the surrounding temperature.

Solution: $T(t) = T_s + (T_0 - T_s)e^{kt}$ where $k < 0$.

Worked Example

A cup of coffee at $200°F$ is placed in a $70°F$ room. After 10 min it's $150°F$.

$150 = 70 + 130e^{10k}$

$80 = 130e^{10k}$

$e^{10k} = \frac{8}{13}$

$T(t) = 70 + 130\left(\frac{8}{13}\right)^{t/10}$

Newton's Cooling 🎯

Key Takeaways — Part 2

1. Newton's Cooling: rate proportional to temperature difference
2. Temperature approaches surroundings exponentially

Exponential Models

Part 3 of 7 — Compound Interest & Continuous Growth

Compound Interest

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

Continuous Compounding

$A = Pe^{rt}$

Connection

As $n \to \infty$: $P(1 + r/n)^{nt} \to Pe^{rt}$

Compound Interest 🎯

Key Takeaways — Part 3

1. Continuous compounding: $A = Pe^{rt}$
2. Doubling time = $\frac{\ln 2}{r}$

Exponential Models

Part 4 of 7 — Derivatives and Integrals of Exponentials

Key Rules

$\frac{d}{dx}[e^{kx}] = ke^{kx} \qquad \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$

$\frac{d}{dx}[a^x] = a^x \ln a \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C$

Exponential Calculus 🎯

Key Takeaways — Part 4

1. $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$
2. $\int a^x\,dx = \frac{a^x}{\ln a} + C$

Exponential Models

Part 5 of 7 — Logistic Growth Preview

The Logistic Model

$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$

where $L$ is the carrying capacity.

Key Features

• Grows exponentially when $P \ll L$
• Fastest growth at $P = L/2$ (inflection point)
• Solution approaches $L$ as $t \to \infty$

Logistic Growth 🎯

$\frac{dP}{dt} = 0.5P(1 - P/1000)$

Key Takeaways — Part 5

1. Logistic growth has a carrying capacity
2. Fastest growth at half the carrying capacity

Exponential Models

Part 6 of 7 — Practice Workshop

Mixed Practice 🎯

Workshop Complete!

Exponential Models — Review

Part 7 of 7 — Final Assessment

Final Assessment 🎯

Exponential Models — Complete! ✅