Unit 7.8 – Exponential Models with Differential Equations

AP® Calculus AB & BC | Real-World Applications of Exponential Growth and Decay

Why This Matters: Exponential models are everywhere in the real world! From population growth to radioactive decay, from compound interest to cooling coffee—the differential equation $\frac{dy}{dt} = ky$ is THE most important DE in applications. This topic brings together everything you've learned: modeling, solving, and interpreting. It's tested extensively on AP® exams and is crucial for science, engineering, economics, and medicine. Master exponential models and you'll understand how calculus explains the natural world!

🎯 The Fundamental Exponential Model

The Exponential Growth/Decay Model

THE DIFFERENTIAL EQUATION:

\[ \frac{dy}{dt} = ky \]

General Solution:

\[ y(t) = y_0 e^{kt} \]

Where:

$y(t)$ = quantity at time $t$
$y_0$ = initial quantity (at $t = 0$)
$k$ = growth/decay constant
$e$ = Euler's number ≈ 2.71828

Understanding the Constant $k$:

If $k > 0$: EXPONENTIAL GROWTH
- Population increasing, money growing, spreading disease
If $k < 0$: EXPONENTIAL DECAY
- Radioactive decay, cooling, drug elimination
If $k = 0$: No change (constant)

📐 Deriving the Solution

Derivation using Separation of Variables:

Start with: $\frac{dy}{dt} = ky$

Separate:

\[ \frac{1}{y} \, dy = k \, dt \]

Integrate:

\[ \ln|y| = kt + C \]

Solve for $y$:

\[ y = Ae^{kt} \]

where $A = \pm e^C$

Apply initial condition $y(0) = y_0$:

\[ y_0 = Ae^{0} = A \]

Therefore: $A = y_0$

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

🔍 Finding the Growth/Decay Constant k

Methods for Finding k

Method 1: Given a data point

If you know $y(t_1) = y_1$ for some time $t_1$:

\[ y_1 = y_0 e^{kt_1} \]

\[ \frac{y_1}{y_0} = e^{kt_1} \]

\[ \ln\left(\frac{y_1}{y_0}\right) = kt_1 \]

\[ k = \frac{1}{t_1}\ln\left(\frac{y_1}{y_0}\right) \]

Method 2: Given growth/decay rate

If told "grows at 5% per year":

$k = 0.05$ (for continuous growth)

Method 3: From half-life or doubling time

See formulas below

⏱️ Half-Life and Doubling Time

SPECIAL TIME PERIODS

Half-Life (for decay, $k < 0$):

Definition: Time for quantity to reduce to half its value

\[ t_{1/2} = \frac{\ln 2}{|k|} = \frac{0.693}{|k|} \]

Or, if half-life is given, find $k$:

\[ k = -\frac{\ln 2}{t_{1/2}} \]

Doubling Time (for growth, $k > 0$):

Definition: Time for quantity to double

\[ t_{\text{double}} = \frac{\ln 2}{k} = \frac{0.693}{k} \]

Or, if doubling time is given, find $k$:

\[ k = \frac{\ln 2}{t_{\text{double}}} \]

🌍 Common Applications

Application 1: Population Growth

Problem: A bacteria population has 1000 cells initially. After 3 hours, there are 8000 cells. Find the exponential model and determine when the population reaches 50,000.

Given:

$y_0 = 1000$ (initial population)
$y(3) = 8000$ (after 3 hours)

Find k:

\[ 8000 = 1000e^{3k} \]

\[ 8 = e^{3k} \]

\[ \ln 8 = 3k \]

\[ k = \frac{\ln 8}{3} \approx 0.693 \text{ per hour} \]

Model:

\[ y(t) = 1000e^{0.693t} \]

Find when $y = 50000$:

\[ 50000 = 1000e^{0.693t} \]

\[ 50 = e^{0.693t} \]

\[ \ln 50 = 0.693t \]

\[ t = \frac{\ln 50}{0.693} \approx 5.64 \text{ hours} \]

Application 2: Radioactive Decay

Problem: Carbon-14 has a half-life of 5730 years. If a sample starts with 100 grams, how much remains after 10,000 years?

Find k from half-life:

\[ k = -\frac{\ln 2}{5730} \approx -0.000121 \text{ per year} \]

Model:

\[ y(t) = 100e^{-0.000121t} \]

Find $y(10000)$:

\[ y(10000) = 100e^{-0.000121(10000)} \approx 29.4 \text{ grams} \]

Application 3: Compound Interest (Continuous)

Problem: $5000 is invested at 6% annual interest compounded continuously. How much after 10 years?

Given:

$y_0 = 5000$ (principal)
$k = 0.06$ (6% = 0.06)
$t = 10$ years

Calculate:

\[ y(10) = 5000e^{0.06(10)} = 5000e^{0.6} \approx 9110.59 \]

Answer: $9,110.59

📋 Problem-Solving Strategy

Step-by-Step Approach

Systematic Method:

Identify the model: Recognize it's exponential ($\frac{dy}{dt} = ky$)
Write general form: $y(t) = y_0 e^{kt}$
Find $y_0$: Initial value (often given directly)
Find $k$: Use given information:
- Data point: solve $y_1 = y_0 e^{kt_1}$ for $k$
- Half-life: $k = -\frac{\ln 2}{t_{1/2}}$
- Doubling time: $k = \frac{\ln 2}{t_{\text{double}}}$
- Rate: $k$ = rate (as decimal)
Write specific model: Substitute $y_0$ and $k$
Answer the question: Evaluate or solve as needed
Check reasonableness: Does answer make sense?

📊 Essential Formulas Summary

Complete Formula Reference

Exponential Model Formulas
Formula	Use
$\frac{dy}{dt} = ky$	Differential equation (the model)
$y(t) = y_0 e^{kt}$	General solution
$k = \frac{1}{t_1}\ln\left(\frac{y_1}{y_0}\right)$	Finding $k$ from data point
$t_{1/2} = \frac{\ln 2}{\|k\|}$	Half-life from $k$
$k = -\frac{\ln 2}{t_{1/2}}$	Finding $k$ from half-life
$t_{\text{double}} = \frac{\ln 2}{k}$	Doubling time from $k$
$k = \frac{\ln 2}{t_{\text{double}}}$	Finding $k$ from doubling time

💡 Essential Tips & Strategies

✅ Success Strategies:

Identify the type: Growth ($k > 0$) or decay ($k < 0$)?
Units matter: Keep track of time units (hours, years, etc.)
Initial value at $t = 0$: Simplifies to $y(0) = y_0$
$\ln 2 \approx 0.693$: Useful for half-life/doubling problems
Check your k: Positive for growth, negative for decay
When solving for t: Take natural log of both sides
Calculator efficiency: Store intermediate values

🔥 Common Shortcuts:

Quick half-life check: After $n$ half-lives, $y = y_0 \left(\frac{1}{2}\right)^n$
Quick doubling check: After $n$ doublings, $y = y_0 (2)^n$
Continuous vs discrete: Continuous uses $e^{kt}$, discrete uses $(1+r)^t$
Percentage to decimal: 5% = 0.05
When $y = 2y_0$: $t = \frac{\ln 2}{k}$ (doubling time)

❌ Common Mistakes to Avoid

Mistake 1: Using wrong sign for $k$ (decay should be negative)
Mistake 2: Confusing discrete and continuous models
Mistake 3: Not converting percentages to decimals
Mistake 4: Using wrong time units (mixing years and days)
Mistake 5: Forgetting initial condition ($y_0$ must be at $t = 0$)
Mistake 6: Arithmetic errors with logarithms
Mistake 7: Not checking if answer makes sense
Mistake 8: Using $e^{-kt}$ when problem says growth
Mistake 9: Confusing half-life formula with doubling time
Mistake 10: Not labeling units in final answer

📝 Practice Problems

Solve these problems:

A population grows from 500 to 2000 in 4 years. Find the exponential model.
A substance has a half-life of 20 days. How long until 25% remains?
Money doubles in 12 years with continuous compounding. What's the annual rate?
If $\frac{dy}{dt} = 0.08y$ and $y(0) = 100$, find $y(10)$.

Answers:

$y(t) = 500e^{0.347t}$ where $k = \frac{\ln 4}{4}$
40 days (two half-lives)
5.78% per year (k = ln 2 / 12 ≈ 0.0578)
$y(10) = 100e^{0.8} \approx 222.55$

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Identify the model: State that it's exponential growth/decay
Write the DE: $\frac{dy}{dt} = ky$ (if asked)
Show finding k: Don't just state it, show calculation
Write complete model: $y(t) = y_0 e^{kt}$ with values
Show work for final answer: All substitutions and calculations
Include units: In context problems
Interpret answer: What does it mean in context?

💯 Exam Strategy:

Read problem carefully—identify what type (growth/decay)
Write down what you know: $y_0$, data points, half-life, etc.
Find $k$ using appropriate method
Write specific model with numerical values
Answer the question asked (don't stop at finding the model)
Check: Does your answer make sense?
Include units and interpret in context

⚡ Quick Reference Guide

EXPONENTIAL MODEL ESSENTIALS

The Model:

\[ \frac{dy}{dt} = ky \quad \Rightarrow \quad y(t) = y_0 e^{kt} \]

Key Concepts:

$k > 0$: Growth (population, interest)
$k < 0$: Decay (radioactive, cooling)
Half-life: $t_{1/2} = \frac{\ln 2}{|k|}$
Doubling: $t_{\text{double}} = \frac{\ln 2}{k}$
Finding k: $k = \frac{1}{t_1}\ln\left(\frac{y_1}{y_0}\right)$

Master Exponential Models! The differential equation $\frac{dy}{dt} = ky$ models exponential growth ($k > 0$) and decay ($k < 0$). The solution is $y(t) = y_0 e^{kt}$ where $y_0$ is the initial value at $t = 0$. To solve problems: (1) identify the model type, (2) find $y_0$ from initial conditions, (3) find $k$ from given information using $k = \frac{1}{t_1}\ln\left(\frac{y_1}{y_0}\right)$, half-life $k = -\frac{\ln 2}{t_{1/2}}$, doubling time $k = \frac{\ln 2}{t_{\text{double}}}$, or rate directly, (4) write complete model, (5) answer the question. Common applications: population growth, radioactive decay, continuous compound interest, Newton's Law of Cooling. Remember $\ln 2 \approx 0.693$. Always check sign of $k$—positive for growth, negative for decay. Include units and interpret answers in context. After one half-life, 50% remains; after two half-lives, 25% remains. This is THE most tested application topic on AP® exams—appears every year! 🎯✨

Formula	Use
\(\frac{dy}{dt} = ky\)	Differential equation (the model)
\(y(t) = y_0 e^{kt}\)	General solution
\(k = \frac{1}{t_1}\ln\left(\frac{y_1}{y_0}\right)\)	Finding \(k\) from data point
\(t_{1/2} = \frac{\ln 2}{\|k\|}\)	Half-life from \(k\)
\(k = -\frac{\ln 2}{t_{1/2}}\)	Finding \(k\) from half-life
\(t_{\text{double}} = \frac{\ln 2}{k}\)	Doubling time from \(k\)
\(k = \frac{\ln 2}{t_{\text{double}}}\)	Finding \(k\) from doubling time