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Exponential Models

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Exponential Models - Complete Interactive Lesson

Part 1: Exponential Growth & Decay

Exponential Models

Part 1 of 7 — Exponential Growth and Decay

In This Topic

Part	Topic
1	Exponential Growth & Decay
2	Newton’s Law of Cooling
3	Compound Interest & Continuous Growth
4	Derivatives & Integrals of Exponentials
5	Logistic Growth
6	Problem-Solving Workshop
7	Comprehensive Assessment

The Fundamental Differential Equation

$\boxed{\frac{dy}{dt} = ky \implies y(t) = y_0 e^{kt}}$

Parameter	Meaning
$y_0$	Initial value $y(0)$
$k > 0$

Key Fact: "Rate proportional to amount" always means $\frac{dy}{dt} = ky$ . This is the most common DE on the AP exam.

Finding the Growth Constant $k$

Step-by-step from two data points:

Step	Action	Example
1	Write model	$y = y_0 e^{kt}$

Exponential Growth & Decay 🎯

Classify each scenario. 🔍

Solve for the growth constant. ✍️

Key Takeaways — Part 1

Concept	Formula
Exponential model	$y = y_0 e^{kt}$
Finding $k$

Part 2: Newton's Law of Cooling

Exponential Models

Part 2 of 7 — Newton’s Law of Cooling

The Differential Equation

$\boxed{\frac{dT}{dt} = k(T - T_s)} \quad \text{where } k < 0$

Part 3: Compound Interest & Continuous Growth

Exponential Models

Part 3 of 7 — Compound Interest & Continuous Growth

Compound Interest Formula

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

Part 4: Derivatives & Integrals of Exponentials

Exponential Models

Part 4 of 7 — Derivatives & Integrals of Exponentials

Derivative Rules

$\boxed{\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[e^{kx}] = ke^{kx}}$

Part 5: Logistic Growth

Exponential Models

Part 5 of 7 — Logistic Growth

The Logistic Differential Equation

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

Part 6: Practice Workshop

Exponential Models

Part 6 of 7 — Problem-Solving Workshop

Model Selection Guide

Verbal Clue	Model	Equation
"Rate proportional to amount"	Exponential	$dy/dt = ky$
"Approaches a limiting value"	Logistic	$dP/dt = kP(1-P/L)$

Part 7: Final Assessment

Exponential Models

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Model	DE	Solution
Exponential growth	$dy/dt = ky$ , $k>0$

Doubling Time and Half-Life

$\boxed{t_{\text{double}} = \frac{\ln 2}{k}} \qquad \boxed{t_{\text{half}} = \frac{\ln 2}{|k|}}$

Quantity	Growth ( $k>0$ )	Decay ( $k<0$ )
Doubling time	$\frac{\ln 2}{k}$	N/A
Half-life	N/A	$\frac{\ln 2}{
Triple time	$\frac{\ln 3}{k}$	N/A
$n$ half-lives	N/A	$y_0 \cdot (\tfrac{1}{2})^n$ remains

AP Tip: The Rule of 70: doubling time $\approx \frac{70}{\text{percent rate}}$ . E.g., 7% growth $\Rightarrow$ doubles in $\approx 10$ years.

Up Next: Part 2 — Newton’s Law of Cooling.

$\boxed{T(t) = T_s + (T_0 - T_s)e^{kt}}$

Variable	Meaning
$T(t)$	Temperature at time $t$
$T_s$	Surrounding (ambient) temperature
$T_0$	Initial temperature $T(0)$
$k$	Cooling constant ( $k < 0$ )
$T_0 - T_s$	Initial temperature difference

Key Fact: The temperature difference $(T - T_s)$ decays exponentially. The object never overshoots the ambient temperature.

Worked Example

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

Step	Computation
Set up	$T(t) = 70 + 130e^{kt}$
Use data point	$150 = 70 + 130e^{10k}$
Solve for $e^{10k}$	$e^{10k} = \frac{80}{130} = \frac{8}{13}$
Find $k$	$k = \frac{1}{10}\ln\frac{8}{13} \approx -0.0486$
Final model	$T(t) = 70 + 130e^{-0.0486t}$

Follow-up: When does the coffee reach $100°F$ ?

$100 = 70 + 130e^{kt}$

$e^{kt} = \frac{30}{130} = \frac{3}{13}$

$t = \frac{\ln(3/13)}{k} = \frac{\ln(3/13)}{\frac{1}{10}\ln(8/13)} \approx 30.2$ min

AP Tip: Always identify $T_s$ first. Then $T_0 - T_s$ is the initial difference. The model is always .

Newton’s Cooling 🎯

Analyze cooling scenarios. 🔍

Apply Newton’s Law. ✍️

Key Takeaways — Part 2

Concept	Formula
Newton’s cooling DE	$\frac{dT}{dt} = k(T - T_s)$
Solution	$T = T_s + (T_0 - T_s)e^{kt}$
Long-term behavior	$T \to T_s$
Warming variant	Same formula when $T_0 < T_s$

Up Next: Part 3 — Compound Interest & Continuous Growth.

Variable	Meaning
$P$	Principal (initial investment)
$r$	Annual interest rate (decimal)
$n$	Compounding periods per year
$t$	Time in years
$A$	Amount after $t$ years

Continuous Compounding

$\boxed{A = Pe^{rt}}$

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

Key Fact: Continuous compounding gives the maximum possible return for a given rate. It arises naturally from $\frac{dA}{dt} = rA$ .

Compounding Frequency Comparison

$1000 at 6% for 10 years:

Frequency	$n$	Formula	Amount
Annual	$1$	$1000(1.06)^{10}$	$\$ 1790.85$
Quarterly	$4$	$1000(1.015)^{40}$	$\$ 1814.02$
Monthly	$12$	$1000(1.005)^{120}$	$\$ 1819.40$
Daily	$365$	$1000(1+.06/365)^{3650}$	$\$ 1822.03$
Continuous	$\infty$	$1000e^{0.6}$	$\$ 1822.12$

Key Formulas

Question	Formula
Doubling time (continuous)	$t = \frac{\ln 2}{r}$
Time to reach amount $A$

AP Tip: On the AP exam, continuous compounding ( $A = Pe^{rt}$ ) appears far more often than discrete compounding.

Compound Interest 🎯

Interest concepts. 🔍

Solve for time. ✍️

Key Takeaways — Part 3

Concept	Formula
Discrete compounding	$A = P(1+r/n)^{nt}$
Continuous compounding	$A = Pe^{rt}$
Doubling time	$\frac{\ln 2}{r}$
Rule of 70	Doubling time $\approx 70/\text{percent rate}$

Up Next: Part 4 — Derivatives & Integrals of Exponentials.

$\boxed{\frac{d}{dx}[a^x] = a^x \ln a \qquad \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \cdot (\ln a) \cdot g'(x)}$

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

Complete Reference Table

Function	Derivative	Integral
$e^x$	$e^x$	$e^x + C$
$e^{kx}$	$ke^{kx}$	$\frac{1}{k}e^{kx} + C$
$e^{g(x)}$	$e^{g(x)} \cdot g'(x)$
$a^x$	$a^x \ln a$	$\frac{a^x}{\ln a} + C$
$\ln x$	$\frac{1}{x}$	$x\ln x - x + C$

Key Fact: $e^x$ is the only function that is its own derivative AND its own antiderivative.

Worked Examples

Derivatives:

Function	Chain Rule Application	Result
$e^{3x^2}$	$e^{3x^2} \cdot 6x$	$6xe^{3x^2}$
$2^{\sin x}$	$2^{\sin x} \cdot \ln 2 \cdot \cos x$
$e^{x^2+1}$	$e^{x^2+1} \cdot 2x$

Integrals:

Integral	Method	Result
$\int e^{-3x}dx$	Direct: $k=-3$	$-$

Exponential Calculus 🎯

Match each integral. 🔍

Evaluate the integral. ✍️

Key Takeaways — Part 4

Rule	Formula
$\frac{d}{dx}[e^{kx}]$	$ke^{kx}$
$\int e^{kx}\,dx$	$\frac{1}{k}e^{kx}+C$
$\frac{d}{dx}[a^x]$	$a^x \ln a$
$\int a^x\,dx$	$\frac{a^x}{\ln a}+C$

Up Next: Part 5 — Logistic Growth.

Variable	Meaning
$P$	Population at time $t$
$k$	Growth rate constant
$L$	Carrying capacity
$kP$	Growth term
$(1-P/L)$	Limiting factor

Condition	Growth Rate	Behavior
$P \ll L$	$\approx kP$	Nearly exponential
$P = L/2$	Maximum: $kL/4$	Inflection point
$P = L$	Zero	Equilibrium
$P > L$	Negative	Population decreases toward $L$
$P = 0$	Zero	No population

Key Fact: The logistic model is the most realistic population model on the AP exam. It accounts for limited resources.

The Logistic Curve (S-Curve)

Phase	$P$ range	Shape	Description
Phase 1	$0 < P < L/2$	Concave up	Accelerating growth
Inflection	$P = L/2$	Changes concavity	Fastest growth rate
Phase 2	$L/2 < P < L$	Concave down	Decelerating growth
Equilibrium	$P = L$	Horizontal	Stable steady state

The Solution (for reference)

$P(t) = \frac{L}{1 + Ae^{-kt}} \quad \text{where } A = \frac{L - P_0}{P_0}$

Maximum Growth Rate

$\boxed{\left(\frac{dP}{dt}\right)_{\max} = \frac{kL}{4} \quad \text{at } P = \frac{L}{2}}$

AP Tip: You do NOT need to memorize the logistic solution formula. AP questions focus on the DE, carrying capacity, inflection point, and qualitative behavior.

Logistic Growth 🎯

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

Logistic analysis. 🔍

Compute the maximum growth rate. ✍️

Key Takeaways — Part 5

Concept	Formula
Logistic DE	$\frac{dP}{dt} = kP(1-P/L)$
Carrying capacity	$L$
Fastest growth at	$P = L/2$
Maximum rate	$kL/4$
Long-term behavior	$P \to L$

Up Next: Part 6 — Problem-Solving Workshop.

d P / d t = k P (1 - P / L)

Key Fact: Read the problem carefully for keywords. The model type determines the entire solution strategy.

AP-Style Worked Problems

Problem 1: Carbon-14 has a half-life of 5730 years. A sample has 30% of its original C-14. How old is it?

Step	Work
Model	$y = y_0 e^{kt}$
Find $k$	$\frac{1}{2} = e^{5730k}$ , $k = -\frac{\ln 2}{5730}$
Use 30%	$0.30 = e^{kt}$
Solve	$t = \frac{\ln(0.30)}{k} = \frac{5730 \ln(0.30)}{-\ln 2} \approx 9953$ years

Problem 2: A lake has 1000 fish. The population follows $\frac{dP}{dt} = 0.2P(1-P/5000)$ .

Question	Answer
Carrying capacity?	$L = 5000$
Currently growing?	Yes: $P = 1000 < 5000$
Current growth rate?	$0.2 (1000) (1 - 1000 / 5000$ fish/year

Mixed Practice 🎯

Identify the model. 🔍

Apply the right model. ✍️

Key Takeaways — Part 6

Keyword	Model
Proportional to amount	Exponential
Approaches limit	Logistic
Proportional to difference	Newton’s cooling
Doubles/halves	Exponential with $2^{t/n}$ or $(1/2)^{t/n}$

Up Next: Part 7 — Comprehensive Assessment.

Mistake	Correction
Using wrong sign for $k$	Decay: $k < 0$ . Growth: $k > 0$ .
Forgetting $T_s$ in Newton’s Law	$T = T_s + (\text{diff})e^{kt}$ , not $T = T_{0} e^{k t}$
Max logistic rate at $P=L$	Max rate at $P = L/2$ , NOT at $L$
Confusing half-life formula	$t_{1/2} = \frac{\ln 2}{
Wrong doubling formula	$t_{double} = \frac{\ln 2}{k}$ , not
Not checking units	Rate units = (quantity units)/(time units)

Quiz — Growth & Decay 🎯

Quiz — Cooling & Logistic 🎯

Final classification. 🔍

Final Challenge ✍️

Exponential Models — Complete!

You’ve mastered:

Part	Topic
1	Exponential growth & decay
2	Newton’s Law of Cooling
3	Compound interest & continuous growth
4	Derivatives & integrals of exponentials
5	Logistic growth
6	Problem-solving workshop
7	Comprehensive assessment

You’re ready for AP-level exponential model problems!

T_s + (\text{difference})e^{kt}

- \frac{1}{3} e^{- 3 x} + C

0.2 (1000) (1 - 1000/5000) = 160

Exponential Models

Doubling Time and Half-Life

The Solution

Worked Example

Key Takeaways — Part 2

Continuous Compounding

Compounding Frequency Comparison

Key Formulas

Key Takeaways — Part 3

Integration Rules

Complete Reference Table

Worked Examples

Key Takeaways — Part 4

Key Behaviors

The Logistic Curve (S-Curve)

The Solution (for reference)

Maximum Growth Rate

Key Takeaways — Part 5

AP-Style Worked Problems

Key Takeaways — Part 6

Top AP Mistakes

Exponential Models — Complete!

Exponential Models

Exponential Models - Complete Interactive Lesson

Part 1: Exponential Growth & Decay

Exponential Models

In This Topic

The Fundamental Differential Equation

Finding the Growth Constant kkk

Key Takeaways — Part 1

Part 2: Newton's Law of Cooling

Exponential Models

The Differential Equation

Part 3: Compound Interest & Continuous Growth

Exponential Models

Compound Interest Formula

Part 4: Derivatives & Integrals of Exponentials

Exponential Models

Derivative Rules

Part 5: Logistic Growth

Exponential Models

The Logistic Differential Equation

Part 6: Practice Workshop

Exponential Models

Model Selection Guide

Part 7: Final Assessment

Exponential Models

Complete Formula Reference

Doubling Time and Half-Life

The Solution

Worked Example

Key Takeaways — Part 2

Continuous Compounding

Compounding Frequency Comparison

Key Formulas

Key Takeaways — Part 3

Integration Rules

Complete Reference Table

Worked Examples

Key Takeaways — Part 4

Key Behaviors

The Logistic Curve (S-Curve)

The Solution (for reference)

Maximum Growth Rate

Key Takeaways — Part 5

AP-Style Worked Problems

Key Takeaways — Part 6

Top AP Mistakes

Exponential Models — Complete!

Finding the Growth Constant $k$