Exponential and Logarithmic Functions: Applications of Exponential Decay

Exponential and Logarithmic Functions

Introduction

This section explores the mathematical modeling of exponential decay, a fundamental concept in Business Calculus. Exponential decay describes processes where quantities decrease at rates proportional to their current value, with applications in physical sciences, life sciences, and finance.

Exponential Decay

Definition and Differential Equation

• Exponential decay occurs when a quantity decreases at a rate proportional to its current value.

• The process is modeled by the differential equation: $\frac{dP}{dt} = -kP$, where $k > 0$

• General solution: $P(t) = P_0 e^{-kt}$ where $P_0$ is the initial amount at time $t = 0$.

Graphical Representation

• For exponential growth: $P = P_0 e^{kt}$, the function increases over time.

• For exponential decay: $P = P_0 e^{-kt}$, the function decreases over time.

• Both models are illustrated by curves that either rise (growth) or fall (decay) from the initial value $P_0$.

Half-Life and Decay Rate

Relationship Between Decay Rate and Half-Life

• Half-life (T): The time required for a quantity to decrease to half its initial value.

• The decay rate (k) and half-life (T) are related by: $T = \frac{\ln 2}{k}$ or equivalently, $k = \frac{\ln 2}{T}$

Table: Relationship Between Decay Rate and Half-Life

Quantity	Formula
Half-life (T)	$T = \frac{\ln 2}{k}$
Decay rate (k)	$k = \frac{\ln 2}{T}$

Applications of Exponential Decay

Physical Science: Half-life Calculation

• Example: Plutonium-239 has a decay rate of 0.0028% per year. Its half-life is calculated as: $T = \frac{\ln 2}{0.000028} \approx 24,755$ years

• Practice: For cesium-137 with a decay rate of 2.3% per year: $T = \frac{\ln 2}{0.023} \approx 30.1$ years

• Practice: For barium-140 with a half-life of 13 days: $k = \frac{\ln 2}{13} \approx 0.053$ (or 5.3% per day)

Life Science: Carbon Dating

• Carbon-14 has a half-life of 5730 years. The decay rate is: $k = \frac{\ln 2}{5730} \approx 0.00012097$ (or 0.012097% per year)

• Example: If a linen wrapping from the Dead Sea Scrolls has lost 22.3% of its carbon-14, the remaining fraction is 77.7% ($0.777$ of the original amount).

• To find the age ($t$): $0.777 = e^{-0.00012097 t}$ $\ln 0.777 = -0.00012097 t$ $t = \frac{\ln 0.777}{-0.00012097} \approx 2086$ years

• Practice: If a skeleton has lost 60% of its carbon-14 (remaining 40%): $0.4 = e^{-0.00012097 t}$ $t = \frac{\ln 0.4}{-0.00012097} \approx 7574$ years

Business Application: Present Value

• Present value is the amount that must be invested now to reach a desired future value, given continuous compounding.

• The formula for present value is: $P_0 = P e^{-kt}$ where $P$ is the future value, $k$ is the interest rate, and $t$ is the time in years.

• Example: To have $10,000 in 20 years at 4% interest compounded continuously: $10,000 = P_0 e^{0.04 \times 20}$ $P_0 = \frac{10,000}{e^{0.8}} \approx 4,493.29$

• Practice: At 6% interest: $P_0 = \frac{10,000}{e^{1.2}} \approx 3,011.94$

Newton's Law of Cooling

• Describes how the temperature $T$ of an object changes over time as it approaches the ambient temperature $C$.

• Modeled by: $\frac{dT}{dt} = -k(T - C)$, where $k > 0$

• General solution: $T(t) = a e^{-kt} + C$ where $a$ is a constant determined by initial conditions.

Forensics: Determining Time of Death

• Example: A body is found at noon with a temperature of 94.6°F. After 1 hour, the temperature is 93.4°F. The room temperature is 70°F.

• Using Newton's Law of Cooling: $T(t) = a e^{-kt} + 70$ At $t = 0$, $T(0) = 94.6$: $94.6 = a + 70 \implies a = 24.6$ At $t = 1$, $T(1) = 93.4$: $93.4 = 24.6 e^{-k} + 70$ $23.4 = 24.6 e^{-k}$ $\frac{23.4}{24.6} = e^{-k}$ $k = -\ln\left(\frac{23.4}{24.6}\right) \approx 0.05$

• Assuming normal body temperature at death (98.6°F), solve for $t$: $98.6 = 24.6 e^{-0.05 t} + 70$ $28.6 = 24.6 e^{-0.05 t}$ $\frac{28.6}{24.6} = e^{-0.05 t}$ $\ln\left(\frac{28.6}{24.6}\right) = -0.05 t$ $t = -\frac{\ln\left(\frac{28.6}{24.6}\right)}{0.05} \approx 3$ hours

• Conclusion: The time of death was approximately 3 hours before noon, or around 9:00 a.m.