Solving Exponential Growth and Decay Problems

Introduction

Key Concepts

Understanding Exponential Functions

Exponential functions are mathematical models that describe processes where the rate of change is proportional to the current value. They are expressed in the form:

$$f(t) = f_0 e^{kt}$$

where:

• f(t) is the quantity at time t,
• f₀ is the initial quantity,
• k is the growth (k > 0) or decay (k < 0) constant, and
• e is the base of the natural logarithm, approximately equal to 2.71828.

This equation forms the basis for modeling various phenomena where changes occur continuously and proportionally to the amount present.

Exponential Growth

Exponential growth occurs when the quantity increases at a rate directly proportional to its current value. This type of growth is common in populations, investments with continuous compounding interest, and certain biological processes.

The general solution for exponential growth is:

$$f(t) = f_0 e^{kt}$$

where k is positive. For example, if a population of bacteria doubles every hour, the growth can be modeled using this equation by determining the appropriate growth constant.

Example: Suppose a population of 500 bacteria grows at a rate of 20% per hour. The population after t hours is:

$$f(t) = 500 e^{0.20t}$$

Exponential Decay

Exponential decay describes a process where the quantity decreases at a rate proportional to its current value. This is observed in radioactive decay, cooling of objects, and depreciation of assets.

The general solution for exponential decay is:

$$f(t) = f_0 e^{-kt}$$

where k is positive. For instance, the half-life of a radioactive substance is the time required for half of the substance to decay and can be modeled using this equation.

Example: If a 1000-gram sample of a radioactive material has a half-life of 3 years, the remaining mass after t years is:

First, find the decay constant k:

$$\frac{1}{2} = e^{-3k} \implies \ln\left(\frac{1}{2}\right) = -3k \implies k = \frac{\ln 2}{3}$$

Thus, the decay model is:

$$f(t) = 1000 e^{-\left(\frac{\ln 2}{3}\right)t}$$

Solve Differential Equations

Exponential growth and decay problems are often framed as differential equations. A simple first-order linear differential equation representing growth or decay is:

$$\frac{df}{dt} = kf$$

where f(t) is the function describing the quantity over time, and k is the constant of proportionality.

To solve this differential equation, separate the variables and integrate:

$$\int \frac{1}{f} df = \int k dt$$

This yields:

$$\ln |f| = kt + C$$

Exponentiating both sides gives:

$$f(t) = Ce^{kt}$$

Where C is the constant of integration determined by initial conditions. If f(0) = f₀, then:

$$C = f₀$$

Thus, the solution is:

$$f(t) = f_0 e^{kt}$$

Applications of Exponential Growth and Decay

Exponential functions apply to various fields. In biology, they model population growth under unrestricted conditions. In finance, they describe compound interest where earnings accumulate over time. Physics utilizes exponential decay to explain radioactive substance decay and cooling processes.

Understanding these applications helps in predicting future states of systems and making informed decisions based on mathematical models.

Logistic Growth: A Contrast

While exponential growth assumes unlimited resources leading to indefinite growth, logistic growth introduces the concept of carrying capacity K, representing the maximum population size an environment can sustain. The logistic growth model is:

$$f(t) = \frac{K}{1 + \left(\frac{K - f_0}{f_0}\right)e^{-kt}}$$

This model shows that as the population grows, the growth rate decreases, eventually stabilizing at the carrying capacity.

Example: If a town has a carrying capacity of 10,000 people and currently has 2,000 residents with a growth rate of 0.3, the population after t years is:

$$f(t) = \frac{10000}{1 + \left(\frac{10000 - 2000}{2000}\right)e^{-0.3t}}$$

Determining Growth and Decay Constants

Calculating the constants in exponential models is essential for accurate predictions. Given data points, methods such as the least squares approximation or utilizing known points (e.g., half-life in decay) help find the appropriate k value.

Sample Problem: A population grows from 500 to 800 in 5 years. Determine the growth constant k.

Using the formula:

$$800 = 500 e^{5k} \implies \frac{800}{500} = e^{5k} \implies \ln\left(\frac{8}{5}\right) = 5k \implies k = \frac{\ln\left(\frac{8}{5}\right)}{5}$$

Calculating:

$$k \approx \frac{0.4700}{5} \approx 0.0940$$

Thus, the growth constant is approximately 0.0940.

Half-Life in Exponential Decay

The half-life is the time required for a quantity to reduce to half its initial value. It is a commonly used parameter in exponential decay problems, especially in radioactive decay.

The relationship between half-life T and the decay constant k is:

$$T = \frac{\ln 2}{k}$$

Rearranging gives:

$$k = \frac{\ln 2}{T}$$

Example: If a radioactive substance has a half-life of 5 years, the decay constant is:

$$k = \frac{\ln 2}{5} \approx 0.1386$$

The decay model is:

$$f(t) = f_0 e^{-0.1386t}$$

Continuous vs. Discrete Growth and Decay

Exponential functions typically represent continuous growth or decay. However, in some scenarios, growth or decay occurs in discrete intervals, such as annual population counts or yearly financial returns. The continuous model provides a smoother and more general approach, whereas discrete models may align more closely with specific real-world data points.

Choosing between continuous and discrete models depends on the nature of the problem and the required precision.

Solving Exponential Equations

Solving exponential equations involves finding the time t when a certain condition is met, such as reaching a specific population or decaying to a particular amount. This typically requires taking natural logarithms to linearize the equation.

Example: Determine when a 1000-gram sample will decay to 250 grams with a decay constant of 0.1.

Starting with:

$$250 = 1000 e^{-0.1t}$$

Divide both sides by 1000:

$$0.25 = e^{-0.1t}$$

Take the natural logarithm:

$$\ln(0.25) = -0.1t \implies t = \frac{\ln(0.25)}{-0.1}$$

Calculating:

$$t \approx \frac{-1.3863}{-0.1} \approx 13.863 \text{ years}$$

Thus, it takes approximately 13.863 years for the sample to decay to 250 grams.

Applications in Financial Mathematics

In finance, exponential growth models are vital for understanding compound interest, investment growth, and loan amortization. The continuous compound interest formula is:

$$A = P e^{rt}$$

where:

• A is the amount of money accumulated after t years, including interest.
• P is the principal amount.
• r is the annual interest rate.

Example: Investing $1000 at an annual interest rate of 5% compounded continuously for 10 years:

$$A = 1000 e^{0.05 \times 10} = 1000 e^{0.5} \approx 1000 \times 1.6487 \approx 1648.72$$

Thus, the investment grows to approximately $1648.72 after 10 years.

Modeling Radioactive Decay

Radioactive decay is a natural example of exponential decay. The number of undecayed nuclei decreases over time, governed by the decay constant and half-life.

The decay model is:

$$N(t) = N_0 e^{-kt}$$

where:

• N(t) is the number of undecayed nuclei at time t,
• N₀ is the initial number of nuclei, and
• k is the decay constant.

This model assists in determining the age of objects through radiometric dating based on the remaining quantity of radioactive isotopes.

Continuous Growth in Biology

In biology, exponential growth models populations under ideal conditions with unlimited resources. The logistic growth model, which incorporates a carrying capacity, provides a more realistic scenario where resources become limited as the population increases.

Exponential growth can be expressed as:

$$P(t) = P_0 e^{rt}$$

where P(t) is the population at time t, P₀ is the initial population, and r is the growth rate.

However, in reality, factors like competition, resource depletion, and environmental changes often lead to deviations from pure exponential growth.

Comparison Table

Summary and Key Takeaways