7 6 Practice Growth And Decay

7.6 Practice: Growth and Decay Problems

This article delves into the practical applications of exponential growth and decay models, focusing on problem-solving strategies and real-world examples. We'll explore various scenarios, from population growth and compound interest to radioactive decay and drug absorption, equipping you with the tools to tackle a wide range of problems involving these fundamental concepts. This detailed guide will cover several key aspects of growth and decay, offering a comprehensive understanding for students and professionals alike.

Understanding Exponential Growth and Decay

Exponential growth and decay describe situations where a quantity changes at a rate proportional to its current value. The general formula for both is:

A = A₀ * e^(kt)

Where:

• A is the final amount
• A₀ is the initial amount
• k is the growth/decay constant (positive for growth, negative for decay)
• t is time
• e is the base of the natural logarithm (approximately 2.71828)

Growth vs. Decay: Identifying the Key Differences

The crucial distinction lies in the sign of the constant k:

• Growth: k is positive. The quantity increases exponentially over time. Examples include population growth, compound interest, and the spread of certain viral infections.

• Decay: k is negative. The quantity decreases exponentially over time. Examples include radioactive decay, drug elimination from the body, and the depreciation of assets.

Problem-Solving Strategies: A Step-by-Step Approach

Tackling growth and decay problems effectively involves a structured approach:

1. Identify the type of problem: Is it growth or decay? This determines the sign of k.

2. Identify the known variables: Determine the values of A₀, A, t, or k, based on the problem statement.

3. Choose the appropriate formula: Use the general formula A = A₀ * e^(kt) and adapt it based on the known and unknown variables.

4. Solve for the unknown variable: Use algebraic manipulation and logarithmic properties to solve for the required value. Remember, to solve for an exponent, use logarithms (often natural logarithms, ln).

5. Interpret the solution: Check if your answer is realistic and makes sense within the context of the problem.

Real-World Applications and Examples:

Let's explore diverse scenarios to illustrate the application of exponential growth and decay models.

1. Population Growth

Example: A city's population grows at a rate of 3% per year. If the current population is 100,000, what will the population be in 10 years?

• Type: Growth
• A₀ = 100,000
• k = 0.03
• t = 10
• A = ?

Solution:

A = A₀ * e^(kt) = 100,000 * e^(0.03 * 10) ≈ 134,986

Therefore, the population will be approximately 134,986 in 10 years.

2. Compound Interest

Example: You invest $1,000 in an account that pays 5% annual interest, compounded continuously. How much will you have after 5 years?

• Type: Growth
• A₀ = 1,000
• k = 0.05
• t = 5
• A = ?

Solution:

A = A₀ * e^(kt) = 1,000 * e^(0.05 * 5) ≈ $1,284.03

Therefore, you will have approximately $1,284.03 after 5 years.

3. Radioactive Decay

Example: A radioactive substance has a half-life of 10 years. If you start with 100 grams, how much will remain after 25 years?

• Type: Decay
• A₀ = 100
• t = 25
• Half-life = 10 years (We need to find k first)

Solution:

First, find k using the half-life:

0.5 = e^(k * 10) ln(0.5) = 10k k = ln(0.5) / 10 ≈ -0.0693

Now, calculate the remaining amount:

A = A₀ * e^(kt) = 100 * e^(-0.0693 * 25) ≈ 9.77 grams

Therefore, approximately 9.77 grams will remain after 25 years.

4. Drug Absorption and Elimination

Example: A patient is given a 200mg dose of a drug that is eliminated from the body at a rate of 15% per hour. How much of the drug remains in the body after 4 hours?

• Type: Decay
• A₀ = 200
• k = -0.15 (15% decay per hour)
• t = 4
• A = ?

Solution:

A = A₀ * e^(kt) = 200 * e^(-0.15 * 4) ≈ 100.52 mg

Therefore, approximately 100.52 mg of the drug remains after 4 hours.

Advanced Concepts and Applications:

Beyond the basic examples, exponential growth and decay models find applications in diverse fields, requiring a deeper understanding of related concepts.

1. Doubling Time and Half-Life

• Doubling Time: The time it takes for a quantity undergoing exponential growth to double its initial value. It's calculated using: t₂ = ln(2) / k

• Half-Life: The time it takes for a quantity undergoing exponential decay to reduce to half its initial value. It's calculated using: t½ = ln(0.5) / k

2. Differential Equations

The core principle behind exponential growth and decay is elegantly expressed using differential equations:

dA/dt = kA

This equation states that the rate of change of the quantity (dA/dt) is directly proportional to the current amount (A). Solving this differential equation yields the familiar formula A = A₀ * e^(kt).

3. Logistic Growth

While simple exponential growth models are useful in many scenarios, they are limited in representing real-world situations where growth is constrained by factors such as resource availability. Logistic growth models incorporate a carrying capacity, representing the maximum sustainable population or quantity. The formula is more complex and involves an S-shaped curve.

4. Applications in Finance

Beyond simple compound interest, exponential growth and decay are crucial in understanding more complex financial models like discounted cash flow analysis, determining bond yields, and valuing options.

Conclusion:

Exponential growth and decay are fundamental mathematical concepts with far-reaching applications in various fields. By understanding the underlying principles, mastering the problem-solving strategies, and exploring the diverse real-world scenarios presented in this guide, you will gain a robust foundation in this important area of mathematics. Remember to always carefully identify the variables, choose the appropriate formula, and interpret your results within the context of the specific problem. Continuous practice and exposure to different problem types will significantly enhance your ability to confidently apply these powerful models.