Exponential Growth & Decay Word Problems Worksheet With Answers
Table of Contents :
Exponential growth and decay are fundamental concepts in mathematics, particularly in the fields of algebra and calculus. These concepts describe how certain quantities change over time in a multiplicative way, often leading to rapid increases (growth) or decreases (decay). This blog post will cover various types of word problems related to exponential growth and decay, provide a structured worksheet for practice, and offer insights on how to solve these problems effectively. 📈📉
Understanding Exponential Growth and Decay
What is Exponential Growth?
Exponential growth occurs when the increase of a quantity is proportional to its current value. This means that as the quantity grows, it grows at a faster rate. A common example is population growth, where the rate of increase is related to the existing population size.
The general formula for exponential growth is:
[ A(t) = A_0 \times e^{kt} ]
Where:
- ( A(t) ) = the amount at time ( t )
- ( A_0 ) = the initial amount
- ( e ) = Euler's number (approximately equal to 2.718)
- ( k ) = growth constant
- ( t ) = time
What is Exponential Decay?
Exponential decay, on the other hand, describes a situation where a quantity decreases at a rate proportional to its current value. This is often seen in contexts such as radioactive decay or depreciation of assets.
The formula for exponential decay is:
[ A(t) = A_0 \times e^{-kt} ]
Where the variables are defined as above, but the decay constant ( k ) is positive.
Word Problems Involving Exponential Growth and Decay
Let’s dive into some common types of word problems involving exponential growth and decay.
Example Problems
Example 1: Population Growth
Problem: A city has a population of 50,000 people and grows at an annual rate of 4%. What will the population be in 5 years?
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Solution: Using the growth formula, ( A(t) = A_0 \times e^{kt} ):
- Initial population ( A_0 = 50000 )
- Growth rate ( k = 0.04 )
- Time ( t = 5 )
[ A(5) = 50000 \times e^{0.04 \times 5} ]
[ A(5) \approx 50000 \times e^{0.2} \approx 50000 \times 1.2214 \approx 61070 ]
So, the population will be approximately 61,070 in 5 years. 🌆
Example 2: Radioactive Decay
Problem: A radioactive substance has a half-life of 3 years. If you start with 80 grams, how much of the substance remains after 9 years?
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Solution: To find the remaining amount, use the decay formula:
- Initial amount ( A_0 = 80 )
- Half-life = 3 years, so ( k = \frac{\ln(0.5)}{3} )
- Time ( t = 9 )
Since 9 years is 3 half-lives:
[ A(9) = 80 \times \left(\frac{1}{2}\right)^{\frac{9}{3}} = 80 \times \left(\frac{1}{2}\right)^{3} ]
[ A(9) = 80 \times \frac{1}{8} = 10 \text{ grams} ]
Therefore, 10 grams of the substance remains after 9 years. ☢️
Practice Worksheet
Below is a worksheet with more problems for practice. Try to solve them using the formulas for exponential growth and decay discussed above.
Exponential Growth & Decay Worksheet
| Problem Number | Problem Description |
|---|---|
| 1 | A culture of bacteria doubles every 3 hours. If there are initially 500 bacteria, how many will there be after 12 hours? |
| 2 | A car depreciates in value by 15% each year. If the car's initial value is $20,000, what will it be worth after 5 years? |
| 3 | A town's population is currently 30,000 and grows at a rate of 5% per year. How many people will be living in the town after 10 years? |
| 4 | A substance has a decay rate of 6% per year. If you start with 200 grams, how much of the substance remains after 4 years? |
| 5 | A bank account earns an interest rate of 3% compounded annually. If you deposit $5,000, how much money will be in the account after 5 years? |
Answers to Practice Problems
Here are the answers to the practice problems:
| Problem Number | Answer |
|---|---|
| 1 | 500 × (2^{4} = 500 × 16 = 8000) bacteria after 12 hours. |
| 2 | (20000 × (0.85)^{5} \approx 20000 × 0.4437 \approx 8874) dollars after 5 years. |
| 3 | (30000 × e^{0.05 × 10} \approx 30000 × e^{0.5} \approx 30000 × 1.6487 \approx 49461) people after 10 years. |
| 4 | (200 × (0.94)^{4} \approx 200 × 0.7787 \approx 155.74) grams remain after 4 years. |
| 5 | (5000 × e^{0.03 × 5} \approx 5000 × 1.1618 \approx 5809) dollars after 5 years. |
Important Notes:
"When dealing with exponential growth and decay, always identify the initial amount, the rate of growth or decay, and the time period involved. It helps to convert percentage rates into decimal form (e.g., 4% becomes 0.04) to maintain accuracy in calculations." ✏️
By practicing these types of problems, you'll gain a better understanding of how to apply the concepts of exponential growth and decay in various real-life situations. Happy learning!