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1.7. Solutions (Selected)
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Today, we'll dive into exponential functions, which are crucial for modeling change in populations and values over time. Who can tell me the general form of an exponential function?
Isn't it something like 𝑦 = 𝑎 ⋅ 𝑏𝑥?
Exactly! In this formula, 𝑎 represents the initial value, and 𝑏 is the growth factor. Remember, if 𝑏 > 1, it's growth; if 0 < 𝑏 < 1, it's decay. A mnemonic to remember this is 'Growth is greater than 1, decay is less.'
Why is it essential to understand the difference between growth and decay?
Great question! Understanding this helps us analyze real-world situations, like populations or finances. Now, let’s apply this to a practical example.
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Let’s say we have a population of 1,000 people, which grows at a rate of 4% per year. How would we calculate the population in 10 years?
We would use the growth formula: 𝑦 = 𝑎(1 + 𝑟)𝑡, right?
Right! We have 𝑎 = 1000, 𝑟 = 0.04, and 𝑡 = 10. Can someone calculate 𝑦 for me?
So, 𝑦 = 1000 * (1.04)^10, which approximately equals 1480.2.
Spot on! The population will be about 1,480 people after 10 years. Remember, this shows how quickly populations can grow exponentially!
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Create a free accountNow let’s look at exponential decay. A car worth $20,000 depreciates by 15% each year. Using the decay formula, how would we find its value after 5 years?
We should use the formula 𝑦 = 𝑎(1 - 𝑟)𝑡.
Correct! What do we have for our values?
Here, 𝑎 = 20000, 𝑟 = 0.15, and 𝑡 = 5.
Can someone calculate the final value?
So, 𝑦 = 20000 * (0.85)^5, which is about $8,874.
Great job! This illustrates the importance of understanding depreciation.
Overview
Short Summary
This section provides the solutions for select exponential growth and decay problems, showcasing the application of formulas in real-world scenarios.
Medium Summary
The section covers specific solutions to selected problems on exponential growth and decay, detailing the formulas used and their applicability in solving real-world mathematical scenarios related to population dynamics and depreciation.
Detailed Summary
Selected Solutions to Exponential Growth and Decay Problems
This section presents worked solutions for selected examples of exponential growth and decay, which are critical in understanding how these functions model real-world phenomena. Two examples are solved here:
- Exponential Growth: Calculating the future population of a town growing at a steady rate.
- Exponential Decay: Determining the depreciated value of an asset over time.
Key Formulas for Solutions
- Exponential Growth Formula: 𝑦 = 𝑎(1 + 𝑟)𝑡
Where:
- 𝑎 = initial amount (population)
- 𝑟 = growth rate (as a decimal)
- 𝑡 = time in years
- 𝑦 = amount after time t.
- Exponential Decay Formula: 𝑦 = 𝑎(1 - 𝑟)𝑡
Where:
- 𝑎 = initial amount (asset value)
- 𝑟 = decay rate (as a decimal)
- 𝑡 = time in years
- 𝑦 = remaining amount after time t.
These formulas reflect how quantities grow or decline exponentially and can be applied across various fields such as biology, finance, physics, and more.
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Create a free account𝑎 = 1000,𝑟 = 0.04,𝑡 = 10 𝑦 = 1000(1.04)10 ≈ 1000×1.4802 = 1480.2 Answer: ~1,480 people
Detailed Explanation
In this problem, we are calculating the future population of a town that grows at a rate of 4% per year. The initial population (𝑎) is 1000 people, the growth rate (𝑟) is 0.04 (which is 4% expressed as a decimal), and the time (𝑡) is 10 years. To find the future population, we apply the formula for exponential growth, 𝑦 = 𝑎(1 + 𝑟)^𝑡. We substitute the values into the formula:
- Calculate (1 + 𝑟) = (1 + 0.04) = 1.04.
- Raise this value to the power of 10: (1.04)^10 ≈ 1.4802.
- Multiply this result by the initial population: 1000 * 1.4802 ≈ 1480.2.
- Since population can’t be a fraction, we round to get approximately 1,480 people.
Examples & Analogies
Consider a small village that has a population of 1,000 people. Each year, more babies are born and families are moving in, which increases the town's population at a steady pace. If every year, about 4% more people are added, we can visualize this growth like a balloon slowly inflating over time—the bigger it gets, the more air (or people) it can hold!
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Create a free account𝑎 = 100,𝑟 = 0.20,𝑡 = 3 𝑦 = 100(0.8)3 = 100×0.512 = 51.2 Answer: 51.2%
Detailed Explanation
In this problem, we are determining how much charge remains in a battery that loses 20% of its charge every hour. The initial charge (𝑎) is 100%, the decay rate (𝑟) is 0.20 (which corresponds to 20% expressed as a decimal), and time (𝑡) is 3 hours. For exponential decay, we use the formula 𝑦 = 𝑎(1 − 𝑟)^𝑡:
- Calculate (1 - 𝑟) = (1 - 0.20) = 0.80.
- Raise this value to the power of 3: (0.8)^3 = 0.512.
- Multiply this result by the initial charge: 100 * 0.512 = 51.2.
- Therefore, after 3 hours, the charge left in the battery is approximately 51.2%.
Examples & Analogies
Imagine you have a new smartphone battery charged to 100%. Every hour, it loses 20% of its charge—kind of like a sponge that keeps soaking up water but with each hour, some of that water gets removed. After three hours, you've still got a good part of your battery left, but it's getting lower each hour, similar to how the sponge would be less full over time.
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Key Concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
Exponential Function: A mathematical function of the form 𝑦 = 𝑎 ⋅ 𝑏𝑥.
Growth and Decay: Growth occurs when 𝑏 > 1 and decay occurs when 0 < 𝑏 < 1.
Key Applications: Used in fields like biology, finance, and physics.
Examples
Memory Aids
Interactive tools to help you remember key concepts
Stories
Flash Cards
Glossary
Exponential Growth
A process where a quantity increases by a fixed percentage over regular intervals.
Exponential Decay
A process where a quantity decreases by a fixed percentage over time.
Growth Rate
The rate at which a quantity increases, expressed as a decimal.
Decay Rate
The rate at which a quantity decreases, expressed as a decimal.
Asymptote
A line that a curve approaches as it heads towards infinity, without ever touching it.