Suppose you start with one liter of vinegar and repeatedly remove 0.13 L, replace with water, mix, and repeat. a. Find a formula for the concentration after n steps. b. After how many steps does the mixture contain less than 5% vinegar?
Suppose you start with one liter of vinegar and repeatedly remove 0.13 L, replace with water, mix, and repeat. a. Find a formula for the concentration after n steps. b. After how many steps does the mixture contain less than 5% vinegar?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
Suppose you start with one liter of vinegar and repeatedly remove 0.13 L, replace with water, mix, and repeat.
#### a. Find a formula for the concentration after \( n \) steps.
[Textbox for answer]
#### b. After how many steps does the mixture contain less than 5% vinegar?
[Textbox for answer]
### Detailed Explanation:
This problem involves a repetitive dilution process, where a fixed volume of vinegar is removed and replaced with water. This process involves geometric progression to model the concentration of vinegar over time.
For part (a), to find the concentration of vinegar after \( n \) steps, we can denote the concentration of vinegar after each step by using the formula:
\[ C_n = C_0 \left(1 - \frac{V_{\text{removed}}}{V_{\text{total}}}\right)^n \]
Where:
- \( C_n \) is the concentration after \( n \) steps.
- \( C_0 \) is the initial concentration of vinegar (which is 1, or 100%).
- \( V_{\text{removed}} \) is the volume removed, which is 0.13 L.
- \( V_{\text{total}} \) is the total volume, which is 1 L.
- \( n \) is the number of steps.
So the formula simplifies to:
\[ C_n = 1 \left(1 - \frac{0.13}{1}\right)^n \]
\[ C_n = (0.87)^n \]
For part (b), we need to determine the number of steps \( n \) such that the concentration \( C_n \) is less than 5% (or 0.05):
\[ (0.87)^n < 0.05 \]
To solve for \( n \):
\[ n \log(0.87) < \log(0.05) \]
\[ n > \frac{\log(0.05)}{\log(0.87)} \]
By calculating the above expression, we can determine the number of steps required for the vinegar concentration to drop below 5%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc31d976c-acd4-451e-8eb9-fde305025b60%2F2f71f7ab-bbaa-4e0a-b0b3-849bdf7f73fc%2F3y4jbq_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Suppose you start with one liter of vinegar and repeatedly remove 0.13 L, replace with water, mix, and repeat.
#### a. Find a formula for the concentration after \( n \) steps.
[Textbox for answer]
#### b. After how many steps does the mixture contain less than 5% vinegar?
[Textbox for answer]
### Detailed Explanation:
This problem involves a repetitive dilution process, where a fixed volume of vinegar is removed and replaced with water. This process involves geometric progression to model the concentration of vinegar over time.
For part (a), to find the concentration of vinegar after \( n \) steps, we can denote the concentration of vinegar after each step by using the formula:
\[ C_n = C_0 \left(1 - \frac{V_{\text{removed}}}{V_{\text{total}}}\right)^n \]
Where:
- \( C_n \) is the concentration after \( n \) steps.
- \( C_0 \) is the initial concentration of vinegar (which is 1, or 100%).
- \( V_{\text{removed}} \) is the volume removed, which is 0.13 L.
- \( V_{\text{total}} \) is the total volume, which is 1 L.
- \( n \) is the number of steps.
So the formula simplifies to:
\[ C_n = 1 \left(1 - \frac{0.13}{1}\right)^n \]
\[ C_n = (0.87)^n \]
For part (b), we need to determine the number of steps \( n \) such that the concentration \( C_n \) is less than 5% (or 0.05):
\[ (0.87)^n < 0.05 \]
To solve for \( n \):
\[ n \log(0.87) < \log(0.05) \]
\[ n > \frac{\log(0.05)}{\log(0.87)} \]
By calculating the above expression, we can determine the number of steps required for the vinegar concentration to drop below 5%.
![### Fish Population Growth and Loss Model
**Problem Statement:**
A lake initially contains 3,000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 6% each month. However, factoring in all causes, 400 fish are lost each month. How many fish will be in the pond after 7 months? (Note: Do not round until the very end.)
**Mathematical Breakdown:**
1. **Initial Population (P0):** 3,000 fish
2. **Monthly Growth Rate (r):** 6% (or 0.06 as a decimal)
3. **Monthly Removal:** 400 fish
**Process to Calculate the Fish Population after 7 Months:**
1. For each month, update the population by first applying the growth rate, then subtracting the fixed removal.
2. Use the following iterative formula:
\[
P_{n+1} = P_{n} \times (1 + r) - 400
\]
Where:
- \( P_{n} \) is the population at month \( n \)
- \( r \) is the growth rate (0.06)
- 400 is the number of fish removed each month
**Calculate for Each Month:**
1. **Month 1:**
\[ P_1 = 3000 \times (1 + 0.06) - 400 = 3000 \times 1.06 - 400 = 3180 - 400 = 2780 \]
2. **Month 2:**
\[ P_2 = 2780 \times (1 + 0.06) - 400 = 2780 \times 1.06 - 400 = 2946.8 - 400 = 2546.8 \]
3. **Month 3:**
\[ P_3 = 2546.8 \times (1 + 0.06) - 400 = 2546.8 \times 1.06 - 400 = 2699.608 - 400 = 2299.608 \]
4. **Month 4:**
\[ P_4 = 2299.608 \times (1 + 0.06) - 400 = 2299.608 \times 1.06 - 400 = 2437.58448 - 400 = 203](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc31d976c-acd4-451e-8eb9-fde305025b60%2F2f71f7ab-bbaa-4e0a-b0b3-849bdf7f73fc%2Fg7q6tds_processed.png&w=3840&q=75)
Transcribed Image Text:### Fish Population Growth and Loss Model
**Problem Statement:**
A lake initially contains 3,000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 6% each month. However, factoring in all causes, 400 fish are lost each month. How many fish will be in the pond after 7 months? (Note: Do not round until the very end.)
**Mathematical Breakdown:**
1. **Initial Population (P0):** 3,000 fish
2. **Monthly Growth Rate (r):** 6% (or 0.06 as a decimal)
3. **Monthly Removal:** 400 fish
**Process to Calculate the Fish Population after 7 Months:**
1. For each month, update the population by first applying the growth rate, then subtracting the fixed removal.
2. Use the following iterative formula:
\[
P_{n+1} = P_{n} \times (1 + r) - 400
\]
Where:
- \( P_{n} \) is the population at month \( n \)
- \( r \) is the growth rate (0.06)
- 400 is the number of fish removed each month
**Calculate for Each Month:**
1. **Month 1:**
\[ P_1 = 3000 \times (1 + 0.06) - 400 = 3000 \times 1.06 - 400 = 3180 - 400 = 2780 \]
2. **Month 2:**
\[ P_2 = 2780 \times (1 + 0.06) - 400 = 2780 \times 1.06 - 400 = 2946.8 - 400 = 2546.8 \]
3. **Month 3:**
\[ P_3 = 2546.8 \times (1 + 0.06) - 400 = 2546.8 \times 1.06 - 400 = 2699.608 - 400 = 2299.608 \]
4. **Month 4:**
\[ P_4 = 2299.608 \times (1 + 0.06) - 400 = 2299.608 \times 1.06 - 400 = 2437.58448 - 400 = 203
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