(10 points) A tank contains 200 gal of liquid. Initially, the tank contains pure wa- ter. At time t = 0, brine containing 3 lb/gal of salt begins to pour into the tank at a rate of 2 gal/min, and the well-stirred mixture is allowed to drain away at the same rate. a. Write down the initial value problem for the amount of salt in the tank a(t) b. Solve for A(t) to determine the amount of salt at any t > 0.
(10 points) A tank contains 200 gal of liquid. Initially, the tank contains pure wa- ter. At time t = 0, brine containing 3 lb/gal of salt begins to pour into the tank at a rate of 2 gal/min, and the well-stirred mixture is allowed to drain away at the same rate. a. Write down the initial value problem for the amount of salt in the tank a(t) b. Solve for A(t) to determine the amount of salt at any t > 0.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![### Problem Statement
**(10 points)** A tank contains 200 gallons of liquid. Initially, the tank contains pure water. At time \( t = 0 \), brine containing 3 lb/gal of salt begins to pour into the tank at a rate of 2 gal/min, and the well-stirred mixture is allowed to drain away at the same rate.
#### Questions
a. Write down the initial value problem for the amount of salt in the tank \( A(t) \).
b. Solve for \( A(t) \) to determine the amount of salt at any \( t \ge 0 \).
### Solution Outline
#### a. Formulating the Initial Value Problem
Let \( A(t) \) be the amount of salt in the tank at time \( t \).
- **Inflow rate of salt:**
\[
\text{Rate of inflow} = 3 \text{ lb/gal} \times 2 \text{ gal/min} = 6 \text{ lb/min}
\]
- **Outflow rate of salt:**
Since the outflow rate is the same as the inflow rate, it is also 2 gal/min. Thus, the amount of salt leaving per minute = \( \frac{A(t)}{200} \times 2 \text{ gal/min} = \frac{2A(t)}{200} = \frac{A(t)}{100} \text{ lb/min} \).
- **Differential equation:**
\[
\frac{dA(t)}{dt} = \text{Rate of inflow} - \text{Rate of outflow} = 6 - \frac{A(t)}{100}
\]
- **Initial condition:**
\[
A(0) = 0 \text{ (since initially, the tank contains pure water)}
\]
Therefore, the initial value problem is:
\[
\frac{dA(t)}{dt} = 6 - \frac{A(t)}{100}, \quad A(0) = 0
\]
#### b. Solving the Differential Equation
We have the first-order linear differential equation:
\[
\frac{dA(t)}{dt} = 6 - \frac{A(t)}{100}
\]
Rewriting it, we get:
\[
\frac{dA(t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8bdf6582-40e2-47b9-bd75-44351bba3f76%2Fd68ebafb-eed4-4a03-9799-0b6692ad3611%2F7tb225q.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**(10 points)** A tank contains 200 gallons of liquid. Initially, the tank contains pure water. At time \( t = 0 \), brine containing 3 lb/gal of salt begins to pour into the tank at a rate of 2 gal/min, and the well-stirred mixture is allowed to drain away at the same rate.
#### Questions
a. Write down the initial value problem for the amount of salt in the tank \( A(t) \).
b. Solve for \( A(t) \) to determine the amount of salt at any \( t \ge 0 \).
### Solution Outline
#### a. Formulating the Initial Value Problem
Let \( A(t) \) be the amount of salt in the tank at time \( t \).
- **Inflow rate of salt:**
\[
\text{Rate of inflow} = 3 \text{ lb/gal} \times 2 \text{ gal/min} = 6 \text{ lb/min}
\]
- **Outflow rate of salt:**
Since the outflow rate is the same as the inflow rate, it is also 2 gal/min. Thus, the amount of salt leaving per minute = \( \frac{A(t)}{200} \times 2 \text{ gal/min} = \frac{2A(t)}{200} = \frac{A(t)}{100} \text{ lb/min} \).
- **Differential equation:**
\[
\frac{dA(t)}{dt} = \text{Rate of inflow} - \text{Rate of outflow} = 6 - \frac{A(t)}{100}
\]
- **Initial condition:**
\[
A(0) = 0 \text{ (since initially, the tank contains pure water)}
\]
Therefore, the initial value problem is:
\[
\frac{dA(t)}{dt} = 6 - \frac{A(t)}{100}, \quad A(0) = 0
\]
#### b. Solving the Differential Equation
We have the first-order linear differential equation:
\[
\frac{dA(t)}{dt} = 6 - \frac{A(t)}{100}
\]
Rewriting it, we get:
\[
\frac{dA(t
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