A 500-liter tank initially contains 10 grams of salt dissolved in 200 liters of water. Starting at t=0, water that contains 1/4 grams of salt per liter is poured into the tank at the rate of 4 liters/min, and the mixture is drained from the tank at the rate of 2 liters/min. Let y(t) represent the amount of salt inside the tank at time t. Find a differential model for y(t) and solve it to find y(t). Graph y(t). What is the amount of salt after 10 minutes? What is the concentration of salt inside the tank after 10 minutes?
A 500-liter tank initially contains 10 grams of salt dissolved in 200 liters of water. Starting at t=0, water that contains 1/4 grams of salt per liter is poured into the tank at the rate of 4 liters/min, and the mixture is drained from the tank at the rate of 2 liters/min. Let y(t) represent the amount of salt inside the tank at time t. Find a differential model for y(t) and solve it to find y(t). Graph y(t). What is the amount of salt after 10 minutes? What is the concentration of salt inside the tank after 10 minutes?
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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![### Problem Statement
A 500-liter tank initially contains 10 grams of salt dissolved in 200 liters of water. Starting at \( t = 0 \), water that contains \( \frac{1}{4} \) grams of salt per liter is poured into the tank at a rate of 4 liters per minute, and the mixture is drained from the tank at the rate of 2 liters per minute. Let \( y(t) \) represent the amount of salt inside the tank at time \( t \).
1. Find a differential model for \( y(t) \) and solve it to find \( y(t) \).
2. Graph \( y(t) \).
3. What is the amount of salt after 10 minutes?
4. What is the concentration of salt inside the tank after 10 minutes?
### Detailed Explanation
**1. Differential Model for \( y(t) \):**
First, let's define the variables:
- \( y(t) \): the amount of salt in the tank at time \( t \) (in grams).
- \( V(t) \): the volume of water in the tank at time \( t \) (in liters).
The rate at which salt enters the tank is:
\[ \text{Rate}_{\text{in}} = 4 \, \text{liters/min} \times \frac{1}{4} \, \text{grams/liter} = 1 \, \text{grams/min} \]
The rate at which salt leaves the tank depends on the concentration of salt in the tank:
\[ \text{Rate}_{\text{out}} = \frac{y(t)}{V(t)} \times 2 \, \text{liters/min} \]
Now, let's model the volume of water in the tank \( V(t) \):
\[ V(t) = 200 + 2t \, \text{liters} \]
Thus, the differential equation modeling the system is:
\[ \frac{dy}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} = 1 - \frac{y(t)}{200 + 2t} \times 2 \]
Simplifying, we get:
\[ \frac{dy}{dt} = 1 - \frac{2y(t)}{200 + 2t} \]
**2. Solution to the Differential](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f7c2325-139a-41dc-93c7-07b552835718%2F7947d277-216a-41eb-9cd6-89dea6e591cf%2F2rn2vql_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A 500-liter tank initially contains 10 grams of salt dissolved in 200 liters of water. Starting at \( t = 0 \), water that contains \( \frac{1}{4} \) grams of salt per liter is poured into the tank at a rate of 4 liters per minute, and the mixture is drained from the tank at the rate of 2 liters per minute. Let \( y(t) \) represent the amount of salt inside the tank at time \( t \).
1. Find a differential model for \( y(t) \) and solve it to find \( y(t) \).
2. Graph \( y(t) \).
3. What is the amount of salt after 10 minutes?
4. What is the concentration of salt inside the tank after 10 minutes?
### Detailed Explanation
**1. Differential Model for \( y(t) \):**
First, let's define the variables:
- \( y(t) \): the amount of salt in the tank at time \( t \) (in grams).
- \( V(t) \): the volume of water in the tank at time \( t \) (in liters).
The rate at which salt enters the tank is:
\[ \text{Rate}_{\text{in}} = 4 \, \text{liters/min} \times \frac{1}{4} \, \text{grams/liter} = 1 \, \text{grams/min} \]
The rate at which salt leaves the tank depends on the concentration of salt in the tank:
\[ \text{Rate}_{\text{out}} = \frac{y(t)}{V(t)} \times 2 \, \text{liters/min} \]
Now, let's model the volume of water in the tank \( V(t) \):
\[ V(t) = 200 + 2t \, \text{liters} \]
Thus, the differential equation modeling the system is:
\[ \frac{dy}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} = 1 - \frac{y(t)}{200 + 2t} \times 2 \]
Simplifying, we get:
\[ \frac{dy}{dt} = 1 - \frac{2y(t)}{200 + 2t} \]
**2. Solution to the Differential
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