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The problem provided involves a mixing tank model where water and salt are flowing into and out of the tank. Here's a breakdown of the solution steps:

### Problem Restatement:
- A tank starts with **100 gallons of fresh water**.
- Saltwater with a concentration of \( \frac{1}{2} \) lb of salt per gallon is poured into the tank at **2 gallons per minute**.
- The mixture leaves the tank at **8 gallons per minute**.
- After **10 minutes**, the flow is stopped, and fresh water flows in at **8 gallons per minute** for an additional **10 minutes**.
- We need to find the amount of salt in the tank at the end of the 20-minute period.

### Solution Walkthrough:

1. **First 10 minutes (saltwater enters and exits)**:
   - The differential equation modeling the amount of salt \( Q(t) \) in the tank is:
     \[
     \frac{dQ}{dt} = \text{rate in} - \text{rate out} = \frac{1}{2} \times 2 - \frac{Q(t)}{100} \times 8
     \]
   - Simplified, this gives:
     \[
     \frac{dQ}{dt} = 1 - \frac{8Q(t)}{100} = 1 - \frac{Q(t)}{12.5}
     \]
   - The initial condition is \( Q(0) = 0 \) (no salt initially).
   - The solution of this linear equation is:
     \[
     Q(t) = 50 \left( 1 - e^{-t/12.5} \right)
     \]
   - At \( t = 10 \) minutes, we substitute into the solution to find:
     \[
     Q(10) = 50 \left( 1 - e^{-10/12.5} \right) \approx 9.063 \text{ lbs of salt}.
     \]

2. **Next 10 minutes (fresh water enters and exits)**:
   - Now, no more salt enters, so we have a new differential equation for the amount of salt \( P(t) \) in the tank:
     \[
     \frac{dP}{dt} = - \frac{8P(t)}{100}
     \]
   - The solution to this equation is:
     \[
     P(t) = 9.063 e^{-0.08t}
     \]
   - After another 10 minutes (\( t = 10 \)), we compute:
     \[
     P(10) = 9.063 e^{-0.8} \approx 4.07 \text{ lbs of salt}.
     \]

Thus, the amount of salt in the tank after 20 minutes is **4.07 lbs**.

### Let me know if you want any further details or clarifications on the steps.

Here are five related questions to expand understanding:
1. How does the rate of inflow and outflow affect the salt concentration over time?
2. Can you derive the differential equation for a different inflow rate?
3. How would the solution change if the outflow rate were higher than the inflow rate?
4. What is the significance of the exponential decay in the solution for the second period?
5. How would the problem change if saltwater continued to flow in during the second period?

**Tip:** When solving mixing problems, always keep track of the rate of change in both salt entering and salt exiting to set up the right differential equation.

A tank originally contains 100 gallons of fresh water. Water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 minutes, the process is stopped, and fresh water is poured into the tank at a rate of 8 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 minutes.

Explore a detailed solution for a mixing tank problem involving differential equations, where saltwater flows into and out of a tank, and the amount of salt is computed over a 20-minute period. Learn key concepts like exponential decay and initial value problems. Ideal for university-level calculus students.

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