Math Problem Statement

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.

Solution

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 12\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)Q(t) in the tank is: dQdt=rate inrate out=12×2Q(t)100×8\frac{dQ}{dt} = \text{rate in} - \text{rate out} = \frac{1}{2} \times 2 - \frac{Q(t)}{100} \times 8
    • Simplified, this gives: dQdt=18Q(t)100=1Q(t)12.5\frac{dQ}{dt} = 1 - \frac{8Q(t)}{100} = 1 - \frac{Q(t)}{12.5}
    • The initial condition is Q(0)=0Q(0) = 0 (no salt initially).
    • The solution of this linear equation is: Q(t)=50(1et/12.5)Q(t) = 50 \left( 1 - e^{-t/12.5} \right)
    • At t=10t = 10 minutes, we substitute into the solution to find: Q(10)=50(1e10/12.5)9.063 lbs of salt.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)P(t) in the tank: dPdt=8P(t)100\frac{dP}{dt} = - \frac{8P(t)}{100}
    • The solution to this equation is: P(t)=9.063e0.08tP(t) = 9.063 e^{-0.08t}
    • After another 10 minutes (t=10t = 10), we compute: P(10)=9.063e0.84.07 lbs of salt.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.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Mixing Problems
Exponential Decay
Initial Value Problems

Formulas

dQ/dt = (rate of salt in) - (rate of salt out)
Q(t) = 50(1 - e^(-t/12.5))
P(t) = 9.063e^(-0.08t)

Theorems

Linear Differential Equations
Exponential Decay Model

Suitable Grade Level

University-level Calculus or Differential Equations