# [Differential equations] Mixing problem.

• Muskyboi
In summary: Therefore, after 300 hours, the concentration of salt in the tank would be approximately 1/5, or 300 pounds. This is a reasonable result based on the given information. In summary, after 300 hours, the final concentration of salt in the tank is approximately 1/5, or 300 pounds. This is a reasonable result based on the given information.
Muskyboi
Homework Statement
9 gallons/hour entering with a salt concentration of 1/5(1+cos(t)) pounds/gallon (this is a function of time t).

15000 gallon capacity tank with 600 gallons of water with 5 pounds of salt dissolved inside.

6 gallons/hour leaving.

How much salt is the tank just before it overflows?
Relevant Equations
salt concentration= 1/5(1+cos(t)) pounds/gallon (this is a function of time t)

v(t)=600+(9-6)t
=600+3t

1500=600+3t
therefore t=300 hrs when tank is full

Cin=1/5(1 + cost)

ds/dt=Rate in - rate out = CinRin - Cout*S(t)/V(t)
=1/5(1 + cost)*9 - 6*S(t)/(600+3t)

S(0)=5 ib

Solving the first order linear ODE we get: https://www.desmos.com/calculator/l7iixzgyll

therefore S(300)=279.797 Ib

does my solution look right?

Last edited:
The differential equation looks right. how did you solve it?

It's a strange problem. After 300 hours the initial concentration is largely irrelevant. The average of the cosine function is 0, so the average concentration entering is ##1/5##. You would expect the final concentration to be close to this. Hence 300lb of salt, approximately.

PeroK said:
The differential equation looks right. how did you solve it?

It's a strange problem. After 300 hours the initial concentration is largely irrelevant. The average of the cosine function is 0, so the average concentration entering is ##1/5##. You would expect the final concentration to be close to this. Hence 300lb of salt, approximately.

Yes, dividing the equation for mass of salt in the tank S(t) by the volume of water in the tank V(t) would give the concentration of salt in the tank. Doing so shows that the concentration approaches 1/5 as time increases.

## 1. What is a mixing problem in differential equations?

A mixing problem in differential equations is a type of problem that involves predicting the concentration of a substance in a solution as it is being mixed or diluted. The rate of change of the concentration is described by a differential equation.

## 2. What types of substances are typically involved in mixing problems?

Mixing problems can involve any type of substance that can be dissolved in a solution, such as salt, sugar, or a chemical compound. These substances can be in liquid or solid form.

## 3. How are mixing problems solved using differential equations?

To solve a mixing problem using differential equations, the initial concentration of the substance, the rate at which it is being mixed or diluted, and the volume of the solution must be known. These values are used to set up a differential equation that describes the rate of change of the concentration over time. The equation is then solved using various methods, such as separation of variables or integrating factors.

## 4. What real-life applications use mixing problems and differential equations?

Mixing problems and differential equations have various applications in fields such as chemistry, biology, and environmental science. For example, they can be used to model the spread of pollutants in a body of water, the rate at which a medication is absorbed into the bloodstream, or the growth of bacteria in a culture.

## 5. What are some common techniques used to analyze and interpret solutions to mixing problems?

Some common techniques used to analyze and interpret solutions to mixing problems include finding the equilibrium concentration, determining the time it takes for the substance to reach a certain concentration, and identifying any critical points or inflection points in the solution curve. Graphing the solution and considering its behavior at different time intervals can also provide insights into the mixing process.

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