Homework help with mixture problem

[Deanna]

I need help setting this problem up. I keep coming up with the wrong answer. I know it is suppose to be either a separable or linear differential equation. Can anyone help me get it set up right? I think its suppose to be a linear 1st order diff equation.

A 100 gallon tank is filled with pure water. At time t=0 a salt solution is added to the container at a rate of 1 gallon per min. The well stirred solution drains from the container at the same rate--1 gallon per min. The concentration of the salt entering the tank is unknown. After 100 minutes it is measured that the amount of salt in the tank is 50 pounds. Determine the concentration of the salt (in pounds per gallon) in the incoming solution.

This is what I do know.
t(0)=0
t(100)=50
dV/dt=rate solution enters the tank - rate solution leaves the tank so
dV/dt=0
so v(t)=0+C and I believe C should = 100 which makes v(t)=100
I am not sure if this is right and I can figure out how to get dx/dt or where t fits into the equation. Help please if you can

 

[Tide]

The concentration in the tank is governed by

$$\frac {dC}{dt} = (C_{in} - C) \frac {R_{in}}{V}$$

where $C_{in}$ is the concentration of the solution flowing into the tank, $R_{in}$ is the rate at which the solution flows in and V is the volume of solution in the tank. You should be able to take it from there.

 

[M98Ranger]

First of all, it's great that you have identified that this problem involves a mixture and that it can be solved using a differential equation. Let's break down the problem and see how we can set it up.

We have a 100 gallon tank filled with pure water at time t=0. At this point, there is no salt in the tank and the concentration of salt in the incoming solution is unknown. Then, at a rate of 1 gallon per minute, a salt solution is added to the tank. At the same rate, 1 gallon per minute, the well-stirred solution drains from the tank. After 100 minutes, the amount of salt in the tank is measured to be 50 pounds. We want to determine the concentration of salt in the incoming solution.

To set up this problem, we need to use the concept of concentration, which is the amount of salt per unit volume of solution. Let's call this concentration x(t) where t is time. We also know that the rate at which salt is added to the tank is 1 gallon per minute and the rate at which salt is draining from the tank is also 1 gallon per minute. This means that the change in concentration over time, dx/dt, is equal to the rate at which salt is added, which is 1 gallon per minute, minus the rate at which salt is draining, also 1 gallon per minute. So we can write the following differential equation:

dx/dt = 1 - 1 = 0

This is a separable differential equation, which means we can separate the variables and solve for x(t). To do this, we need to integrate both sides of the equation with respect to t:

∫dx/dt dt = ∫0 dt

This gives us:

x(t) = 0t + C

Since we know that at t=0, x(t)=0, we can substitute these values into the equation and solve for C:

0 = 0(0) + C

C = 0

So our final equation for concentration is:

x(t) = 0t + 0

But this doesn't give us any information about the concentration of the salt in the incoming solution. To find that, we need to use the information given in the problem. We know that after 100 minutes, the amount of salt in the tank is 50 pounds. So we can set up another equation using this information