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#### alane1994

##### Active member

- Oct 16, 2012

- 126

How on earth...

This is so similar I feel like I should be able to do it, yet I am drawing a big blank.

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- Thread starter
- #1

- Oct 16, 2012

- 126

How on earth...

This is so similar I feel like I should be able to do it, yet I am drawing a big blank.

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- #2

- Jan 26, 2012

- 995

Again, we need to make a list of the things we know and then end up using the differential equation $\dfrac{dx}{dt} = r_ic_i - r_oc_o$ from the previous thread.

How on earth...

This is so similar I feel like I should be able to do it, yet I am drawing a big blank.

So, we know that $x(0)=100\text{ lb}$, $c_i = 1\text{ lb/gal}$, $V(0) = 200\text{ gal}$, $r_i = 3\text{ gal/min}$ and $r_o=2\text{ gal/min}$.

The fun part is figuring out the amount of concentration leaving the tank. Just as before, it's a quantity that depends on $t$: $c_o(t) = \dfrac{x(t)}{V(t)}$. However, the rate of change of the volume of solution in the tank is the difference of the entry/exiting rates. Thus, we see that

\[\frac{dV}{dt} = r_i-r_o\implies V(t) = (r_i-r_o)t+C\]

Since $V(0)=V_0$, we have that $C=V_0$ and thus $V(t) = (r_i-r_o)t+V_0$. In the context of this problem, that means $V(t)= t+200$ and thus $c_o(t) = \dfrac{x(t)}{t+200}$.

Therefore, the differential equation this time is

\[\frac{dx}{dt} = 3(1) - 2\cdot\frac{x}{t+200}\implies \frac{dx}{dt} = 3-\frac{2x}{t+200}\implies \frac{dx}{dt} +\frac{2}{t+200}x = 3\]

which is a linear first order equation.

To solve it, you'll need to use the integrating factor. Once you find $x(t)$, you need to compute the amount of concentration of salt in the tank right when it's about to overflow (i.e. when $V(t)=500\implies t=t_0$) by computing $c(t_0)=\dfrac{x(t_0)}{500}\text{ lb/gal}$.

The limiting concentration amount is then $\displaystyle\lim_{t\to\infty}\frac{x(t)}{t+200}$.

I hope this makes sense!

Last edited:

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\(\displaystyle \text{time rate of change of salt = time rate of salt coming in minus time rate of salt going out}\)

\(\displaystyle \text{time rate of salt coming in=concentration of brine coming in times the time rate of volume coming in}\)

\(\displaystyle \text{time rate of salt going out=concentration of brine going out times the time rate of volume going out}\)

Hence, stated mathematically, we may write:

\(\displaystyle \frac{dA}{dt}=1\frac{\text{lb}}{\text{gal}}\cdot3 \frac{\text{gal}}{\text{min}}-\frac{A(t)}{V(t)}\frac{\text{lb}}{\text{gal}}\cdot2\frac{\text{gal}}{\text{min}}\)

And so, in \(\displaystyle \frac{\text{lb}}{\text{min}}\), we have:

\(\displaystyle \frac{dA}{dt}=3-\frac{2A(t)}{V(t)}\)

Now, how can we determine $V(t)$? What kind of ODE do we have, and how should we solve it?

Hahaha...this time I got ninja'd...

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- #4

- Jan 26, 2012

- 995

Touché.... Hahaha...this time I got ninja'd...

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- #5

- Oct 16, 2012

- 126

Couple questions mentioned above. If they are in fact true, doesn't that alter all of what you have said? Just input different values in?Again, we need to make a list of the things we know and then end up using the differential equation $\dfrac{dx}{dt} = r_ic_i - r_oc_o$ from the previous thread.

So, we know that $\color{red}{x(0)=200\text{ lb}}$, $\color{red}{c_i = 1\text{ lb/gal}}$, $V(0) = 200\text{ gal}$, $r_i = 3\text{ gal/min}$ and $r_o=2\text{ gal/min}$.

\(\color{red}{\text{I believe this is supposed to be something else}}\)

The fun part is figuring out the amount of concentration leaving the tank. Just as before, it's a quantity that depends on $t$: $c_o(t) = \dfrac{x(t)}{V(t)}$. However, the rate of change of the volume of solution in the tank is the difference of the entry/exiting rates. Thus, we see that

\[\frac{dV}{dt} = r_i-r_o\implies V(t) = (r_i-r_o)t+C\]

Since $V(0)=V_0$, we have that $C=V_0$ and thus $V(t) = (r_i-r_o)t+V_0$. In the context of this problem, that means $V(t)= t+200$ and thus $c_o(t) = \dfrac{x(t)}{t+200}$.

Therefore, the differential equation this time is

\[\frac{dx}{dt} = 3(1) - 2\cdot\frac{x}{t+200}\implies \frac{dx}{dt} = 3-\frac{2x}{t+200}\implies \frac{dx}{dt} +\frac{2}{t+200}x = 3\]

which is a linear first order equation.

To solve it, you'll need to use the integrating factor. Once you find $x(t)$, you need to compute the amount of salt in the tank right when it's about to overflow (i.e. when $V(t)=500\implies t=\ldots$).

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- #6

- Jan 26, 2012

- 995

I noticed I had the wrong x(0) originally (read the problem a little too fast) and edited my original post. It should be x(0)=100, not x(0)=200.Couple questions mentioned above. If they are in fact true, doesn't that alter all of what you have said? Just input different values in?

Sorry about that!