Dilution Problem (Probably Separable)

**ragnar** · January 19th 2011, 11:32 AM

So I'm doing this problem 3 on page 125 of Tenenbaum's ODE and I'm not getting the same answer, not sure why.

It says, suppose you have a tank of 100 gallons of water with 30 lbs of salt, evenly diluted, and 3 gallons are drained per minute while four gallons of fresh water are fed in each minute while constant stirring ensures homogeneity of the fluid's salt content. What is the amount of salt contained after 10 minutes?

I figure $x, t$ are total quantity of salt and time respectively so $\Delta x = -3\Delta t \cdot \frac{x}{100+t}$ since $-3\Delta t$ will represent the number of gallons of fluid leaving in $\Delta t$ minutes, and $\frac{x}{100+t}$ the quantity of salt per gallon. If that were so, though, when I separate variables and integrate, I wouldn't have any logarithmic functions in my equation, whereas the answer given in the book has an exponential.

**dwsmith** · January 19th 2011, 12:17 PM

Originally Posted by ragnar

So I'm doing this problem 3 on page 125 of Tenenbaum's ODE and I'm not getting the same answer, not sure why.

It says, suppose you have a tank of 100 gallons of water with 30 lbs of salt, evenly diluted, and 3 gallons are drained per minute while four gallons of fresh water are fed in each minute while constant stirring ensures homogeneity of the fluid's salt content. What is the amount of salt contained after 10 minutes?

I figure $x, t$ are total quantity of salt and time respectively so $\Delta x = -3\Delta t \cdot \frac{x}{100+t}$ since $-3\Delta t$ will represent the number of gallons of fluid leaving in $\Delta t$ minutes, and $\frac{x}{100+t}$ the quantity of salt per gallon. If that were so, though, when I separate variables and integrate, I wouldn't have any logarithmic functions in my equation, whereas the answer given in the book has an exponential.

$\displaystyle\frac{dA}{dt}=R_{\text{in}}-R_{\text{out}}$

$A(0)=30$

$R_{\text{in}}=4\text{gals/min of water}\cdot 0\text{lb/gal of salt}=0$

$\displaystyle R_{\text{out}}=\frac{A(T)}{100+t}\text{lb/gal}\cdot 3\text{gal/min}$

$\displaystyle\frac{dA}{dt}=-\frac{3A}{100+t}\text{lb/gal}$

Is this your equation?

$\displaystyle\frac{dA}{3A}=-\frac{dt}{100+t}$

$\displaystyle\int\frac{dA}{3A}=-\int\frac{dt}{100+t}$

$\displaystyle\frac{\ln |A|}{3}=-\ln |100+t|+C\Rightarrow\exp{(\ln (A))}=\exp{(-3\ln (100+t))+C_2)}$

$\displaystyle A(t)=C_3\exp{(\ln(100+t)^{-3})}=\frac{C_3}{(100+t)^{3}}$

**ragnar** · January 19th 2011, 02:03 PM

Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of $-3$ rather than $\frac{-1}{3}$ since at the previous step we had $-3\ln(...)$ .)

But the book says my function should be such that $A(10) = 30e^{-.286}$ !

Also, with our function, $A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow$
$A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) =$ something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.

**dwsmith** · January 19th 2011, 02:05 PM

Originally Posted by ragnar

Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of $-3$ rather than $\frac{-1}{3}$ since at the previous step we had $-3\ln(...)$ .)

But the book says my function should be such that $A(10) = 30e^{-.286}$ !

Also, with our function, $A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow$
$A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) =$ something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.

It shouldn't be 1/3. That was an error on my part.

**dwsmith** · January 19th 2011, 02:21 PM

Originally Posted by ragnar

Well gosh-golly god-damnit that's what I got too! (Except at the last step I get an exponent of $-3$ rather than $\frac{-1}{3}$ since at the previous step we had $-3\ln(...)$ .)

But the book says my function should be such that $A(10) = 30e^{-.286}$ !

Also, with our function, $A(0) = 30 = \frac{C}{(100+t)^{3}} \Rightarrow C = 3/100000 \Rightarrow$
$A(T) = \frac{3}{100000(100+t)^{3}} \Rightarrow A(10) =$ something freaking tiny, which doesn't seem right (even if we used a cube-root it'd be tiny). So it makes me suspect our solution is wrong.

$\displaystyle A(0)=\frac{C_3}{(100)^3}=30\Rightarrow C_3=30000000$

$\displaystyle A(t)=\frac{30000000}{(100+t)^3}$

**ragnar** · January 19th 2011, 02:24 PM

AHHHH, I'm just retarded. That mystery's solved.

So do you suppose the book is wrong in having an exponential function as part of their solution? Or is there something I'm missing about how our function is related to $e$ ?

**dwsmith** · January 19th 2011, 02:31 PM

Originally Posted by ragnar

AHHHH, I'm just retarded. That mystery's solved.

So do you suppose the book is wrong in having an exponential function as part of their solution? Or is there something I'm missing about how our function is related to $e$ ?

A(10) = 22.5394

30 exp(-2.86) = 22.5379

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