# Dilution Problem (Probably Separable)

• Jan 19th 2011, 11:32 AM
ragnar
Dilution Problem (Probably Separable)
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.
• Jan 19th 2011, 12:17 PM
dwsmith
Quote:

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.

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

$A(0)=30$

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

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

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

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

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

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

$\displaystyle A(t)=C_3\exp{(\ln(100+t)^{-3})}=\frac{C_3}{(100+t)^{3}}$
• Jan 19th 2011, 02:03 PM
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.
• Jan 19th 2011, 02:05 PM
dwsmith
Quote:

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.
• Jan 19th 2011, 02:21 PM
dwsmith
Quote:

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.

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

$\displaystyle A(t)=\frac{30000000}{(100+t)^3}$
• Jan 19th 2011, 02:24 PM
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 $e$?
• Jan 19th 2011, 02:31 PM
dwsmith
Quote:

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