# Is this done correctly?

• Sep 16th 2007, 02:44 PM
circuscircus
Is this done correctly?
A tank contains 300 gal of salt free water. A brine containing .5lb of salt per gallon of water runs into the tank at the rate of 2 gal/min and the well stirred mixtur runs out at the rate of 2 gal/min. What is the concentration of the salt in the tank at the end of the 10 min?

$\displaystyle A' = 2gal/min * .5lb/gal - 2gal/min * Alb/300gal$

$\displaystyle A' = 1 - (1/150)A$

$\displaystyle \frac{1}{1-1/150A} dA/dt = 1$

$\displaystyle \int t dt = \int \frac{150}{150-A} dA$

$\displaystyle t + c = 150 * ln |150-A|$

$\displaystyle \frac{t + c}{150} = ln |150-A|$

$\displaystyle +-e^{\frac{t + c}{150}} = 150-A$

$\displaystyle +-Ke^{\frac{t}{150}} = 150-A$

$\displaystyle +-150e^{\frac{t}{150}} = 150-A$

$\displaystyle A(t) = 150 +- 150 e^{t/150}$

Are my steps correct?
• Sep 16th 2007, 03:03 PM
TKHunny
Your first integral has a problem. Why is there a 't' in that argument? You ignored that error and mamaged the right antiderivative, but that was sort of magic.

Your right-hand antiderivative has the wrong sign. You can always check by simply finding the derivative. The chain rule will show the sign error.

When you lose the absolute values you introduce that very odd symbol "+-". I realize this is an attempt to retain the meaning of the absolute value, but this is no good. Make up your mind. Is it positive or negative? Think about the behavior of the salt. Does it increase or decrease? Does it have a maximum? Often you can resolve absolute values by determining the sign of the argument.

You didn't show how you determined that K = 150.

Note: GREAT WORK at showing your work! Very, very, very good. I didn't have to guess at anything. Did I mention this is GREAT?!