# Mixtures

• Jan 30th 2011, 04:50 PM
VonNemo19
Mixtures
A large tank is filled to capacity with 500 gallons of pure water. Brine (a mixture of salt and water) containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gallons per minute. The well-mixed solution is pumped out at a rate of 10 gallons per minute. Find the number $\displaystyle A(t)$ of pounds of salt in the tank at time $\displaystyle t$.

OK. I know that the rate at which the amount of salt is changing inside the tank with respect to time - $\displaystyle \frac{dA}{dt}$ - is simply the net rate of the amount being pumped in and the amount being pumped out, that is

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

Now, $\displaystyle R_{in}=(2\frac{lbs}{gal})\cdot(5\frac{gal}{min})=1 0\frac{lbs}{min}$

and I think that $\displaystyle R_{out}=\frac{A(t)lbs}{(500-5t)gal}\cdot10\frac{gal}{min}$

Assuming my $\displaystyle R_{out}$ is correct, this would imply the DE

$\displaystyle \frac{dA}{dt}+\frac{5}{100-t}A=10$ which is linear with the integrating factor

$\displaystyle e^{\int\frac{5}{100-t}dt}=e^{-5\ln|100-t|}=e^{\ln(100-t)^{-5}}=(100-t)^{-5}$.

So, I have

$\displaystyle \int\frac{d}{dt}\left[(100-t)^{-5}A\right]dt=\int10(100-t)^{-5}dt$

Am I on the right track?
• Jan 30th 2011, 04:55 PM
dwsmith
Quote:

Originally Posted by VonNemo19
A large tank is filled to capacity with 500 gallons of pure water. Brine (a mixture of salt and water) containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gallons per minute. The well-mixed solution is pumped out at a rate of 10 gallons per minute. Find the number $\displaystyle A(t)$ of pounds of salt in the tank at time $\displaystyle t$.

OK. I know that the rate at which the amount of salt is changing inside the tank with respect to time - $\displaystyle \frac{dA}{dt}$ - is simply the net rate of the amount being pumped in and the amount being pumped out, that is

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

Now, $\displaystyle R_{in}=(2\frac{lbs}{gal})\cdot(5\frac{gal}{min})=1 0\frac{lbs}{min}$

and I think that $\displaystyle R_{out}=\frac{A(t)lbs}{(500-5t)gal}\cdot10\frac{gal}{min}$

Assuming my $\displaystyle R_{out}$ is correct, this would imply the DE

$\displaystyle \frac{dA}{dt}+\frac{5}{100-t}A=10$ which is linear with the integrating factor

$\displaystyle e^{\int\frac{5}{100-t}dt}=e^{-5\ln|100-t|}=e^{\ln(100-t)^{-5}}=(100-t)^{-5}$.

So, I have

$\displaystyle \int\frac{d}{dt}\left[(100-t)^{-5}A\right]dt=\int10(100-t)^{-5}dt$

Am I on the right track?

You should have

$\displaystyle \displaystyle A'+\frac{2A}{100-t}=10$
• Jan 30th 2011, 05:00 PM
VonNemo19
Oh, duh. But everything else is OK?
• Jan 30th 2011, 05:01 PM
dwsmith
Quote:

Originally Posted by VonNemo19
Oh, duh. But everything else is OK?

Yeah once you change your P(t) numbers.