- #1

- 7

- 0

## Main Question or Discussion Point

I was hoping that I could find help solving this problem. I've been working on it for a while and haven't been able to solve it correctly.

[tex]

\frac{dQ}{dt} = r - \frac{rQ}{100}

[/tex]

Where Q is the amount of salt, r is the rate in and out.

I think that this may work for the first 10 minutes while the salt is part of what is getting mixed in, but after that I'm lost. Any help will be appriciated.

Thank you.

So far I have:A tank originally contains 100 gal of fresh water. Then water containing 0.5 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After ten minutes the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min.

[tex]

\frac{dQ}{dt} = r - \frac{rQ}{100}

[/tex]

Where Q is the amount of salt, r is the rate in and out.

I think that this may work for the first 10 minutes while the salt is part of what is getting mixed in, but after that I'm lost. Any help will be appriciated.

Thank you.

Last edited: