Thread: Mixture problem "A tank contains..."

1. Mixture problem "A tank contains..."

This is a first order differential equation, but it's taught in calc 2. If I'm in the wrong forum, please move me!

It is in verbose detail because I'm having trouble with both problems ( 2 problems, 2 hours... ) and hope that it's a similar issue so I can fix both.

"A tank contains 100 gal of fresh water. A solution containing 1 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min and the mixture is pumped out of the tank at the rate of 3 gal/min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum."

The book's given answer is "y(27.8)=14.8 lb, t = 27.8 min" verbatim.

So, I start out by writing dy/dt = 1 gal/min * 1 lb/gal - 3 gal/min * Y/(100+(1-3)*t) lb/gal.

I simplified that to dy/dt = 1 lb/min - 3*y/(100-2t) lb/min.

Then I chucked my units of measurement and put it into standard form.

dy/dt + 3Y/(100-2t)=1

I used the integrating factor e^integral(3/(100-2t) = e^(3*ln(100-2t) =
(100-2t)^-3

This gave me d((100-2t)^-3*y)=(100-2t)^-3.

I integrated, got (100-2t)^-3*y=1/4 * (100-2t)^-2 + C

I multiplied across by (100-2t)^3 and ended up with y=1/4 * (100-2t) + C.

y(0)=100, which means 100=100/4+C, or C=75.

And now I'm completely cornered. Taking the derivative of what I have gets me -1/2, so I can't find out when the slope is 0 and the tank is full, nor go any further with the problem. Where did I go wrong in it?

2. For the integral

$\displaystyle \displaystyle\int\frac{3\,dt}{100-2t},$

you have to do the $\displaystyle u=100-2t$ substitution before you can use the logarithm rule. That is, $\displaystyle du=-2\,dt,$ and the integral becomes

$\displaystyle \displaystyle-\frac{3}{2}\int\frac{du}{u}=-\frac{3}{2}\,\ln|u|=-\frac{3}{2}\,\ln|100-2t|.$

And yes, this post does belong in Differential Equations, despite being taught in Calc II. This forum is more organized around subject matter than courses. Report your own post, and a mod will move it for you.

Cheers.

3. Even with that ( and I used the calculator and wolfram alpha given answers too ), I'm solving and finding that I have my integrating factor's inverse (100-2t)^(3/2) multipled to C, and when solved for C with initial value y(0)=100, C=0, and the only thing I have left is something to the first power which gives me a constant when I differentiate which means I can't solve the problem.

I need a bit more in depth help here!

4. Ok, so the integrating factor is

$\displaystyle e^{-(3/2)\ln|100-2t|}=(100-2t)^{-3/2}.$ Note the negative exponent.

Multiplying through by that factor yields

$\displaystyle \displaystyle\frac{y'(t)}{(100-2t)^{3/2}}+\frac{3y(t)}{(100-2t)^{5/2}}=\frac{1}{(100-2t)^{3/2}},$ which implies that

$\displaystyle \displaystyle d\left(\frac{y(t)}{(100-2t)^{3/2}}\right)=\frac{dt}{(100-2t)^{3/2}},$ and hence

$\displaystyle \displaystyle\frac{y(t)}{(100-2t)^{3/2}}=\frac{1}{(100-2t)^{1/2}}+C.$

Therefore,

$\displaystyle y(t)=100-2t+C(100-2t)^{3/2}.$

Is that what you got? Can you finish from here?

5. Nope, still lost. I get to there in the problem and that's my issue. The book's answer is y(27.8)=14.8 lb, t = 27.8 min.

y(0)=100 if I have my initial value right, which gives me 100=100-0+C(100-0)^3/2 which when solved gives me C=0. Substituting back for C I get y(t)=100-2t+0*...

The instructor told us to remember the maximum amount of fertilizer is when y'(t)=0, however, differentiating this leaves me with y'(t)=0-2 or y'(t)=-2.

Now I'm getting worried, was I supposed to look for the slope = 0 somewhere else in the problem?

6. Your initial condition can't possibly be right. Is $\displaystyle \displaystyle y$ representing the amount of fresh water or the amount of fertiliser?

7. The tank contains 100 gal of fresh water and a solution containing 1 lb/gal of fertilizer runs into the tank at the rate of 1 gal/min. It's pumped out at a rate of 3 gal/min. I have to find the maximum amount of fertilizer in it and the time required to reach that maximum.

dy/dt = 1 gal/min * 1 lb/gal - 3 gal/min * Y/(100+(1-3)*t) lb/gal

So Y is the amount of fertilizer. . .

:\ It's 1:10 AM on a Saturday night somewhere in an Oklahoman city. A resounding titanic thud shakes the ground as one calculus student realizes he's blown any number of hours fretting over a set of differential equations problems because his initial condition was wrong and he didn't think things through despite writing out all of his units of measurement when he set up the problem.

y(0)=0?

8. Yes, $\displaystyle \displaystyle y(0) = 0$ is the initial condition.