(a) How much salt will be in the tank after 25 minutes?

(b) How long will it take for there to be 100 pounds of salt in the tank? (NOTE: there will be two answers.)

(c) How much salt will be in the tank after 100 minutes have passed?

- Thread starter AmyA
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(a) How much salt will be in the tank after 25 minutes?

(b) How long will it take for there to be 100 pounds of salt in the tank? (NOTE: there will be two answers.)

(c) How much salt will be in the tank after 100 minutes have passed?

\(\displaystyle 200+(2-4)t=2(100-t)\)

Now the amount coming in is:

\(\displaystyle \left(2\frac{\text{gal}}{\text{min}}\right) \left(2\frac{\text{lb}}{\text{gal}}\right)= 4\frac{\text{lb}}{\text{min}}\)

And the amount going out is:

\(\displaystyle \left(4\frac{\text{gal}}{\text{min}}\right) \left(\frac{A}{2(100-t)}\frac{\text{lb}}{\text{gal}}\right)=\frac{2}{100-t}\frac{\text{lb}}{\text{min}}\)

And so this leads to the IVP:

\(\displaystyle \frac{dA}{dt}=4-2\frac{A}{100-t}\) where \(\displaystyle A(0)=20\)

Can you proceed?

Start by setting it up as a "differential equation". Let X(t) be the amount of salt, in pounds, in the tank at time t in minutes. The rate of change of X, dX/dt, is the rate at which salt is coming into the tank minus the rate at which salt is going out of the tank. The rate at which salt is going into the tank is 2 pounds per gallon times 2 gallons per minute= 4 pounds per minute.

The rate at which salt is going out of the tank is a little more complicated. It is the number of pounds per gallon times 4 gallons per minute.

One complication is that, because water is coming in at 2 gallons per minute but going out of the tank at 4 gallons per minute so the amount of water in the tank is reducing at 2- 4= -2 gallons per minute. The tank originally contains 200 gallons so after t minutes there are 200- 2t gallons of water in the tank. With X(t) pounds of salt in the tank there are \(\displaystyle \frac{X(t)}{200- 2t}\) "pounds per gallon". The salt is going out of the tank at the rate of \(\displaystyle -4\frac{X(t)}{200- 2t}\).

Putting those together, \(\displaystyle \frac{dX}{dt}= 4- \frac{X}{200- 2t}\). Solve that differential equation with the initial condition X(0)= 20.

Once you know X(t)

a) What is X(25)?

b) Solve X(t)= 100 for t.

c) What is X(100)?

I think your ODE should be:triedthis yourself? You have posted several problems where you just state the problem. That violates the rules of this forum.

Start by setting it up as a "differential equation". Let X(t) be the amount of salt, in pounds, in the tank at time t in minutes. The rate of change of X, dX/dt, is the rate at which salt is coming into the tank minus the rate at which salt is going out of the tank. The rate at which salt is going into the tank is 2 pounds per gallon times 2 gallons per minute= 4 pounds per minute.

The rate at which salt is going out of the tank is a little more complicated. It is the number of pounds per gallon times 4 gallons per minute.

One complication is that, because water is coming in at 2 gallons per minute but going out of the tank at 4 gallons per minute so the amount of water in the tank is reducing at 2- 4= -2 gallons per minute. The tank originally contains 200 gallons so after t minutes there are 200- 2t gallons of water in the tank. With X(t) pounds of salt in the tank there are \(\displaystyle \frac{X(t)}{200- 2t}\) "pounds per gallon". The salt is going out of the tank at the rate of \(\displaystyle -4\frac{X(t)}{200- 2t}\).

Putting those together, \(\displaystyle \frac{dX}{dt}= 4- \frac{X}{200- 2t}\). Solve that differential equation with the initial condition X(0)= 20.

Once you know X(t)

a) What is X(25)?

b) Solve X(t)= 100 for t.

c) What is X(100)?

\(\displaystyle \frac{dX}{dt}= 4-4\frac{X}{200- 2t}\)

And this is equivalent to what I posted.

Hello, Amy!I havetiredand yes I did post 2 problems that I had questions on. I think I have the problems figured out.

Thank you for your response and it appears that I was correct in my thought process.

What most math helpers want from people posting questions is to see what they've tried so they can see where the issue preventing them from completing the exercise might be. Unfortunately, there are so many people who just want someone else to do their homework, that when someone posts a question without any effort shown, that's what it might look like. A student who posts the question along with what they've tried, or what their thoughts are on how to begin is much more likely to receive timely help.