Solving Mixture Problem ODE Homework

[dipset24]

Homework Statement

A 500 gallon tank is filled with pure water. A solution containing 4lbs of salt per gallon is added at a rate of 2 gallons per minute. The well-mixed solution is drained at a rate of 3 gallons per minute.

A) How long does it take for the container to achieve a concentration of 2lbs/gal?

B) What is the domain of x(t)?

Homework Equations

ODE

The Attempt at a Solution

I found x(t) to be (500-t)^2*(4000-t^2)

then I tried using x(t)/v(t)=2 lbs/gal and this is where I am stuck

 

[Dick]

If x(t)=(500-t)^2*(4000-t^2), then if t=0 then x(0)=500^2*4000. What does that mean? What is x(t) supposed to be? What ODE did you solve? How did you get that strange answer?

 

[dipset24]

x'+(3/(500-t))x=8 was the DE I used

 

[Dick]

I'm going to guess x(t) is the amount of salt. The ODE looks right. The solution doesn't. Can you show us more of how you solved it?

 

[dipset24]

Integrating Factor

e^$$\int(3/(500-t))$$ dt = (500-t)^(-3)

Dt$$\int [(500-t)^(-3)*x]dt$$=8$$\int(500-t)^(-3)$$


(500-t)^(-3) X=4/(t-500)^2 + C

 

[Dick]

Integrating Factor

e^$$\int(3/(500-t))$$ dt = (500-t)^(-3)

Dt$$\int [(500-t)^(-3)*x]dt$$=8$$\int(500-t)^(-3)$$(500-t)^(-3) X=4/(t-500)^2 + C

That seems ok again. Now what's C? What is x(0) supposed to be?

 

[HallsofIvy]

Homework Statement

A 500 gallon tank is filled with pure water. A solution containing 4lbs of salt per gallon is added at a rate of 2 gallons per minute. The well-mixed solution is drained at a rate of 3 gallons per minute.

A) How long does it take for the container to achieve a concentration of 2lbs/gal?

B) What is the domain of x(t)?

Surely this is not the exact statement of the problem! There is NO "x(t)" in the original statement. What does x(t) mean here?

Homework Equations

ODE

The Attempt at a Solution

I found x(t) to be (500-t)^2*(4000-t^2)

then I tried using x(t)/v(t)=2 lbs/gal and this is where I am stuck

 

[dipset24]

Integrating Factor

e^$$\int(3/(500-t))$$ dt = (500-t)^(-3)

Dt$$\int [(500-t)^(-3)*x]dt$$=8$$\int(500-t)^(-3)$$(500-t)^(-3) X=4/(t-500)^2 + C

The initial conditions are x(0)=0 since it is all pure water at time 0

After solving using the integrating factor my x(t)(pounds of salt)= -4(t-500)+c(t-500)^3

 

[Dick]

The initial conditions are x(0)=0 since it is all pure water at time 0

After solving using the integrating factor my x(t)(pounds of salt)= -4(t-500)+c(t-500)^3

Ok, but now you have to find C using x(0)=0. Then find where the concentration x(t)/(500-t)=2lbs/gallon, right?

 

[dipset24]

Ok, but now you have to find C using x(0)=0. Then find where the concentration x(t)/(500-t)=2lbs/gallon, right?

Yes that is where I am caught up I tried several ways to get c and then plug it into x(t)/(500-t)=2lbs/gallonThe c I come up with is .000016

 

[Dick]

Yes that is where I am caught up I tried several ways to get c and then plug it into x(t)/(500-t)=2lbs/gallon


The c I come up with is .000016

That's gives you x(0)=0 doesn't it? What's wrong with that?

 

[dipset24]

That's gives you x(0)=0 doesn't it? What's wrong with that?

Yes but when I plug that into my X(t)/v(t)

(-4(t-500)+.000016(t-500)^3)/((500-t)=2

which I can reduce to 4-.000016(t-500)^2=2

but the answer t=853.55 is not correct I just need some help verifying my answer

 

[Dick]

Yes but when I plug that into my X(t)/v(t)

(-4(t-500)+.000016(t-500)^3)/((500-t)=2

which I can reduce to 4-.000016(t-500)^2=2

but the answer t=853.55 is not correct I just need some help verifying my answer

It's a quadratic equation. It has two solutions, not just one. Can you find the other one?

 

[dipset24]

It's a quadratic equation. It has two solutions, not just one. Can you find the other one?

Yea i am sorry t=146.44 t= 853.553
However doesn't t have to be less then 500 going all the way back to the integrating factor.

e^(-3*ln(500-t)) t<500?

 

[Dick]

Yea i am sorry t=146.44 t= 853.553
However doesn't t have to be less then 500 going all the way back to the integrating factor.

e^(-3*ln(500-t)) t<500?

Sure, t has to be less than 500. At t=500 the tank is drained. I think that's the sense to the question about what is the domain of x(t). The ODE has a solution outside of that range, but it doesn't make physical sense.

 

[dipset24]

Sure, t has to be less than 500. At t=500 the tank is drained. I think that's the sense to the question about what is the domain of x(t). The ODE has a solution outside of that range, but it doesn't make physical sense.

Yes thank you so the Domain is [0,500] and when t=146.446609 min our concentration is 2lbs/gal