Question

Differential Equations, Mixing problem.

A tank contains 1160 L of pure water. A solution that contains 0.03 kg of sugar per liter enters a tank at the rate 7 L/min The solution is mixed and drains from the tank at the same rate.

a.) How much sugar is in the tank initially?
b.) Find the amount of sugar in the tank after t minutes.
c.) Find the concentration of sugar in the solution in the tank after 72 minutes.

Answer 1

I made s(t) be the amount of salt at time 't'

ds/dt = rate in - rate out

Well the rate in would be the concentration times the rate... so

ds/dt = (.03kg)(7L/min) - (s/1160)(7L/min)

ds/dt = .21 - (7s/1160)

Kind of get confused from there...I figure I'll have to get the 's' on one side right? separable?

I wanted to let 7/1160 = a...so I would just have

ds/dt +as = .21


Am I even on the right track here? Just starting out in the class.

Answer 2

I am a bit rusty, but try these notes.

http://tutorial.math.lamar.edu/Classes/DE/Modeling.aspx

Answer 3

Hmm I don't see much that explains what the next step would be...but thanks for the link!

Answer 4

This is a linear Diff. eq correct?

Answer 5

I believe so.

Answer 6

Hmm I'm going to try and follow those notes then...

Integration Factor
\[\large u(t) = e^{\frac{7}{1160}t}\]

Multiply everything by that

\[\large e^{\frac{7}{1160}t} \frac{ds}{dt} + \frac{7}{1160}e^{\frac{7}{1160}t}s = .21e^{\frac{7}{1160}t}\]
Simplify

\[\large (e^{\frac{7}{1160}t}s)^t = .21e^{\frac{7}{1160}t}\]

Integrate both sides

\[\large e^{\frac{7t}{1160}} s + k = 34.8e^{\frac{7t}{1160}} + c\]

Constants on 1 side

\[\large e^{\frac{7t}{1160}} s= 34.8e^{\frac{7t}{1160}} + c\]

Divide by \[\large e^{\frac{7t}{1160}}\]

\[\large s(t) = 34.8 + ce^{\frac{-7t}{1160}}\]

Hows that look? lol

Answer 7

Since we know then that s(0)- 0 we can solve for c

0 = 34.8 + c(1)

so C = 34.8

so my final equation would be

\[\large s(t) = 34.8 + 34.8e^{\frac{-7t}{1160}} \]

Hows this look?

Answer 8

And apparently that isn't correct...sigh

Answer 9

seems okay

Answer 10

as t->infinity ... the solution must converge to input rate. and at t=0, the solution must be in it's initial condition. just note these to know if you are correct or not.

Answer 11

Well i guess that makes sense it isnt correct...when I plug t = 0 into here I don't get 0

Answer 12

yeah and this approaches infinity if I'm correct...so I guess start over lol

Answer 13

did you apply initial condition?

Answer 14

you should get,
http://www.wolframalpha.com/input/?i=y%27+%3D+0.03%2F7+-+x%2F1160%2C+y%280%29+%3D+0

Answer 15

Well yeah, isnt tht what you do to solved for 'c' ?

Since s(0) = 0 ...plugging that into the formula I have give you

0 = 34.8 + c

Which means c would = -34.8

Omg -_- hang on that's it isnt it...I dropped that negative sign before

Answer 16

woops!! something went wrong with MMA.

Answer 17

there you go
http://www.wolframalpha.com/input/?i=y%27+%3D+0.03%2F7+-+y%2F1160%2C+y%280%29+%3D+0

Answer 18

http://www.wolframalpha.com/input/?i=y%27+%3D+0.03%2F7+-+y%2F1160%2C+y%280%29+%3D+0

Answer 19

Well then at 72 minutes that would give me .3 which is incorrect

Answer 20

\[
\large s(t) = 34.8 + ce^{\frac{-7t}{1160}} \\
0 = 34.8 + ce^{0} \implies
-34.8 = c\]
rest is okay.

Answer 21

probably i made miscalculation on concentration ... rest is okay.

Answer 22

Yeah I finally got the equation down...just now at 72 minutes my equation is messing me up

I get 12.26 ...that seems correct to me...but says incorrect online...sigh I'm going to hate this class lol

Answer 23

Wait a minute...I finding the concentration of sugar in the tank after 72 minutes, Is that different than the AMOUNT of sugar in the tank after 72 minutes?

Answer 24

amount of sugar = volume times concentration ... of course.

Answer 25

oh crap that's right...lol thank you for walking me through dumb mistakes like that @experimentX

Answer 26

yw

Answer 27

And that's correct! Thank's again!