Problem 1 (Cascading tanks). Consider the following system...

Problem 1 (Cascading tanks). Consider the following system of two tanks: Tank A Tank B At t = 0, Tank A contains 100 gallons of a salt solution with concentration 0.5 lb/gal, and Tank B initially contains 100 gallons of pure water. Fresh water is poured into Tank A at a rate of 4 gal/min, and the resulting solution flows from Tank A to Tank B at a rate of 6 gal/min. Part of the solution contained in Tank B then flows back into Tank A at a rate of 2 gal/min, while the rest flows out of Tank B at a rate of 4 gal/min. (a) Write down a system of differential equations, together with an initial condition, that describes the amount of salt in each of the tanks. (b) Solve the system from part (a). (c) Describe the behavior of the salt content in each tank as t ? ?.

Transcript

00:01 Hello students, here we have tang a which contains 100 gallons of salt solution.

00:07 And we have tang b, which contains 100 gallons of pure water.

00:12 Okay.

00:13 So four gallons per matter, rate of four gallons per meter, water is poured to tang a.

00:20 From it is tang a to tang b.

00:23 It is poured at the rate of 6 gas per meter.

00:27 And from tang b to tang a, it is two gallons.

00:30 Gallon per meter and the left is node through tang b which is 4 gallons per meter.

00:37 We have to find a differential equation for this and we have to solve the system and we have to find the behavior of the salt content in each tank as t tends to infinity.

00:49 So let me start again.

00:52 So let dx over dt be the concentration of the salt rate in tang a.

01:03 So it will be salt in minus salt out.

01:12 And we have the equation for the salt which is concentration, concentration multiplied by rate.

01:29 Correct.

01:31 So here it will be equal to y, again, y minus 2 y divided by 100 minus 6x, divided by 100.

01:49 Similarly, we have for the tag b, let it be y d .y over d b.

01:57 D .y over d .y over d .t is equal to we have x by 100 multiplied by 6 minus y by 100 multiplied by 6 again.

02:15 Here x is the tag a and y is that of tag b.

02:19 So this is the answer to the first part, part a.

02:25 Okay, now we have just all this differential equation.

02:29 So here we can write this equation in the matrix form as dx over dt.

02:36 Here it's d .y over dt equal to minus 0 .06, 0 .02, 0 .06 .0 .6 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0.

02:51 Minus 0 .06 multiplied by x x x and y.

03:01 X, comma, okay, x, comma, y.

03:04 Okay, x comma y.

03:08 So here we have the equal values.

03:13 Solving this equation, we get the equal values as negative 0 .02 .54 and negative 0 .4 .2 .54 and negative 0 .4 .2...

Question

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