Question

Mixing Problem... I know I have derived the correct equation from the paragraph. I just need help with the differentiation.

Answer 1

\[\frac{ dx }{ dt }=.4-\frac{ 2x }{ 25 }\rightarrow \frac{ 10-2x }{ 25 }\]

Answer 2

I need it in the form x(t)

Answer 3

Im thinking separation of variables

Answer 4

\[\int\limits_{}^{}\frac{ dx }{ 10-2x }=\int\limits_{}^{}\frac{ dt }{ 25 }\]

Answer 5

\[-\frac{ 1 }{ 2 }\ln (10-2x)=\frac{ t }{ 25 }+C\]

Answer 6

If you see a mistake stop me or else I'm going to keep going

Answer 7

thats correct, go on...

Answer 8

x(0) = 0.005

Answer 9

based on the problem.

Answer 10

Now this is where i get tripped up. Do I have to get it in the form x(t)=, or can I just plug .005 and 0 in right now?

Answer 11

u can plug it right now...

Answer 12

u will get C

Answer 13

ok C = ln 9.99

Answer 14

nopes.

Answer 15

I need to find t when x(t)=.02

Answer 16

ok let me try again

Answer 17

-(ln 9.99)/2

Answer 18

the first time I multiplied the right side by -2 and assumed C could just absorb the -2 because it is a constant times a constant. That is breaking a rule?

Answer 19

now u are correct and yes.

Answer 20

So now I just simply plug in .02 for x and solve for t

Answer 21

exactly!

Answer 22

The back of the book says (0.4)(100-t)-(3.9E-7)(100-t)^4; t=19.96mins. 100L is the volume of the tank, .4 is the initial concentration. But I dont see how they got the rest of their equation

Answer 23

I do feel like we did it right

Answer 24

in flow and out flow are the same

Answer 25

if there's 100-t^4 then there is some thing missing in your original equation....if your original equation is correct, then we are correct.

Answer 26

in flow rate is 8L/min and is equal to the out flow. The volume of the tank is kept constant at 100L. Initially there is 0.5kg of salt in the tank. Concentration of salt entering the tank is 0.05kg/L.

Answer 27

So what I did was dy/dx= Conc(initial)*InFlow rate - outFlow rate*x(t) over 100L

Answer 28

where x(t) is the kg of salt. So when divide by the Volume gives me the final concentration.

Answer 29

i donno how to form the equation with your given word problem.....but if your first equation is correct, then all your next math part is correct...

Answer 30

I copied the setup from a very similar problem in the book so I think my first equation is right. The book went from \[\frac{ dy }{ dx }=0.6-\frac{ 3x }{ 500 }\rightarrow x(t)=100(1-e ^{-3t/500)}\]This said it gives them the weight of the salt then they divided the equation by the volume to get the equation for the concentration of salt at time t. I forgot to do that part, but my integration doesn't look like there's

Answer 31

Thanks for your help. If you want to pull the rip cord on this question and help others I'd understand

Answer 32

thanks and sorry....

Answer 33

my example above is from a different but similar problem

Answer 34

no big