Applications of Calculas Question :) (1 Viewer)

[xionethic]

This is annoying me because partially, the example i can find disincludes some factors in the actual question im doing, can any of you explain me how you get the intial conditions for:

"A tank contains 100L of brine whose concentration is 3grams/litre. Three litres of brine whos concentration is 2grams/litre flow into the tank each minute and at the same time 3 ltres of mixture flow out each minute."
a) Show that the quantity of salt, Qgrams, in the tank at anytime t is given by

Q = 200 + 100e^(-0.03t)
Notes- for any of you who don't know, brine is salty water.

 

[xionethic]

I did the working process backwards, i figured out most things but i dont understand how initially t=0 Q=100 so the c = 100/3 ln(300)

 

[Studentleader]

Q = 200 + 100e^(-0.03t)

alright so when t = 0
Q = 300 and since 100,000mL x 0.03grams = 300 we know it works for initial.

since we have 3L (2 grams/L) flowing in and 3L of it flowing out (100(L)e^(-0.03xt).

I give up :l

I think you'd needa use t=0 to solve for that constant k, and i dont understand how the 2 grams/L is in the equation.

Hope I helped, and sorry

 

[xionethic]

Nvm i think i figured it out, initiall t=0 Q=300 (i stopped after that because you can't logbase e a negative number) so if you continue it will become evident that you just use your logarithm rules. Damn sunday night and im doing maths god, it's annoying.

 

[xionethic]

Here is the working=

we know that rate of inflow is 3L and 2g per litre, hence 6grams total inflow.
Outflow is Q/100 x 3 (which means Q unit of salt per litre is following out)
because, in the tank it contains 100L, for now Q is the amount of salt at any time. In addition 3Litres is following out, that is why we multiply it by 3.



dQ/dt = 6 - 3Q/100 (rate of inflow - rate of outflow) rate of flow is 6grams in (hence 3L or 2g density)
= (600 - 3Q)/100

dt/dQ = 100/(600 - 3Q)

t = -100/3 x ln(600-3Q) + c

When t = 0 Q = 300 (Initial salt in tank is 3grams x 100 litres)

therefore c = 100/3 x ln(-300) (ignoring the fact that that cannot occur for now)

t = -100/3 x ln(600-3Q) + 100/3 x ln(-300)
t = 100/3 x ln( -300/(600 - 3Q) )

e^(0.03t) = -300/(600 - 3Q)
-- Simplify and you get
Q = 200 + 100e^(-0.03t)


TADAAAAAAAAAAAAAAAAAAAAAAA (I Think)

 

[Studentleader]

Nvm i think i figured it out, initiall t=0 Q=300 (i stopped after that because you can't logbase e a negative number) so if you continue it will become evident that you just use your logarithm rules. Damn sunday night and im doing maths god, it's annoying.

Looked at your working and well done, didn't look at flows as a derivative :S

Well done, i'm really rusty on exponentials

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Ignore?


(My maths here may be a bit wrong)

Doesn't e^(k.t)
k < 0 = Decay
k > 0 = The opposite to decay (w/e it is)

and since we have 6 grams of salt coming into the tank (in 3L) and initially 9 grams of salt coming out, we can see that the system will limit 100L of 2 grams per litre when all the 3 grams per L is removed?

 

[xionethic]

Looked at your working and well done, didn't look at flows as a derivative :S



-----------------------


Ignore?


(My maths here may be a bit wrong)

Doesn't e^(k.t)
k < 0 = Decay
k > 0 = The opposite to decay (w/e it is)

and since we have 6 grams of salt coming into the tank (in 3L) and initially 9 grams of salt coming out, we can see that the system will limit 100L of 2 grams per litre when all the 3 grams per L is removed?

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Thanks, i kinda figured that out, but when i took that into account, things didn't work out as i got Q = 200 - 100e(-0.03t) , in this case that is incorrect according to the solution. But you do make a good point, that's what i was thinking before i used the other method.