# Homework Help: Help ?

1. Jan 14, 2012

### sam_0017

Help.... ???

A 200 liter tank initially contains 100 liters of water with a salt concentration of 0.1 grams per liter.
Water with a salt concentration of 0.5 grams per liter flows into the tank at a rate of 20 liters per
minute. Assume that the fluid is mixed instantaneously and that this well-mixed fluid is pumped out
at a rate of 10 liters per minute. Let c (t) and
v(t), be the concentration of salt and the volume of
water in the tank at time t (in minutes), respectively. Then,
v(t)=10
v(t) c(t) +20c(t)=10

a) Solve these differential equations to find the particular solutions for v(t) and c(t).
b) What is the concentration of salt in the tank when the tank first overflows?

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v(0)= 100 L , c(0) = 0.1 gm/L , c(t)=0.5 gm/L

v= 10 ==> v (t) = 10t + k
==> v(0) = 0 + k =100 ==> k=100
so : v(t) = 10 t + 100
(mass) m= c(t) * v(t)
m=c(t)v(t) + c(t) v(t)
10 = c(t)(10t+100)+ c(t) *10
c(t) + c(t) (10/(10t+100)) = 10/(10t+100)
by solving this DE :
e^∫(1/ t+10) dx = t+10
Multiplying through by both sides gives:
(t+10)c`(t)+c(t)=1
∂/∂t{c(t) (t+10)} =1

==> by integral
c(t) ( t+ 10) = t
==> c(t) = t/(t+10) +k
c(0) = 0.1
==>
0.1= 0 + k
so c(t) = t/(t+10) +0.1

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