Chapter 3
and so the mass of salt in the tank after
t
minutes is
x
(
t
)=5
−
4
.
5
e
−
2
t/
25
.
If the concentration of salt in the tank is 0
.
02 kg/L, then the mass of salt is 0
.
02
×
100 = 2 kg,
and, to Fnd this moment, we solve
5
−
4
.
5
e
−
2
t/
25
=2
⇒
e
−
2
t/
25
=
2
3
⇒
t
=
25 ln(3
/
2)
2
≈
5
.
07 (min)
.
3.
Let
x
(
t
) be the volume of nitric acid in the tank at time
t
. The tank initially held 200 L
of a 0
.
5% nitric acid solution; therefore,
x
(0) = 200
×
0
.
005 = 1. Since 6 L of 20% nitric
acid solution are ﬂowing into the tank per minute, the rate at which nitric acid is entering
is 6
×
0
.
2=1
.
2 L/min. Because the rate of ﬂow out of the tank is 8 L/min and the rate
of ﬂow in is only 6 L/min, there is a net loss in the tank of 2 L of solution every minute.
Thus, at any time
t
, the tank will be holding 200
−
2
t
liters of solution. Combining this
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Nitric acid

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