IV) 17. 9.3 exercise 45
A tank contains 1000L of brine with 15kg of dissolved salt. Pure water enters the tank at
a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at
same rate. How much salt is in the tank:
a) after t minutes?
b) after 20 minutes?
I first referred back to example 6 for the equation:
dy = (rate in) – (rate out)
dt
And then drew a sketch of the tank:
*Let y(t) be the amount of salt in the tank at time t
10L/min ________ 10L/min
=====| 1000L |=====
0 kg/L | brine | y(t) kg
|15 kg salt| 1000 L
-------------dy = (0) – y(t)
dt
100

rate in = ( 0 kg / L ) (10 L / min) = 0 kg / min
rate out = ( y(t) /1000 kg/L ) ( 10 L / min) = y(t) / 100
dy = - y(t)
dt
100

dy = - 1 dt
y(t)
100
∫ dy / y(t) = ∫ - (1 / 100) dt
ln y(t) = - t + C
100
Since 15 kg of brine was in the tank originally at time t = 0; y(0) = 15.
thus: ln y(0) = - 0 + C  C = ln 15
100
So, ln y = ln 15 – (t/100)
y = 15e^(-t / 100) kg
After 20 minutes:
y = 15e^(-0.2)
y = 12.281
y ≈ 12.3
exponentiation to the e