# Thread:

1. ## Function of Time Rate Problem

I have a claculus word problem that looks like this....
A tank contains 2220L of pure water. A solution that contains 0.02kg of sugar per liter enters a tank at the rate 6L/min. The solution is mixed and drains from the tank at the same rate. Initially there is no sugar in the solution.

Find an equation for the amount of sugar in the tank after t minutes (A function of t)

I have calculuated so far:

(0.02kg/l)(6L/min)-(y/2220L)(6L/min)
dy/dt= 0.12kg/min -(64y/2220)
The from here I know to take the antiderivatives of both sides:
(This is where I begin to get confused)
-34.68ln(0.12-64y/2220) = t + C
ln(0.12-64y/2220)=-t/34.68 + C
This is where I am not sure how to get any farther....
If anyone could help me out I would greatly appreciate it.

2. The rate of change of the amount of sugar in the tank at time t is given by

dy/dt=rate in - rate out.

$\text{rate in} = (\frac{1}{50})(6)=\frac{3}{25} \;\ \text{kg/min}$

$\text{rate out} = \frac{y}{2220}(6)=\frac{y}{370} \;\ \text{kg/min}$

Therefore, $\frac{dy}{dt}+\frac{y}{370}=\frac{3}{25}$

The integrating factor is $e^{\int\frac{1}{370}dt}=e^{\frac{t}{370}}$

$ye^{\frac{t}{370}}=\frac{222}{5}e^{\frac{t}{370}}+ C$

$y=\frac{222}{5}+Ce^{\frac{-t}{370}}$

Since y(0)=0, then $C=\frac{-222}{5}$

$\boxed{\frac{222}{5}-\frac{222}{5}e^{\frac{-t}{370}}}$

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