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hocuspocus102

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## Homework Statement

A tank contains 2860 L of pure water. A solution that contains 0.04 kg of sugar per liter enters a tank at the rate 5 L/min The solution is mixed and drains from the tank at the same rate. Find the amount of sugar in the tank after t minutes.

## Homework Equations

## The Attempt at a Solution

Obviously it starts with 0 kg sugar and an initial condition at time t = 0.

The rate then would be equal to (the rate entering) - (the concentration exiting).

Entering = .04*5 = .2

Exiting = (Y*5)/2860 = Y/572

dY/dt = .2 - Y/527

dY/(.2 - Y/572) = dt

integrate both sides

-572ln(.2-Y/572) = t + C

simplify

ln(.2 - Y/572) = -t/572 + C

.2 - Y/572 = e^(-t/572+C)

-Y/572 = Ce^(-t/572) - .2

Y = Ce^(-t/572) + 114.4

initial value t = 0 and Y = 0

0 = Ce^(-0/572) + 114.4

C = -114.4

so Y = 114.4 - 114.4e^(-t/572)

Is this right because I've done it a bunch of times and keep getting the same answer, but it's an online problem and it tells me that it's wrong and I can't figure out what I did incorrectly. Thanks!