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Unformatted text preview: same rate. What is the amount of alcohol in the vat at time t ? Let y ( t ) be the amount of alcohol at time t . Then dy ( t ) dt = rate in-rate out = . 06 5 gal min-y ( t ) 500 5 gal min = 30-y ( t ) 100 gal min . Separating variables we obtain Z 1 30-y ( t ) dy = Z 1 100 dt. Thus-ln | 30-y | = 1 100 t + C. The initial condition is y (0) = 0 . 04 500 = 20 gal. Plugging this in the equation above allows us to nd that C =-ln 10. Therefore we have that ln | 30-y | =-1 100 t + ln 10 . Taking the exponent of this equation we get that | 30-y | = 10 e-1 100 t . Now observe that the y (0) = 0, y ( t ) is continuous and the right hand side is always positive. Thus y = 30-10 e-1 100 t ....
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