document.write( "Question 1072004: Please help me solve this problem .
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document.write( "A 2000-L tank is initially filled with a copper sulfate solution with a concentration of 60g/L . A copper sulfate solution with a concentration of 20g/L flows into the tank at rate of 8L/min . The thoroughly mixed solution is drained from the tank at rate 8L/min .
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document.write( "A)write an initial value problem for the mass of the copper sulfate .
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document.write( "B)Solve the initial value problem . Set up .\r
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document.write( "This is how i solved it but i might be 90% wrong .Please feel free to correct anything i did wrong .\r
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document.write( "Let y(t) be the amount of sulfate .
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document.write( "y(0)=60
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document.write( "dy/dt=(rate in)-(rate out)
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document.write( "rate in =(20g/L)(8L/min)=160g/min
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document.write( "rate out =y(t)g/2000L)(8L/min)=y(t)g/250min
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document.write( "dy/dt=160-(y(t)/250)
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document.write( "dy/dt=(40000-y(t))/250
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document.write( "taking integral both side
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document.write( "int(dy/40000-y)=int(dt/250)
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document.write( "-ln(40000-y)=t/250+c
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document.write( "since y(0)=60 , we have -ln39940=c
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document.write( "-ln(40000-y)=t/250-ln39940
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document.write( "since y(t)is continuous and y(0)=60 , so 40000-y(t) is positive . Thus 40000-y=40000-y
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document.write( "y(t) 40000-39940e^(-t/250).\r
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document.write( "*Please any help will be very appreciate . Thank
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document.write( "
Algebra.Com's Answer #686930 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! You can't integrate both sides. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "General solution, \n" ); document.write( " \n" ); document.write( "Particular solution, \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |