q3 - Q ( t ) = inow - outow = (0 . 2)(2)-(2)( Q 200 ) or Q...

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MA366 Sathaye Quiz 3 Name/ Sec./Row A tank originally contains 200 liters of fresh water. Water containing 0 . 2kg salt/liter is pumped in at a rate of 2 liters per minute and the mixture is allowed to leave the tank at the same rate. Answer the following: 1. Find the formula for the amount of salt in the tank at time t for 0 t 30. 2. Find the amount of salt in the tank at the end of 50 minutes. 3. If the pumping is continued indefinitely, what is the limiting amount of salt? 4. When will the resulting amount of salt equal half the limiting amount found above? 1
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Key A tank originally contains 200 liters of fresh water. Water containing 0 . 2kg salt/liter is pumped in at a rate of 2 liters per minute and the mixture is allowed to leave the tank at the same rate. Answer the following: 1. Find the formula for the amount of salt in the tank at time t for 0 t 30. Answer: Let Q ( t ) be the quantity of salt at time t . Then the equation is:
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Unformatted text preview: Q ( t ) = inow - outow = (0 . 2)(2)-(2)( Q 200 ) or Q + Q 100 = 0 . 4 . Integrating factor is = exp( Z dt 100 ) = exp( t 100 ) . Then d dt (exp( t 100 ) Q ) = Z . 4 exp( t 100 ) dt. This gives exp( t 100 ) Q = . 4 1 / 100 exp( t 100 ) + c and thus Q ( t ) = 40 + c exp(-t 100 ) . Initial condition Q (0) = 0 gives c =-40 so the nal answer is Q ( t ) = 40(1-exp(-t 100 )) . 2. Find the amount of salt in the tank at the end of 50 minutes. Answer: Calculate Q (50) = 40(1-exp(-50 / 100)) = 15 . 7387761. 3. If the pumping is continued indenitely, what is the limiting amount of salt? Answer: The limit is clearly 40(1-0) = 40. 4. When will the resulting amount of salt equal half the limiting amount found above? Answer: Solve Q ( t ) = 20 which gives 1 2 = 1-exp(-t 100 ) . It gives t = 100 ln(2) = 69 . 31471806. 2...
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q3 - Q ( t ) = inow - outow = (0 . 2)(2)-(2)( Q 200 ) or Q...

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