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Unformatted text preview: MA 2930, Feb 9, 2011 Worksheet 3 1. A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon enters at a rate of 3 gal/min, and the mixture flows out at a rate of 2 gal/min. In order to find the amount of salt in the tank at any time prior to its overflow (a) set up the differential equation for the quantity of interest, (b) identify the initial condition, and (c) solve the equation. (d) What is the concentration of salt at the moment of overflow? (a) Its clear that the quantity of interest is the amount of salt in the tank at any time, so let that amount be q ( t ) lb at time t min. The problem tells us how q ( t ) is changing with time. The rate at which salt is coming into the tank at time t is (1 lb/gal)(3 gal/min) = 3 lb/min. The rate at which salt is leaving the tank is, however, not constant with time. It depends on what the concentration of salt in the tank is at time t . The concentration of salt, c ( t ), can be calculated as follows: the volume of the salt solution in the tank at time t , V ( t ), is equal to the initial volume + added solution in t minutes = 200+(3- 2) t = 200+ t gal. The amount of salt at time t is q ( t ), by definition. So, the concentration, c ( t ) = q ( t ) /V ( t ) = q ( t ) / (200+ t ) lb/gal. So, rate at which salt leaves the tank at time t is c ( t ) lb/gal 2 gal/min = 2 q ( t ) / (200 + t ) lb/min. Therefore, since in rate - out rate = rate of change of amount of salt in the tank, dq dt = 3- 2 q 200 + t (b) The initial condition is q (0) = 100, which is the initial amount of salt in the tank. (c) The differential equation for q is first-order linear, written in the standard form as dq dt + 2 200 + t q = 3 So we can solve it using the integrating factor ( t ) = exp( R 2 200+ t dt ) = exp(2 ln(200 + t )) = exp(ln(200 + t ) 2 ) = (200 + t ) 2 ....
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