Whitman MATH 244 - Modeling (8 pages)

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Modeling



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Modeling

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Pages:
8
School:
Whitman College
Course:
Math 244 - Differential Equations
Differential Equations Documents

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Section 2 3 Modeling 1 Problem 3 In this model of salt in a tank of water let Q t be the amount of salt in pounds at time t measured in minutes Then dQ dt will be measured in pounds per minute The rate in at least initially 1 lbs gal 2 1 lbs min 2 gal min The rate out at least initially 2 gal Q t lbs 1 Q lbs min min 100 gal 50 The model equation valid for 0 t 10 1 dQ 1 Q Q 0 0 dt 50 The initial condition is zero since we started with fresh water You can solve this either as a linear equation or a separable equation We give the solution here as if it were linear 1 1 Q 0 Q 1 e 1 50 t Q 0 Q e 1 50 t Q t 50 Ce 1 50 t 50 50 With the initial condition 0 50 C C 50 The amount of salt in the tank at time 0 t 10 minutes is Q t 50 50e 1 50 t At time t 10 the dynamics change Suddenly no salt comes in but water still does Now we have 1 valid for t 10 Q0 Q 50 Note We could have left this as t 10 but it was convenient to do it this way That s because the initial condition for this DE is where we ended with the last DE Q 10 50 50e 1 5 9 06343 For t 10 the solution is Q t Ae 1 50 t Putting in Q 10 9 06343 9 06343 Ae 1 5 A 11 07 Therefore for t 10 Q t 11 07e 1 50 t And now substitute t 20 to find the final amount of salt in the tank Q 20 7 421 pounds 2 Problem 5 As usual let Q t be the amount of salt in ounces at time t measured in minutes Then dQ dt will be measured in ounces per minute The rate in 1 1 oz gal 1 1 oz 1 sin t 2 1 sin t 4 2 gal min 2 2 min The rate out 2 gal Q t lbs 1 Q lbs min min 100 gal 50 The model equation dQ 1 1 1 1 sin t Q dt 2 2 50 Q 0 50 The initial condition was that we started with 50 ozs of salt in a tank of 100 gallons This is a linear differential equation and the integrating factor is the same as computed earlier in Problem 3 e 1 50 t 1 1 1 Q Q e 1 50 t e 1 50 t sin t 50 2 4 0 Note that we need to integrate by parts twice so that Z e 1 50 t sin t dt 2500 1 50 t 50 1 50 t e cos t e sin t 2501 2501 Now we write the



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