Transcription of Di erential Equations Water Tank Problems
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Di erential Equations Water Tank Problems
Differential Equations Water Tank ProblemsChapter #3 VariationA tank originally contains 100 gal of fresh Water . Then Water containing12lb of salt per 2 gallonis poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the samerate. What is the amount of salt at any instant?dQdt= rate in - rate out= (salt concentration in) x (flow rate in) - (tank salt concentration) x (flow rate out)*NotedQdtshould be in terms oflbminParametersQ(t) : the amount of salt at time t (lbs)Qo: initial amount of salt in the tank Q(0) =Qo= 0salt concentration in =12lbgaltank salt concentration =Qcurrent amount of Water solution=Q100lbgalIn this question flow rate in is the same as flow rate out so we will let the rates be defined as:rate in = rate out = 2galminSubstituting everything into our differential equation, we arrive atdQdt=12lbgal 2galmin Q100lbgal 2galmin= 1lbmin Q50lbminRewriting our equation, we havedQdt+Q50= 11 Multiplying by the integrating factoret/50followed by applying the product rulefor derivatives we havedQdt et/50+Q et/5050= 1 et/50ddt[Q et/50] =et/50 ddt[Q et/50]dt= et/50dtAfter integrating both sides we are left withQ et/50= 50 et/50+CQ= 50 +Ce t/50 From our initial conditionQ(0) = 0we haveC= 50.
Di erential Equations Water Tank Problems Chapter 2.3 Problem #3 Variation A tank originally contains 100 gal of fresh water. Then water containing 1 2 lb of salt per 2 gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same
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