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if the vat starts with 0 sugar, and we wish to find the approximate time that the amount of sugar approximately becomes the equilibrium solution (2000/3) how does one do this?

if the vat starts with 0 sugar, and we wish to find the approximate time that the amount of sugar approximately becomes the equilibrium solution (2000/3) how does one do this?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hi, if the vat starts with 0 sugar, and we wish to find the approximate time that the amount of sugar approximately becomes the equilibrium solution (2000/3) how does one do this?

1.2 Analytic Technique: Separation of Variables 33 where c2 = -3c1. Exponentiating we obtain |2000 - 3S| = e-0.03r+c2) = C3e-0.03t -0.031 where c3 = ec2. Note that this means that c3 is a positive constant. Now we must be careful. Removing the absolute value signs yields 2000 – 3S = ±c3e¬0.03t where we choose the plus sign if S(t) < 2000/3 and the minus sign if S(t) > 2000/3. Therefore we may write this equation more simply as 2000 – 3S = cae¬0.03t where c4 is an arbitrary constant (positive, negative, or zero). Solving for S yields the general solution 2000 S(t) = ce-0.03t 3 where c = -c4/3 is an arbitrary constant. We can determine the precise value of c if we know the exact amount of sugar that is initially in the vat. Note that, if c = 0, the solution is simply S(t) = 2000/3, an equilibrium solution. | %3D SES FOR SECTION 1.2 1. Bob, Glen, and Paul are once again sitting around enjoying their nice, cold glasses of iced cappucino when one of their students asks them to come up with solutions to the differential equation dy y +1 dt t+ 1 After much discussion, Bob says y(t) = t, Glen says y(t) = 2t + 1, and Paul says y(t) = t² – 2. (a) Who is right? (b) What solution should they have seen right away? 2. Make up a differential equation of the form dy = 2y – t+g(y) dt that has the function y(t) = e2r as a solution. %3D 3. Make un a differential equation of the form dy/dt = f(t, y) that has y(t) = e as a
Transcribed Image Text:1.2 Analytic Technique: Separation of Variables 33 where c2 = -3c1. Exponentiating we obtain |2000 - 3S| = e-0.03r+c2) = C3e-0.03t -0.031 where c3 = ec2. Note that this means that c3 is a positive constant. Now we must be careful. Removing the absolute value signs yields 2000 – 3S = ±c3e¬0.03t where we choose the plus sign if S(t) < 2000/3 and the minus sign if S(t) > 2000/3. Therefore we may write this equation more simply as 2000 – 3S = cae¬0.03t where c4 is an arbitrary constant (positive, negative, or zero). Solving for S yields the general solution 2000 S(t) = ce-0.03t 3 where c = -c4/3 is an arbitrary constant. We can determine the precise value of c if we know the exact amount of sugar that is initially in the vat. Note that, if c = 0, the solution is simply S(t) = 2000/3, an equilibrium solution. | %3D SES FOR SECTION 1.2 1. Bob, Glen, and Paul are once again sitting around enjoying their nice, cold glasses of iced cappucino when one of their students asks them to come up with solutions to the differential equation dy y +1 dt t+ 1 After much discussion, Bob says y(t) = t, Glen says y(t) = 2t + 1, and Paul says y(t) = t² – 2. (a) Who is right? (b) What solution should they have seen right away? 2. Make up a differential equation of the form dy = 2y – t+g(y) dt that has the function y(t) = e2r as a solution. %3D 3. Make un a differential equation of the form dy/dt = f(t, y) that has y(t) = e as a
Mixing in a vat Consider a laege vat containing sugar water that is to be made into soft drinks (see Fig. ure 1.9) Suppose + The vut contains 100 gallons of liquid. Moreover, the amount flowing in is she same as the amount Howing out, so there are always 100 gallons in the vat. • The vat is kept well mixed, so the sugar concentration is uniform throughout the vat Sugar water containing 5 tablespoons of sugar per gallon enters the vat through pipe A at a rte of 2 gallons per minute. • Sugar water containing 10 tablespoons of sugar per gallon enters the vat through pipe B at a rate of i gallon per minute. • Sugar water leaves the vat through pipe Cat a rate of 3 gallons per minute. To make the model, we let f be time measured in minutes (the independent vari able). Foe the dependent variable, we have two choices. We could choose either the total amount of sugar, S(0), in the vat at time t measured in tablespoons, or Cir), the concentration of sugar in the vat at time t measured in tablespoons per gallon. We de- velop the model for S, leaving the model for C as an exercise for the reader. Using the total sugar S(1) in the vat as the dependent variable, the rate of change of S is the difference between the amount of sugar being added and the amount of sugar being removed. The sugar entering the vat comes from pipes A and B and can be easily computed by multiplying the mumber of gallons per minute of sugar mixture entering the vat by the amount of sugar per gallon. The amount of sugar leaving the val through pipe Cat any given moment depends on the concentration of sugar in the vat at that mo- ment. The concentration is given by S/100, so the sugar leaving the vat is the product of the number of gallons leaving per minute (3 galons per minute) and the concentra- tion (S/100). The model is gure 1.9 Ting vt. 2.3 1. 10 !! dt 100 sugar in from pipe A sugar in Irom pipe B sugar out from pipe C That is. ds = 20- dt 35 2000 - 35 100 100 To solve this equation analytically, we separate and integrate. We find ds dr 2000 - 35 100 In |2000- 351 -3 100 3t In 2000 - 351 =- -3c1 100 In |2000 - 3S--0.03r +ey.
Transcribed Image Text:Mixing in a vat Consider a laege vat containing sugar water that is to be made into soft drinks (see Fig. ure 1.9) Suppose + The vut contains 100 gallons of liquid. Moreover, the amount flowing in is she same as the amount Howing out, so there are always 100 gallons in the vat. • The vat is kept well mixed, so the sugar concentration is uniform throughout the vat Sugar water containing 5 tablespoons of sugar per gallon enters the vat through pipe A at a rte of 2 gallons per minute. • Sugar water containing 10 tablespoons of sugar per gallon enters the vat through pipe B at a rate of i gallon per minute. • Sugar water leaves the vat through pipe Cat a rate of 3 gallons per minute. To make the model, we let f be time measured in minutes (the independent vari able). Foe the dependent variable, we have two choices. We could choose either the total amount of sugar, S(0), in the vat at time t measured in tablespoons, or Cir), the concentration of sugar in the vat at time t measured in tablespoons per gallon. We de- velop the model for S, leaving the model for C as an exercise for the reader. Using the total sugar S(1) in the vat as the dependent variable, the rate of change of S is the difference between the amount of sugar being added and the amount of sugar being removed. The sugar entering the vat comes from pipes A and B and can be easily computed by multiplying the mumber of gallons per minute of sugar mixture entering the vat by the amount of sugar per gallon. The amount of sugar leaving the val through pipe Cat any given moment depends on the concentration of sugar in the vat at that mo- ment. The concentration is given by S/100, so the sugar leaving the vat is the product of the number of gallons leaving per minute (3 galons per minute) and the concentra- tion (S/100). The model is gure 1.9 Ting vt. 2.3 1. 10 !! dt 100 sugar in from pipe A sugar in Irom pipe B sugar out from pipe C That is. ds = 20- dt 35 2000 - 35 100 100 To solve this equation analytically, we separate and integrate. We find ds dr 2000 - 35 100 In |2000- 351 -3 100 3t In 2000 - 351 =- -3c1 100 In |2000 - 3S--0.03r +ey.
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