Chapter 3Integrating, we obtainBu(t)=2Z(e−t−e−2t)dt=−2e−t+e−2t+C,and the initial condition Bu(0) = 0 givesC= 1. Therefore,Bu(t)=−2e−t+e−2t+1.EXERCISES 3.3:Heating and Cooling of Buildings, page 1071.LetT(t) denote the temperature of coFee at timet(in minutes). According to the Newton’sLaw (1) on page 102 of the text,dTdt=K[21−T(t)],wherewehavetakenH(t)≡U(t)≡0,M(t)≡21◦C, with the initial conditionT(0) = 95◦C.Solving this initial value problem yieldsdT21−T=Kdt⇒−ln|T−21|=Kt+C1⇒T(t)=21+Ce−;95 =T(0) = 21 +−K(0)⇒C=74⇒T(t)=21+74e−.To ±ndK, we use the fact that after 5 min the temperature of coFee was 80◦C. Thus80 =T(5) = 21 + 74e−K(5)⇒K=ln(74/59)5,and so
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.