nagle_differential_equations_ISM_Part19

nagle_differential_equations_ISM_Part19 - Chapter 3 We are...

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Chapter 3 We are given that T (0) = 110, and that the temperature M ( t ) outside the tank is constantly 80 F. Hence, the temperature in the tank is governed by dT dt = 1 72 [80 - T ( t )] + 2 = - 1 72 T ( t ) + 28 9 , T (0) = 110 . Solving this separable equation gives T ( t ) = 224 - Ce - t/ 72 . To find C , we use the initial condition and find that T (0) = 110 = 224 - C C = 114 . This yields T ( t ) = 224 - 114 e - t/ 72 . So, after 12 hours of sunlight, the temperature will be T (12) = 224 - 114 e - 12 / 72 127 . 5 ( F) . 16. Let A := p C 2 1 + C 2 2 . Then C 1 cos ωt + C 2 sin ωt = A ± C 1 p C 2 1 + C 2 2 cos ωt + C 2 p C 2 1 + C 2 2 sin ωt ! = A ( α 1 cos ωt + α 2 sin ωt ) . (3.3) We note that α 2 1 + α 2 2 = ± C 1 p C 2 1 + C 2 2 ! 2 + ± C 2 p C 2 1 + C 2 2 ! 2 = 1 . Therefore, α 1 and α 2 are the values of the cosine and sine functions of an angle φ , namely, the angle satisfying cos φ = α 1 , sin φ = α 2 tan φ = α 2 α 1 = C 2 C 1 . (3.4) Hence, (3.3) becomes C 1 cos ωt + C 2 sin ωt = A (cos φ cos ωt + sin φ sin ωt ) = A cos ( ωt - φ ) . (3.5) In the equation (7) of the text, F ( t ) = cos ωt + ( ω/K ) sin ωt 1 + ( ω/K ) 2 = C 1 cos ωt + C 2 sin ωt 86
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Exercises 3.4 with C 1 = 1 1 + ( ω/K ) 2 , C 2 = ( ω/K ) 1 + ( ω/K ) 2 A = s ± 1 1 + ( ω/K ) 2 ² 2 + ± ( ω/K ) 1 + ( ω/K ) 2 ² 2 = ( 1 + ( ω/K ) 2 ) - 1 / 2 . Thus, (3.4) and (3.5) give us F ( t ) = ( 1 + ( ω/K ) 2 ) - 1 / 2 cos ( ωt - φ ) , where tan φ = ω K . EXERCISES 3.4: Newtonian Mechanics
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part19 - Chapter 3 We are...

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