124_pdfsam_math 54 differential equation solutions odd

124_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 Also, we are assuming that there is no heating or air conditioning in the warehouse. There- fo r e ,w ehav etha t U ( t ). We are also given that the outside temperature has a sinusoidal fluctuation. Thus, as in Example 2, page 103, we see that M ( t )= M 0 B cos ωt, where M 0 is the average outside temperature, B is a positive constant for the magnitude of the temperature shift from this average, and ω = π/ 2 radians per hour. To Fnd M 0 and B , we are given that at 2 : 00 a.m. , M ( t ) reaches a low of 16 C and at 2 : 00 p.m. it reaches a high of 32 C. This gives M 0 = 16 + 32 2 =24 C . By letting t = 0 at 2 : 00
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Unformatted text preview: − B cos 0 = 24 − B ⇒ B = 8 . Therefore, we see that M ( t ) = 24 − 8 cos ωt, where ω = π/ 12. As in Example 2, using the fact that B = M + H /K = M + 0 /K = M , we see that T ( t ) = 24 − 8 F ( t ) + Ce − Kt , where F ( t ) = cos ωt + ( ω/K ) sin ωt 1 + ( ω/K ) 2 = ± 1 + ² ω K ³ 2 ´ − 1 / 2 cos( ωt − α ) . In the last expression, α is chosen such that tan α = ω/K . By assuming that the exponential term dies o±, we obtain T ( t ) = 24 − 8 ± 1 + ² ω K ³ 2 ´ − 1 / 2 cos( ωt − α ) . This function will reach a minimum when cos( ωt − α ) = 1 and it will reach a maximum when cos( ωt − α ) = − 1. 120...
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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.

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