123_pdfsam_math 54 differential equation solutions odd

123_pdfsam_math 54 differential equation solutions odd - 11...

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Exercises 3.3 7. The temperature function T ( t ) changes according to Newton’s law of cooling (1) on page 102 of the text. Similarly to Example 1 we conclude that, with H ( t ) U ( t ) 0 and the outside temperature M ( t ) 35 C, a general solution formula (4) on page 102 becomes T ( t )=35+ Ce Kt . To Fnd C , we use the initial condition, T (0) = T (at noon) = 24 C , and get 24 = T (0) = 35 + Ce K (0) C =24 35 = 11 T ( t )=35 11 e Kt . The time constant for the building 1 /K =4hr;so K =1 / 4and T ( t )=35 11 e t/ 4 . At 2 : 00 p.m. t =2 ,and t = 6 at 6 : 00 p.m. Substituting this values into the solution, we obtain that the temperature at 2 : 00 p.m. will be T (2) = 35 11 e 2 / 4 28 . 3 C; at 6 : 00 p.m. will be T
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Unformatted text preview: 11 e 6 / 4 32 . 5 C . inally, we solve the equation T ( t ) = 35 11 e t/ 4 = 27 to Fnd the time when the temperature inside the building reaches 27 C. 35 11 e t/ 4 = 27 11 e t/ 4 = 8 t = 4 ln 11 8 1 . 27 . Thus, the temperature inside the building will be 27 C at 1 . 27 hr after noon, that is, at 1 : 16 : 12 p.m. 9. Since we are evaluating the temperature in a warehouse, we can assume that any heat gener-ated by people or equipment in the warehouse will be negligible. Therefore, we have H ( t ) = 0. 119...
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