Ordinary Diff Eq Exam Review Solutions 14

# Ordinary Diff Eq Exam Review Solutions 14 - 16 1 Solutions...

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16 1 Solutions T ( t ) = 70 + 110 e rt , with r as calculated. The temperature requested is T (5) = 70 + 110 7 11 5 3 121 : 8 . 36. The ambient temperature is T a = 65 . Equation (12) gives T ( t ) = 65 + ke rt for the temperature at time t . Since the initial temperature of the thermometer is T (0) = 90 we get 90 = T (0) = 65 + k . Thus k = 25 . The constant r is determined from the temperature at a second time: 85 = T (2) = 65 + 25 e 2 r so r = 1 2 ln 4 5 . Thus T ( t ) = 65 + 25 e rt , with r = 1 2 ln 4 5 . To answer the first question we solve the equation 75 = T ( t ) = 65 + 25 e rt for t . We get t = 2 ln 2 ln 5 ln 4 ln 5 8 : 2 minutes. The temperature at t = 20 is T (20) = 65 + 25 4 5 10 67 : 7 . 37. The ambient temperature is T a = 70 . Equation (12) gives T ( t ) = 70 + ke rt for the temperature of the soda at time t . Since the initial temperature of the soda is T (0) = 40 we get 40 = T (0) = 70 + k . Thus k = 30 . The constant r is determined from the temperature at a second time: 60 = T (2) = 70 30 e 2 r so r = 1 2 ln 1 3 . Thus T ( t ) = 70 30 e rt , with r = 1 2 ln 1 3 . The temperature at t = 1 is T (1) = 70 30 e 1 2 ln 1 3 = 70 30 p 3 52 : 7 . 38. The ambient temperature is T a = 70 . Equation (12) gives T ( t ) = 70 + ke rt for the temperature of the coffee at time t . We are asked to determine the initial temperature of the coffee so T (0) is unknown. However, we have
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