Explanation Given: The initial temperature of the roast is 20 ° F , while the oven is with a constant temperature 350 ° F . Theorem used: Separation of Variables: The solution of the differential equation, f ( y ) y ′ = g ( x ) is given implicitly by the equation, ∫ f ( y ) d y = ∫ g ( x ) d x . Formula used: Limited growth model: If the rate of change with respect to time t of a quantity y is proportional to the difference between y and a limiting value M , then y satisfies the differential equation, d y d t = k ( M − y ) , k > 0 , t > 0 Calculation: Let y represents the temperature of the roast at time t . Given that, the oven is maintained at a constant temperature of 350 ° F . Thus, the rate is proportional to the difference of the temperature and the limit 350 ° F . That is, d y d t = k ( y − 350 ) , k > 0 , t > 0 . Thus, the problem follows a limited growth model. Rewrite the differential equation d y d t = k ( y − 350 ) as follows. d y d t = k ( y − 350 ) 1 y − 350 d y = k d t The given differential equation d y d t = k ( y − 350 ) is in the form of f ( y ) y ′ = g ( x ) where, f ( y ) = 1 ( y − 350 ) and g ( t ) = k . By the above mentioned theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( t ) d t ∫ 1 ( y − 350 ) d y = ∫ k d t ln ( y − 350 ) = k t + C 1 e ln ( y − 350 ) = e k t + C 1 Simplify further as, y − 350 = e k t + C 1 y = 350 + e k t + C 1 y = 350 + C e k t , where C = e C 1 The roast is at 20 ° F and heated to 185 ° F after 1 hour

14th Edition

ISBN: 9780135961438

Author: Barnett

Publisher: PEARSON CO

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Differential Equations

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Chapter 9.2, Problem 74E

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To find: The temperature of the roast after 3 hours in the oven.

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Chapter 9.2, Problem 73E

Chapter 9.2, Problem 75E

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The temperature of the roast after 3 hours in the oven.

Solution Summary: Theorem used: Separation of Variables: The solution of the differential equation is given implicitly by the equation.

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