Explanation Given information: Assume that the outside temperature varies as T ( t ) = 15 + 5 s i n ( π t 12 ) Where t = 0 is 12 noon. A house is heated to 25 o C at t = 0 and after that, its temperature y ( t ) varies according to Newton’s Law of Cooling: d y d t = − 0.1 ( y ( t ) − T ( t ) ) Formula used: If a, b and r be constants and if d y d t = − k ( y − a − b sin r t ) then its general solution is y = C e − k t + a + b k ( k sin r t − r cos r t k 2 + r 2 ) . Calculation: From the given data, we have T ( t ) = 15 + 5 s i n ( π t 12 ) (1) d y d t = − 0.1 ( y ( t ) − T ( t ) ) (2) Using (1) in (2), we get d y d t = − 0.1 ( y ( t ) − ( 15 + 5 s i n ( π t 12 ) ) ) = − 0.1 ( y ( t ) − 15 − 5 s i n ( π t 12 ) ) The above equation is of the form d y d t = − k ( y − a − b sin r t ) with k = 0.1 , a = 15 , b = 5 , r = π 12 ,then its general solution will be y ( t ) = C e − 0.1 t + 15 + 5 ( 0.1 ) ( 0.1 sin ( π 12 ) t − ( π 12 ) cos ( π 12 ) t ( 0.1 ) 2 + ( π 12 ) 2 ) = C e − 0

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 9.5, Problem 46E

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The differential equation dydt=−0.1(y(t)−T(t)) for y(t).

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Chapter 9.5, Problem 45E

Chapter 9.5, Problem 47E

Chapter 9 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 4PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E

Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Prob. 32E Ch. 9.1 - Prob. 33E Ch. 9.1 - Prob. 34E Ch. 9.1 - Prob. 35E Ch. 9.1 - Prob. 36E Ch. 9.1 - Prob. 37E Ch. 9.1 - Prob. 38E Ch. 9.1 - Prob. 39E Ch. 9.1 - Prob. 40E Ch. 9.1 - Prob. 41E Ch. 9.1 - Prob. 42E Ch. 9.1 - Prob. 43E Ch. 9.1 - Prob. 44E Ch. 9.1 - Prob. 45E Ch. 9.1 - Prob. 46E Ch. 9.1 - Prob. 47E Ch. 9.1 - Prob. 48E Ch. 9.1 - Prob. 49E Ch. 9.1 - Prob. 50E Ch. 9.1 - Prob. 51E Ch. 9.1 - Prob. 52E Ch. 9.1 - Prob. 53E Ch. 9.1 - Prob. 54E Ch. 9.1 - Prob. 55E Ch. 9.1 - Prob. 56E Ch. 9.1 - 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The differential equation d y d t = − 0.1 ( y ( t ) − T ( t ) ) for y ( t ) .

Solution Summary: The author calculates the differential equation y(t) from the given data.

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To solve:

The differential equation dydt=−0.1(y(t)−T(t)) for y(t).

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Chapter 9 Solutions

The differential equation d y d t = − 0.1 ( y ( t ) − T ( t ) ) for y ( t ) .

Solution Summary: The author calculates the differential equation y(t) from the given data.

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To solve: The differential equation dydt=−0.1(y(t)−T(t)) for y(t).

Want to see the full answer?

Chapter 9 Solutions

To solve:

The differential equation dydt=−0.1(y(t)−T(t)) for y(t).