Explanation Given information: Objects A and B are placed in a warm bath at temperature. T 0 = 40 o C . Object A has initial temperature − 20 o C and cooling constant k = 0.004 s − 1 . Object B has initial temperature 0 o C and cooling constant k = 0.002 s − 1 . Formula used: General form of linear differential equation. d y d t + P y = Q Solution to the above differential equation is. y ⋅ ( I . F . ) = C + ∫ [ Q . ( I . F . ) ] d t Where I . F . = e ∫ P d t Calculation: To plot the temperature A and B for 0 ≤ t ≤ 1000 with the details provided. Recall that y ' = − k ( y − T 0 ) where T 0 is the ambient temperature and k is called the cooling constant. First consider object A Here initial temperature is − 20 0 C and cooling constant k = 0.004 s − 1 Thus, y ' = − 0.004 ( y − 40 ) Simplify to obtain y ' + 0.004 y = 0.160 Observe that the above equation is a liner differential equation of the form d y d t + P y = Q where P = 0.004 and Q = 0.160 To obtain the solution, find its integrating factor ( I . F . ) first that is I . F . = e ∫ P d t = e ∫ ( 0.004 ) d t = e 0.004 t Solution to the above differential equation is y . ( I . F . ) = C + ∫ [ Q . ( I . F . ) ] d t Here, I . F . = e 0.004 t and Q = 0.160 Therefore, y . e 0.004 t = C + ∫ 0.160. e 0.004 t d t = C + 0.160. e 0.004 t ( 0.004 ) As ∫ e a t d t = e a t a + C = C + 40 e 0.004 t Simplify to obtain. y = C e − 0.004 t + 40............ ( 1 ) Now calculate the integration constant C with the help of initial condition y ( 0 ) = − 20 . Write the above solution ( 3 ) as y ( t ) = C e − 0.004 t + 40 And substitute t = 0 Then y ( 0 ) = C e 0 ( − 0.004 ) + 40 = − 20 As y ( 0 ) = − 20 , thus C = − 20 − 40 = − 60 Hence, complete solution to the differential equation y ' + 0

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 9.2, Problem 11E

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Plot the temperatures of A and B for 0≤t≤1000. The time the objects will have the same temperature.

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Chapter 9.2, Problem 10E

Chapter 9.2, Problem 12E

Chapter 9 Solutions

Calculus: Early Transcendentals

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

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Plot the temperatures of A and B for 0 ≤ t ≤ 1000 . The time the objects will have the same temperature.

Solution Summary: The author explains how to plot the temperature of objects AandB for 0 tle 1000. The equation is a liner differential equation.

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Plot the temperatures of A and B for 0≤t≤1000. The time the objects will have the same temperature.

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Chapter 9 Solutions