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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

The constant temperature of the kitchen by the method of Newton’s law of cooling. The constant temperature of the kitchen is 20 ° C _ . Given: The water temperature in a pot is 100 ° C at the time of removed from a stove. Take t = 0 when the water temperature in a pot is 100 ° C . The water temperature has decreased to 80 ° C after 5 minutes. The water temperature has decreased to 65 ° C after 10 minutes. Calculation: Let T be the water temperature and M be the constant temperature of the kitchen. The Problem 34 states that, “According to Newton’s law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M then the rate of change of T is proportional to the difference of temperature M − T This gives the differential equation is, d T d t = k ( M − T ) ”. Assume Newton’s law of cooling (Problem 34) applies in this problem. Separate the variables and rewrite the differential equation as follows. d T d t = k ( M − T ) 1 ( M − T ) d T = k d t Integrate on both sides of the equation. ∫ 1 ( M − T ) d T = ∫ k d t − ln | M − T | = k t + C 1 ln | ( M − T ) − 1 | = k t + C 1 ln | 1 M − T | = k t + C 1 Take exponential on both side of the equation. e ln ( 1 M − T ) = e k t + C 1 1 M − T = e C 1 e k t 1 M − T = C 2 e k t M − T = C 3 e − k t Simplify further as follows. − T = − M + C 3 e − k t T = M − C 3 e − k t T = M + C e − k t ( − C 3 = C , where C is an arbitrary constant ) T = M + C e − k t ... ( 1 ) Now find the constant temperature of the kitchen M by using the above equation (1). Since the water temperature is 100 ° C at t = 0 , substitute t = 0 and T = 100 in the above solution. T = M + C e − k t 100 = M + C e − k ( 0 ) 100 = M + C C = 100 − M ... ( 2 ) Since the water temperature has decreased to 80 ° C after 5 minutes, substitute t = 5 and T = 80 in equation (1). 80 = M + C e − k ( 5 ) 80 = M + C e − 5 k e − 5 k = 80 − M C ( e − 5 k ) 2 = ( 80 − M C ) 2 ( on squaring both sides ) e − 10 k = ( 80 − M ) 2 C 2 ... ( 3 ) Since the water temperature has decreased to 65 ° C after 10 minutes, substitute t = 10 and T = 65 in equation (1). T = M + C e − k t 65 = M + C e − 10 k C e − 10 k = 65 − M e − 10 k = 65 − M C ... ( 4 ) Equate the equations (3) and (4). ( 80 − M ) 2 C 2 = 65 − M C ( 80 − M ) 2 65 − M = C 2 C ( 80 − M ) 2 65 − M = C Substitute C = 100 − M in the above result. ( 80 − M ) 2 ( 65 − M ) = ( 100 − M ) ( 80 − M ) 2 = ( 100 − M ) ( 65 − M ) 6400 − 160 M + M 2 = 6500 − 100 M − 65 M + M 2 − 160 M + 100 M + 65 M = 6500 − 6400 + M 2 − M 2 Simplify further as follows. 5 M = 100 M = 100 5 M = 20 ° C Thus, the constant temperature of the kitchen is 20 ° C _ .

The constant temperature of the kitchen by the method of Newton’s law of cooling. The constant temperature of the kitchen is 20 ° C _ . Given: The water temperature in a pot is 100 ° C at the time of removed from a stove. Take t = 0 when the water temperature in a pot is 100 ° C . The water temperature has decreased to 80 ° C after 5 minutes. The water temperature has decreased to 65 ° C after 10 minutes. Calculation: Let T be the water temperature and M be the constant temperature of the kitchen. The Problem 34 states that, “According to Newton’s law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M then the rate of change of T is proportional to the difference of temperature M − T This gives the differential equation is, d T d t = k ( M − T ) ”. Assume Newton’s law of cooling (Problem 34) applies in this problem. Separate the variables and rewrite the differential equation as follows. d T d t = k ( M − T ) 1 ( M − T ) d T = k d t Integrate on both sides of the equation. ∫ 1 ( M − T ) d T = ∫ k d t − ln | M − T | = k t + C 1 ln | ( M − T ) − 1 | = k t + C 1 ln | 1 M − T | = k t + C 1 Take exponential on both side of the equation. e ln ( 1 M − T ) = e k t + C 1 1 M − T = e C 1 e k t 1 M − T = C 2 e k t M − T = C 3 e − k t Simplify further as follows. − T = − M + C 3 e − k t T = M − C 3 e − k t T = M + C e − k t ( − C 3 = C , where C is an arbitrary constant ) T = M + C e − k t ... ( 1 ) Now find the constant temperature of the kitchen M by using the above equation (1). Since the water temperature is 100 ° C at t = 0 , substitute t = 0 and T = 100 in the above solution. T = M + C e − k t 100 = M + C e − k ( 0 ) 100 = M + C C = 100 − M ... ( 2 ) Since the water temperature has decreased to 80 ° C after 5 minutes, substitute t = 5 and T = 80 in equation (1). 80 = M + C e − k ( 5 ) 80 = M + C e − 5 k e − 5 k = 80 − M C ( e − 5 k ) 2 = ( 80 − M C ) 2 ( on squaring both sides ) e − 10 k = ( 80 − M ) 2 C 2 ... ( 3 ) Since the water temperature has decreased to 65 ° C after 10 minutes, substitute t = 10 and T = 65 in equation (1). T = M + C e − k t 65 = M + C e − 10 k C e − 10 k = 65 − M e − 10 k = 65 − M C ... ( 4 ) Equate the equations (3) and (4). ( 80 − M ) 2 C 2 = 65 − M C ( 80 − M ) 2 65 − M = C 2 C ( 80 − M ) 2 65 − M = C Substitute C = 100 − M in the above result. ( 80 − M ) 2 ( 65 − M ) = ( 100 − M ) ( 80 − M ) 2 = ( 100 − M ) ( 65 − M ) 6400 − 160 M + M 2 = 6500 − 100 M − 65 M + M 2 − 160 M + 100 M + 65 M = 6500 − 6400 + M 2 − M 2 Simplify further as follows. 5 M = 100 M = 100 5 M = 20 ° C Thus, the constant temperature of the kitchen is 20 ° C _ .

Solution Summary: The author calculates the constant temperature of the kitchen by the method of Newton's law of cooling.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Fundamentals of Differential Equations (9th Edition)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. 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In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. 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