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’ FontSize ’ ,12) ; 62 63 fontlabs = ’ Times New Roman ’ ; 64 xlabel ( xlab , ’ FontSize ’ ,14 , ’FontName ’ , fontlabs , ’ i n t e r p r e t e r ’ , ’ latex ’ ) ; 65 ylabel ( ylab , ’ FontSize ’ ,14 , ’FontName ’ , fontlabs , ’ i n t e r p r e t e r ’ , ’ latex ’ )
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; 66 t i t l e ( mytitle2 , ’ FontSize ’ ,16 , ’FontName ’ , ’ Times New Roman ’ , i n t e r p r e t e r ’ , ’ latex ’ ) ; 67 set ( gca , ’ FontSize ’ ,12) ; 68 69 print - depsc l i n r a t 1 8 . eps This program outputs: -2 -1 0 1 2 3 4 5 x -20 -10 0 10 20 30 y Linear and Cubic Functions f ( x ) g ( x ) -10 -5 0 5 10 x -10 -5 0 5 10 y Linear and Rational Functions f ( x ) g ( x ) 19. a. With Newton’s Law of Cooling and a constant environmental temperature the data provided gives the following problem: dH dt = - k 1 ( H - 15 . 5) , with H (0) = 21 . 7 and H (1) = 19 . 4 . This is a linear first order equation, which can be written: dH dt + k 1 H = 15 . 5 k 1 . This has an integrating factor μ ( t ) = exp ( R k 1 dt ) = e k 1 t , so d dt h e k 1 t H i = 15 . 5 k 1 e k 1 t , which upon integration gives: e k 1 t H ( t ) = 15 . 5 k 1 Z e k 1 t dt = 15 . 5 e k 1 t + C or H ( t ) = 15 . 5 + Ce - k 1 t . where with the initial condition gives C = 6 . 2, so H ( t ) = 15 . 5 + 6 . 2 e - k 1 t . The condition H (1) = 19 . 4 gives H (1) = 19 . 4 = 15 . 5 + 6 . 2 e - k 1 , so e k 1 = 6 . 2 3 . 9 or k 1 = 0 . 463573 . Since the body temperature of the live cat is 38 . 8 C, the time of death is found solving 38 . 8 = 15 . 5 + 6 . 2 e - k 1 t d , so e - k 1 t d = 23 . 3 6 . 2 or t d = - 2 . 855871 .
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Since t = 0 is 7 AM, the time of death is about 4:08.6 AM. b. With Newton’s Law of Cooling and a declining environmental temperature the data provided gives the following problem: dH dt = - k 2 ( H - (15 . 5 - 0 . 4 t )) , with H (0) = 21 . 7 and H (1) = 19 . 4 . This is a linear first order equation, which can be written: dH dt + k 2 H = k 2 (15 . 5 - 0 . 4 t ) . This again has an integrating factor μ ( t ) = exp ( R k 2 dt ) = e k 2 t , so d dt h e k 2 t H i = k 2 (15 . 5 - 0 . 4 t ) e k 2 t , which upon integration gives: e k 2 t H ( t ) = k 2 Z (15 . 5 - 0 . 4 t ) e k 2 t dt = 15 . 5 - 2 t 5 + 2 5 k 2 e k 2 t + C. Thus, H ( t ) = 15 . 5 - 2 t 5 + 2 5 k 2 + Ce - k 2 t . where with the initial condition gives C = 6 . 2 - 2 5 k 2 , so H ( t ) = 15 . 5 - 2 t 5 + 2 5 k 2 + 6 . 2 - 2 5 k 2 e - k 2 t . The condition H (1) = 19 . 4 gives H (1) = 19 . 4 = 15 . 5 - 2 5 + 2 5 k 2 + 6 . 2 - 2 5 k 2 e - k 2 , which is a transcendental equation and can only be solved numerically. This is done in Maple (shown below), and we find k 2 = 0 . 444022. Again with Maple, the time of death is found t d = - 3 . 117410, which is equivalent to 3:53 AM or about 16 min earlier. The Maple commands are given below: de := diff(H(t), t) = -k*(H(t)-15.5+.4*t); # Define ODE dsolve({de, H(0) = 21.7}, H(t)); # Solve ODE h := unapply(rhs(%), t); # Make ODE into function K := fsolve(h(1) = 19.4, k); k := K; # Find value of k td := fsolve(h(t) = 38.8, t); # Find time of death c. Though the last part is more complicated, the Maple commands are effectively the same
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