Respuesta :
A) The differential equation comes from the fact that the rate of temperature change is proportional to the difference in temperatures.
[tex]\frac{dT}{dt} = \alpha(T -52) [/tex]
B) Find general solution by separating variables and integrating[tex]\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52[/tex]:
C) Initial condition is t=0, T = 87
[tex]87 = C +52 \\ C = 35[/tex]
D) Total time elapsed is 10 minutes, new temperature is 84
[tex]84 = 35 e^{10\alpha} +52[/tex]
solve for alpha
[tex] e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896[/tex]
E) Temperature function is:
[tex]T = 35 e^{-.00896 t} +52[/tex]
solving for t
[tex]t = \frac{ln (\frac{T-52}{35})}{-.00896}[/tex]
Plug in T = 98.6
[tex]t = -31.95[/tex]
This is approximately 32 minutes before he arrived or about 1:20 AM
[tex]\frac{dT}{dt} = \alpha(T -52) [/tex]
B) Find general solution by separating variables and integrating[tex]\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52[/tex]:
C) Initial condition is t=0, T = 87
[tex]87 = C +52 \\ C = 35[/tex]
D) Total time elapsed is 10 minutes, new temperature is 84
[tex]84 = 35 e^{10\alpha} +52[/tex]
solve for alpha
[tex] e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896[/tex]
E) Temperature function is:
[tex]T = 35 e^{-.00896 t} +52[/tex]
solving for t
[tex]t = \frac{ln (\frac{T-52}{35})}{-.00896}[/tex]
Plug in T = 98.6
[tex]t = -31.95[/tex]
This is approximately 32 minutes before he arrived or about 1:20 AM