Answer:
See Below.
Step-by-step explanation:
We are given that:
[tex]\displaystyle \frac{dT}{dt} = -k(T - T_0)[/tex]
And we want to show that:
[tex]\displaystyle T = T_0+Ae^{-kt}[/tex]
From the original equation, divide both sides by (T - T₀) and multiply both sides by dt. Hence:
[tex]\displaystyle \frac{dT}{T-T_0}= -k\, dt[/tex]
Take the integral of both sides:
[tex]\displaystyle \int \frac{dT}{T- T_0} = \int -k \, dt[/tex]
Integrate. For the left integral, we can use u-substitution. Note that T₀ is simply a constant. Hence:
[tex]\displaystyle \ln\left|T - T_0\right| = -kt+C[/tex]
Raise both sides to e:
[tex]\displaystyle e^{\ln\left|T-T_0\right|} = e^{-kt+C}[/tex]
Simplify:
[tex]\displaystyle \begin{aligned} \left| T- T_0\right| &= e^{-kt} \cdot e^C \\ \\ &= e^C\left(e^{-kt}\right) \\ \\ &=Ae^{-kt} & \text{Let $e^C = A$}\end{aligned}[/tex]
Since the temperature T will always be greater than or equal to the surrounding medium T₀, we can remove the absolute value. Hence:
[tex]\left(T - T_0\right) = Ae^{-kt}[/tex]
Therefore:
[tex]\displaystyle T = T_0+Ae^{-kt}[/tex]
