Q. 5 - Calculus

Question

Theorems: Fill in the blanks to complete each of the following theorem statements.

If an object with temperature $T\left(t\right)$ is in a location with constant ambient temperature $A$, then the temperature of the object can be modeled by a differential equation of the form $\frac{dT}{dt}=\_\_\_\_\_$, with solution $T\left(t\right)=\_\_\_\_\_$.

Step-by-Step Solution

Answer

If an object with temperature $T\left(t\right)$ is in a location with constant ambient temperature $A$, then the temperature of the object can be modeled by a differential equation of the form $\frac{dT}{dt}=k(A-T)$, with solution $T\left(t\right)=A-(A-T_0)e^{-kt}$.

Step 1. Given information

If an object with temperature $T\left(t\right)$ is in a location with constant ambient temperature $A$, then the temperature of the object can be modeled by a differential equation of the form $\frac{dT}{dt}=\_\_\_\_\_$, with solution  $T\left(t\right)=\_\_\_\_\_$.

Step 2. Filling the blanks

If an object with temperature $T\left(t\right)$ is in a location with constant ambient temperature $A$, then the temperature of the object can be modeled by a differential equation of the form $\frac{dT}{dt}=k(A-T)$, with solution  $T\left(t\right)=A-(A-T_0)e^{-kt}$.

The differential equation is: $\frac{dT}{dt}=k(A-T)$

By separation of variables and integrating, we get,

$$\begin{gathered} \int \frac{dT}{A-T}=\int k dt \\ \Rightarrow -\ln \left(A-T\right)=kt+C \\ \Rightarrow \ln \left(A-T\right)=-kt-C \\ \Rightarrow \left(A-T\right)=e^{-kt-C} \\ \Rightarrow \left(A-T\right)=B{e}^{-kt} \\ \Rightarrow T=A-B{e}^{-kt} \end{gathered}$$

Since, $T\left(0\right)=T_0$ and $T_0=A-B{e}^{-k\left(0\right)}$

$\Rightarrow B=A-T_0$

Therefore, $T\left(t\right)=A-(A-T_0)e^{-kt}$