Question

Help with number 6 please. thank you.​

Answer 1

Mandatory minimum character count of 20.

Answer 2

Answer:

See Below.

Step-by-step explanation:

We are given that:

$\displaystyle \frac{dT}{dt} = -k(T - T_0)$

And we want to show that:

$\displaystyle T = T_0+Ae^{-kt}$

From the original equation, divide both sides by (T - T₀) and multiply both sides by dt. Hence:

$\displaystyle \frac{dT}{T-T_0}= -k\, dt$

Take the integral of both sides:

$\displaystyle \int \frac{dT}{T- T_0} = \int -k \, dt$

Integrate. For the left integral, we can use u-substitution. Note that T₀ is simply a constant. Hence:

$\displaystyle \ln\left|T - T_0\right| = -kt+C$

Raise both sides to e:

$\displaystyle e^{\ln\left|T-T_0\right|} = e^{-kt+C}$

Simplify:

$\displaystyle \begin{aligned} \left| T- T_0\right| &= e^{-kt} \cdot e^C \\ \\ &= e^C\left(e^{-kt}\right) \\ \\ &=Ae^{-kt} & \text{Let e^C = A }\end{aligned}$

Since the temperature T will always be greater than or equal to the surrounding medium T₀, we can remove the absolute value. Hence:

$\left(T - T_0\right) = Ae^{-kt}$

Therefore:

$\displaystyle T = T_0+Ae^{-kt}$