Question

Help with number 6 please. thank you.​

Answer 1

Mandatory minimum character count of 20.

Answer 2

See Below.

Explanation:

We are given that:

`\displaystyle (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 (dT)/(T-T_0)= -k\, dt`

Take the integral of both sides:

`\displaystyle \int (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)`