Question

Hi I have a meeting at my house in about

Answer 1

The derivative is the instantaneous rate of change of a function with respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a point.

The function is given to be:

`T(t)=Ate^(-kt)`

where A and k are positive constants.

We can find the derivative of the function as follows:

`T^(\prime)(t)=(d)/(dt)(Ate^(-kt))`

Step 1: Pull out the constant factor

`T^(\prime)(t)=A\cdot(d)/(dt)(te^(-kt))`

Step 2: Apply the product rule

`(d(uv))/(dx)=u (dv)/(dx)+v (du)/(dx)`

Let

`\begin{gathered} u=t \\ v=e^(-kt) \\ \therefore \\ (du)/(dt)=1 \\ (dv)/(dt)=-ke^(-kt) \end{gathered}`

Therefore, we have:

`T^(\prime)(t)=A(t\cdot(-ke^(-kt))+e^(-kt)\cdot1)`

Step 3: Simplify

`T^(\prime)(t)=A(-kte^(-kt)+e^(-kt))`

QUESTION A

At t = 0, the instantaneous rate of change is calculated to be:

`\begin{gathered} t=0 \\ \therefore \\ T^(\prime)(0)=A(-k(0)e^(-k(0))+e^(-k(0))) \\ T^(\prime)(0)=A(0+e^0) \\ Recall \\ e^0=1 \\ \therefore \\ T^(\prime)(0)=A \end{gathered}`

The rate of change is:

`rate\text{ }of\text{ }change=A`

QUESTION B

At t = 30, the instantaneous rate of change is calculated to be:

`\begin{gathered} t=30 \\ \therefore \\ T(30)=A(-k(30)e^(-k(30))+e^(-k(30))) \\ T(30)=A(-30ke^(-30k)+e^(-30k)) \\ Collecting\text{ }common\text{ }factors \\ T(30)=Ae^(-30k)(-30k+1) \end{gathered}`

The rate of change is:

`rate\text{ }of\text{ }change=Ae^(-30k)(-30k+1)`

QUESTION C

When the rate of change is equal to 0, we have:

`0=A(-kte^(-kt)+e^(-kt))`

We can make t the subject of the formula using the following steps:

Step 1: Apply the Zero Factor principle

`\begin{gathered} If \\ ab=0 \\ a=0,b=0 \\ \therefore \\ -kte^(-kt)+e^(-kt)=0 \end{gathered}`

Step 2: Collect common terms

`e^(-kt)(-kt+1)=0`

Step 3: Apply the Zero Factor Principle:

`\begin{gathered} e^(-kt)=0 \\ \ln e^(-kt)=\ln0 \\ -kt=\infty \\ t=\infty \end{gathered}`

or

`\begin{gathered} -kt+1=0 \\ -kt=-1 \\ t=(-1)/(-k) \\ t=(1)/(k) \end{gathered}`

The time will be:

`t=(1)/(k)`