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:
[tex]T(t)=Ate^{-kt}[/tex]where A and k are positive constants.
We can find the derivative of the function as follows:
[tex]T^{\prime}(t)=\frac{d}{dt}(Ate^{-kt})[/tex]Step 1: Pull out the constant factor
[tex]T^{\prime}(t)=A\cdot\frac{d}{dt}(te^{-kt})[/tex]Step 2: Apply the product rule
[tex]\frac{d(uv)}{dx}=u \frac{dv}{dx}+v \frac{du}{dx}[/tex]Let
[tex]\begin{gathered} u=t \\ v=e^{-kt} \\ \therefore \\ \frac{du}{dt}=1 \\ \frac{dv}{dt}=-ke^{-kt} \end{gathered}[/tex]Therefore, we have:
[tex]T^{\prime}(t)=A(t\cdot(-ke^{-kt})+e^{-kt}\cdot1)[/tex]Step 3: Simplify
[tex]T^{\prime}(t)=A(-kte^{-kt}+e^{-kt})[/tex]QUESTION A
At t = 0, the instantaneous rate of change is calculated to be:
[tex]\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}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=A[/tex]QUESTION B
At t = 30, the instantaneous rate of change is calculated to be:
[tex]\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}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=Ae^{-30k}(-30k+1)[/tex]QUESTION C
When the rate of change is equal to 0, we have:
[tex]0=A(-kte^{-kt}+e^{-kt})[/tex]We can make t the subject of the formula using the following steps:
Step 1: Apply the Zero Factor principle
[tex]\begin{gathered} If \\ ab=0 \\ a=0,b=0 \\ \therefore \\ -kte^{-kt}+e^{-kt}=0 \end{gathered}[/tex]Step 2: Collect common terms
[tex]e^{-kt}(-kt+1)=0[/tex]Step 3: Apply the Zero Factor Principle:
[tex]\begin{gathered} e^{-kt}=0 \\ \ln e^{-kt}=\ln0 \\ -kt=\infty \\ t=\infty \end{gathered}[/tex]or
[tex]\begin{gathered} -kt+1=0 \\ -kt=-1 \\ t=\frac{-1}{-k} \\ t=\frac{1}{k} \end{gathered}[/tex]The time will be:
[tex]t=\frac{1}{k}[/tex]