Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 62

Suppose that the function $c(t) = k (e^{-at} - e^{-bt})$

a.) Show that $\displaystyle \lim_{t \to \infty} c(t) = 0$


$
\begin{equation}
\begin{aligned}

\lim_{t \to \infty} c(t) =& \lim_{t \to \infty} k (e^{-at} - e^{-bt}) = \lim_{t \to \infty} k \left( \frac{1}{e^{at}} - \frac{1}{e^{bt}} \right)
\\
\\
=& k \left( \frac{1}{e^{a(\infty)} - \frac{1}{e^{b(\infty)}}} \right)
\\
\\
=& k \left( \frac{1}{\infty} - \frac{1}{\infty} \right)
\\
\\
=& k (0- 0)
\\
\\
=& 0

\end{aligned}
\end{equation}
$



b.) Find $c'(t)$


$
\begin{equation}
\begin{aligned}

\text{If } c(t) =& k (e^{-at} - e^{-bt}) \text{ then}
\\
\\
c'(t) =& k (e^{-at} (-a) - e^{-bt} (-b))
\\
\\
c'(t) =& k (be^{-bt} - ae^{-at})

\end{aligned}
\end{equation}
$


c.) When is the rate equal to 0?

When $c'(t) = 0$


$
\begin{equation}
\begin{aligned}

0 =& k (be^{-bt} - ae^{-at})
\\
\\
0 =& be^{-bt} - ae^{-at}
\\
\\
ae^{-at} =& be ^{-bt}
\\
\\
\frac{a}{b} =& \frac{e^{-bt}}{e^{-at}}
\\
\\
\frac{a}{b} =& e^{t(-b - (-a))}
\\
\\
\frac{a}{b} =& e^{t(a - b)}

\end{aligned}
\end{equation}
$


If we take the natural logarithm, we have


$
\begin{equation}
\begin{aligned}

\ln \left( \frac{a}{b} \right) =& t (a - b)(\ln e)
\\
\\
\ln \left( \frac{a}{b} \right) =& t(a - b)(1)
\\
\\
t =& \frac{\displaystyle \ln \left( \frac{a}{b} \right)}{a - b}

\end{aligned}
\end{equation}
$