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


$\begin{equation} \begin{aligned} \ln \left( \frac{a}{b} \right) =& t (a - b)(\ln e) \\ \\ \ln \left( \frac{a}{b} \right) =& t(a - b)(1) \\ \\ t =& \frac{\displaystyle \ln \left( \frac{a}{b} \right)}{a - b} \end{aligned} \end{equation}$