Chapter 1.7, Problem 70E

(a)

To determine

The graphs to check the given conditions

Explanation of Solution

Consider the following functions:

  $\begin{array}{l}{S}_1\left(t\right)=3.4e^{0.2t}\\ S_2\left(t\right)=0.2e^{3.4t}\\ H_1\left(t\right)=3.4e^{-0.2t}\\ H_2\left(t\right)=0.2e^{-3.4t}\end{array}$

The given condition is $S_1\left(t\right)=S_2\left(t\right)$

  $\begin{array}{l}3.4e^{0.2t}=0.2e^{3.4t}\\ \frac{e^{3.4t}}{e^{0.2t}}=\frac{3.4}{0.2}\\ e^{3.2t}=17\\ \text{Taking logarithm on both the sides,}\\ t=\frac{\ln \left(17\right)}{3.2}\\ t=0.8853\end{array}$

The graph of $S_1\left(t\right)\text{ and }{S}_2\left(t\right)$

is shown below:

  

From the graph, it is observed that the point of intersection of the graphs shows the time when both the functions have equal value.

(b)

To determine

The graphs to check the given conditions

Explanation of Solution

Consider the following functions:

  $\begin{array}{l}{S}_1\left(t\right)=3.4e^{0.2t}\\ S_2\left(t\right)=0.2e^{3.4t}\\ H_1\left(t\right)=3.4e^{-0.2t}\\ H_2\left(t\right)=0.2e^{-3.4t}\end{array}$

The given condition is $H_1\left(t\right)=2H_2\left(t\right)$

  $\begin{array}{l}3.4e^{-0.2t}=2\left(0.2e^{-3.4t}\right)\\ \frac{e^{-3.4t}}{e^{-0.2t}}=\frac{3.4}{0.4}\\ e^{-3.2t}=8.5\\ \text{Taking logarithm on both the sides,}\\ t=\frac{\ln \left(8.5\right)}{-3.2}\\ t=-0.6687\end{array}$

The graph of $H_1\left(t\right)\text{ and 2}{H}_2\left(t\right)$ is shown below:

  

From the graph, it is observed that the point of intersection of the graphs shows the time when both the functions have equal value.

(c)

To determine

The graphs to check the given conditions

Explanation of Solution

Consider the following functions:

  $\begin{array}{l}{S}_1\left(t\right)=3.4e^{0.2t}\\ S_2\left(t\right)=0.2e^{3.4t}\\ H_1\left(t\right)=3.4e^{-0.2t}\\ H_2\left(t\right)=0.2e^{-3.4t}\end{array}$

The given condition is $H_1\left(t\right)=0.5H_2\left(t\right)$

  $\begin{array}{l}3.4e^{-0.2t}=\frac{1}{2}\left(0.2e^{-3.4t}\right)\\ \frac{e^{-3.4t}}{e^{-0.2t}}=\frac{3.4}{0.1}\\ e^{-3.2t}=34\\ \text{Taking logarithm on both the sides,}\\ t=\frac{\ln \left(34\right)}{-3.2}\\ t=-1.1019\end{array}$

The graph of $H_1\left(t\right)\text{ and 0}\text{.5}{H}_2\left(t\right)$ is shown below:

  

From the graph, it is observed that the point of intersection of the graphs shows the time when both the functions have equal value.