Chapter 9.2, Problem 59E

a.

To determine

To find: The limit of $\underset{t\to \infty }{\lim }A\left(t\right)$ .

Answer to Problem 59E

The limit is $1$ .

Explanation of Solution

Given:

The area of the region in the first quadrant is $A\left(t\right)$ , the curve $y=e^{-x}$ and the line $x=t>0$ .

  

Calculation:

Here, $A\left(t\right)$ and $V\left(t\right)$ can be calculated as,

  $\begin{array}{c}A\left(t\right)=\underset{0}{\overset{t}{\int }}{e}^{-x}dx\\ =-e^{-x}{0}^t\\ =1-e^{-t}\\ \end{array}$

And,

  $\begin{array}{c}V\left(t\right)=\pi \underset{0}{\overset{t}{{\displaystyle \int }}}{e}^{-2x}dx\\ =-\frac{\pi }{2}{e}^{-2x}{0}^t\\ =-\frac{\pi }{2}\left(e^{-2t}-1\right)\end{array}$

Now,

The limit of $\underset{t\to \infty }{\lim }A\left(t\right)$ can be calculated by integrating the given curve,

  $\begin{array}{c}A\left(t\right)=\underset{0}{\overset{t}{\int }}{e}^{-x}dx\\ =-e^{-x}{0}^t\\ =1-e^{-t}\\ \end{array}$

Now, on simplifying,

  $\begin{array}{c}\underset{t\to \infty }{\lim }A\left(t\right)=1-\underset{t\to \infty }{\lim }{e}^{-t}\\ =1-e^{-\infty }\\ =1-0\\ =1\end{array}$

Therefore, the limit is $1$ .

b.

To determine

To find: The limit of $\underset{t\to \infty }{\lim }\frac{V\left(t\right)}{A\left(t\right)}$ .

Answer to Problem 59E

The limit is $\frac{\pi }{2}$ .

Explanation of Solution

Given:

The information is given above.

Calculation:

Here,

  $A\left(t\right)=1-e^{-t}$ as calculated in the above part and limit of $V\left(t\right)$ can be calculated as,

  $\begin{array}{c}\underset{t\to \infty }{\lim }V\left(t\right)=\pi \underset{0}{\overset{t}{{\displaystyle \int }}}{e}^{-2x}dx\\ =-\frac{\pi }{2}\underset{t\to \infty }{\lim }{e}^{-2t}-1\\ =-\frac{\pi }{2}\left(0-1\right)\\ =\frac{\pi }{2}\end{array}$

Thus,

  $\begin{array}{c}\underset{t\to \infty }{\lim }\frac{V\left(t\right)}{A\left(t\right)}=\frac{\pi }{2}\underset{t\to \infty }{\lim }\frac{e^{-2t}-1}{e^{-t}-1}\\ =\frac{\pi }{2}\underset{t\to \infty }{\lim }\frac{\left(e^{-t}-1\right)\left(e^{-t}-1\right)}{e^{-t}-1}\\ =\frac{\pi }{2}\left(e^{-\infty }+1\right)\\ =\frac{\pi }{2}\end{array}$

Therefore, the limit is $\frac{\pi }{2}$ .

c.

To determine

To find: The limit $\underset{t\to 0}{\lim }\frac{V\left(t\right)}{A\left(t\right)}$ .

Answer to Problem 59E

The limit is $\pi$ .

Explanation of Solution

Given:

The information is given above.

Calculation:

  $\begin{array}{c}\underset{t\to 0}{\lim }\frac{V\left(t\right)}{A\left(t\right)}=\frac{\pi }{2}\underset{t\to 0}{\lim }\frac{e^{-2t}-1}{e^{-t}-1}\\ =\frac{\pi }{2}\underset{t\to 0}{\lim }\frac{\left(e^{-t}-1\right)\left(e^{-t}-1\right)}{e^{-t}-1}\\ =\frac{\pi }{2}\left(e^{-0}+1\right)\\ =\pi \end{array}$

Therefore, the limit is $\pi$ .