Chapter 4.7, Problem 110E

(a).

To determine

To graph:

The $\beta$ as a function of $x$ .

Explanation of Solution

Given:

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$ .

Graph:

Consider the provided function

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$

The graph of the function $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$ obtained with the help of graphing utility is shown below:

  

(b).

To determine

To find:

The approximate distance from the picture when $\beta$ is maximum.

Answer to Problem 110E

The required maximum distance is $2\text{ft}$ .

Explanation of Solution

Given:

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$ .

Calculation:

Consider the provided function

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$

The graph of the function $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$ obtained with the help of graphing utility is shown below:

  

Move the cursor along the graph to approximate the distance from the picture when $\beta$ is maximum.

  $\max \approx 2.0$

Hence, the required maximum distance is $2\text{ft}$ .

(c).

To determine

To find:

The asymptote of the graph and discuss its meaning in the context of the problem.

Answer to Problem 110E

The asymptote is $\beta =0$ .

Because as the distance increases the $x$ gets smaller and smaller.

Explanation of Solution

Given:

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$ .

Calculation:

Consider the provided function

  $\beta =\arctan \left(\frac{3x}{x^2+4}\right)$ , $x>0$

  $\begin{array}{c}\underset{x\to \infty }{\lim }\beta =\underset{x\to \infty }{\lim }\arctan \left(\frac{3x}{x^2+4}\right)\\ =0\end{array}$

So, the asymptote is $\beta =0$ .

Because as the distance increases the $x$ gets smaller and smaller.