Chapter 10.1, Problem 3E

To determine

To find: The equation that relates $x$ and $y$ directly.

Answer to Problem 3E

The equation is $y=\sqrt{1+x^2}$

Explanation of Solution

Given information:

The equations $x=\tan t$ and

  $y=\sec t$ for $t$ in the interval $[0,\pi /4]$ .

Calculation:

The given equations

  $\begin{array}{c}x=\tan t,\\ y=\sec t\text{,}\end{array}$

And

  $t\in [0,\pi /4]$

To use the previous result, the parametric curve in the following graph.

  

Thus may infer from the graph that the parametric curve generates a line segment, which indicates that it defines $y$ as a function of $x$ .

Since for $t\in [0,\pi /4]$

So,

  $\sec t\ge 0$ and $0\le \tan t\le 1$

To use the trigonometric identity

  ${\sec }^{2}t=1+{\tan }^{2}t$ .

So,

  $\sec t=\sqrt{1+{\tan }^{2}t}$

Where,

  $0\le t\le \pi /4$

Thus can conclude from the previous findings that the equation connecting $x$ and $y$ is equivalent to:

  $y=\sqrt{1+x^2}$

Therefore, the required parametric curve defines $y$ as a function of $x$ , while the equation that relates $x$ and $y$ is equal to $y=\sqrt{1+x^2}$ .