Chapter 1.4, Problem 8E

(a)

To determine

To plot: The graph of the parametric equations, the initial and terminal points. Also, indicate the direction in which the curve is traced.

Answer to Problem 8E

The graph of the parametric equations with the direction is shown in Figure (1), there are no initial or terminal points.

Explanation of Solution

Given information: The equations are $x=\left({\sec }^{2}t\right)-1$ and $y=\tan t$ , $-\pi /2<t<\pi /2$.

Calculation:

Use the following step to graph the parametric equations by graphing calculator.

Step 1: First press the “ON” button graphical calculator.

Step 2: Press the $\boxed{\text{MODE}}$ button, under FUNC and choose PAR option.

Step 2: Press $\boxed{\text{Y=}}$ , and set the parametric equations under ${\text{X}}_{\text{1T}}$ and ${\text{Y}}_{\text{1T}}$.

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-3,9\right]$ by $\left[-4,4\right]$ as shown below.

  

Figure (1)

From the graph it can be seen that there are no initial or terminal points.

Therefore, the graph of the parametric equations with the direction is shown in Figure (1), there are no initial or terminal points.

(b)

To determine

To find: The Cartesian equation and the portion of the graph of the Cartesian equation that is traced by the parameterized curve.

Answer to Problem 8E

The Cartesian equation is $y^2=x$ and the entire portion of the graph is traced by parameterized curve.

Explanation of Solution

Given information: The equations are $x=\left({\sec }^{2}t\right)-1$ and $y=\tan t$ , $-\pi /2<t<\pi /2$.

Calculation:

Simplify the equation $x=\left({\sec }^{2}t\right)-1$.

  $\begin{array}{c}x=\left({\sec }^{2}t\right)-1\\ =1+{\tan }^{2}t-1\\ ={\tan }^{2}t\end{array}$

Square both sides of the equation $y=\tan t$.

  $y^2={\tan }^{2}t$

Substitute $x$ for ${\tan }^{2}t$ in the equation $y^2={\tan }^{2}t$.

  $y^2=x$

The above Cartesian form is an equation of parabola. From the graph shown in part (a), the entire portion of the graph is traced by parameterized curve.

Therefore, the Cartesian equation is $y^2=x$ and the entire portion of the graph is traced by parameterized curve.