Chapter 1.4, Problem 45E

(a)

To determine

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

Answer to Problem 45E

The graph of the parametric equations is shown in figure (1), there is no initial and terminal point.

Explanation of Solution

Given information: The equations are $x=-\sec t$ and $y=\tan t$ , $-\frac{\pi }{2}<t<\frac{\pi }{2}$.

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 as ${\text{X}}_{\text{1T}}=\boxed{(-)}\boxed{1}\boxed{÷}\boxed{\text{COS}}\boxed{\text{T}}\boxed{)}$ and ${\text{Y}}_{\text{1T}}=\boxed{\text{SIN}}\boxed{\text{T}}\boxed{)}\boxed{÷}\boxed{\text{COS}}\boxed{\text{T}}\boxed{)}$.

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-3,3\right]$ by $\left[-2,2\right]$ as shown below.

  

Figure (1)

There are no initial and terminal points because $\sec t$ and $\tan t$ both undefined at $t=-\frac{\pi }{2}$ and $t=\frac{\pi }{2}$.

Therefore, the graph of the parametric equations is shown in figure (1), there is no initial and terminal point.

(b)

To determine

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

Answer to Problem 45E

The Cartesian equation for a curve that contains the parameterized curve is $x^2-y^2=1$ and parameterized curve traces the left branch of the hyperbola defined by $x^2-y^2=1$.

Explanation of Solution

Given information: The equations are $x=-\sec t$ and $y=\tan t$ , $-\frac{\pi }{2}<t<\frac{\pi }{2}$.

Calculation:

Subtract the equations $x=-\sec t$ , and $y=\tan t$ after taking square.

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

This equation is the graph of a horizontal hyperbola centered at the origin.

The parameterized curve traces the left branch of the hyperbola defined by $x^2-y^2=1$ or all of the curve defined by $x=-\sqrt{y^2+1}$.

Therefore, the Cartesian equation for a curve that contains the parameterized curve is $x^2-y^2=1$ and parameterized curve traces the left branch of the hyperbola defined by $x^2-y^2=1$.