Chapter 1.4, Problem 46E

(a)

To determine

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

Answer to Problem 46E

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 $x=\tan t$ and $y=-2\sec t$ , $-\frac{\pi }{2}<t<\frac{\pi }{2}$.

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 ${\text{X}}_{\text{1T}}=\boxed{\text{SIN}}\boxed{\text{T}}\boxed{)}\boxed{÷}\boxed{\text{COS}}\boxed{\text{T}}\boxed{)}$ and ${\text{Y}}_{\text{1T}}=\boxed{(-)}\boxed{2}\boxed{)}\boxed{÷}\boxed{\text{COS}}\boxed{\text{T}}\boxed{)}$.

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-6,6\right]$ by $\left[-5,1\right]$ 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 46E

The Cartesian equation for a curve that contains the parameterized curve is ${\left(\frac{y}{2}\right)}^2-x^2=1$ and parameterized curve traces the lower branch of the hyperbola defined by ${\left(\frac{y}{2}\right)}^2-x^2=1$.

Explanation of Solution

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

Calculation:

Rewrite the equation $y=-2\sec t$ as $\frac{y}{2}=-\sec t$.

Subtract the equations $x=\tan t$ , and $\frac{y}{2}=-\sec t$ after taking square.

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

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

Therefore, the Cartesian equation for a curve that contains the parameterized curve is ${\left(\frac{y}{2}\right)}^2-x^2=1$ and parameterized curve traces the lower branch of the hyperbola defined by ${\left(\frac{y}{2}\right)}^2-x^2=1$.