Chapter 1.4, Problem 21E

(a)

To determine

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

Answer to Problem 21E

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

Explanation of Solution

Given information: The equations are $x=\sin t$ and $y=\cos 2t$ , $-\infty <t<\infty$.

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 3: Press $\boxed{\text{Y=}}$ , and set the parametric equations equal to ${\text{X}}_{\text{1T}}$ and ${\text{Y}}_{\text{1T}}$.

Step 4: 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, since the interval has no beginning or end point and curve is traced in both the directions.

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

(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 21E

The Cartesian equation for a curve that contains the parameterized curve is $y=1-2x^2$ and parameterized curve traces the portion of the parabola defined by $y=1-2x^2$ corresponding to $-1\le x\le 1$.

Explanation of Solution

Given information: The equations are $x=\sin t$ and $y=\cos 2t$ , $-\infty <t<\infty$.

Calculation:

Simplify the equation $y=\cos 2t$.

  $\begin{array}{c}y=\cos 2t\\ ={\cos }^{2}t-{\sin }^{2}t\\ =1-2{\sin }^{2}t\\ =1-2x^2\end{array}$

As shown in the graph, the parameterized curve traces the portion of the parabola defined by $y=1-2x^2$ corresponding to $-1\le x\le 1$.

Therefore, the Cartesian equation for a curve that contains the parameterized curve is $y=1-2x^2$ and parameterized curve traces the portion of the parabola defined by $y=1-2x^2$ corresponding to $-1\le x\le 1$.