Chapter 10.6, Problem 29E

To determine

To find: the rectangular equations of the curve and graph the equation by using arrows to indicate orientation.

Answer to Problem 29E

  $\frac{x^2}{1^2}+\frac{y^2}{2^2}=1$

Counter-clockwise

Explanation of Solution

Given information:

The parametric equations are as follows:

  $\begin{array}{l}x=-\sin 2t\mathrm{...}\left(1\right)\\ y=2\cos 2t\left[\text{where }0\le t\le \pi \right]\mathrm{...}\left(2\right)\end{array}$

Calculation:

Solve equation (1) as follows:

  $\begin{array}{c}x=-\sin 2t\\ x^2={\sin }^2\left(2t\right)\end{array}$

Solve equation (2) as follows:

  $\begin{array}{c}y=2\cos 2t\\ y^2=4{\cos }^2\left(2t\right)\\ \frac{y^2}{4}={\cos }^2\left(2t\right)\end{array}$

Now evaluate the equation by using the trigonometric identity as shown below.

  $\begin{array}{c}{\cos }^{2}2t+{\sin }^{2}t=1\\ \frac{y^2}{4}+x^2=1\left[\text{Put the value of cosine and sine function}\right]\\ \frac{x^2}{1}+\frac{y^2}{4}=1\\ \frac{x^2}{1^2}+\frac{y^2}{2^2}=1\left[\text{Using }\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\right]\end{array}$

This equation represents the ellipse. The center $\left(h,k\right)$ is $\left(0,0\right)$ , semi-minor axis $\left(a\right)$ is 1 and semi-major axis $\left(b\right)$ is 2. The curve and its orientation for the given interval are shown below.

  

Thus, the orientation is counter-clockwise.