Chapter 10.6, Problem 31E

To determine

To find: the rectangular equations of the curve.

Answer to Problem 31E

  $x^2+y^2=2^2$

Explanation of Solution

Given information:

The parametric equations are as follows:

  $\begin{array}{l}x=-3\cos 2t\mathrm{...}\left(1\right)\\ y=3sin2t\left[\text{where }0\le t\le 2\pi \right]\mathrm{...}\left(2\right)\end{array}$

Calculation:

Solve equation (1) as follows:

  $\begin{array}{c}y=3\sin 2t\\ y^2=9{\sin }^{2}2t\left[\text{Squaring both sides}\right]\\ {\sin }^{2}2t=\frac{y^2}{9}\left[\text{Divide each side by 9}\right]\end{array}$

Solve equation (2) as follows:

  $\begin{array}{c}x=-3\cos 2t\\ x^2=9{\cos }^{2}2t\left[\text{Squaring both sides}\right]\\ {\cos }^{2}2t=\frac{x^2}{9}\left[\text{Divide each side by 9}\right]\end{array}$

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

  $\begin{array}{c}{\cos }^{2}2t+{\sin }^{2}2t=1\\ \frac{x^2}{9}+\frac{y^2}{9}=1\left[\text{Put the value of cosine and sine function}\right]\\ x^2+y^2=9\left[\text{Multiply each side by 9}\right]\\ x^2+y^2=2^2\left[\text{Using }{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\right]\end{array}$

Thus, the equation is $x^2+y^2=2^2$ .