Chapter 6, Problem 28RE

Solution

To find: eliminate the parameter $t$ and identify the graph.

After eliminating $t$ , the obtained graph is a circle of radius 3 centered at $\left(0,0\right)$ , and the equation of circle is $x^2+y^2=9$ .

Given information:

Given parametric equations: $x=3\cos t,$

  $y=3\sin t$ .

Formula used:

• Parametric curve: The graph of the ordered pairs $\left(x,y\right)$ , $x=f\left(x\right),$ $y=g\left(x\right)$ where are functions $f,$ $g$ defined on an interval $I$ of $t$ values, is a parametric curve.
• The trigonometric identity: ${\cos }^{2}t+{\sin }^{2}t=1$ .

Calculation:

Compute $x^2+y^2:$

  $\begin{array}{c}{x}^2+y^2={\left(3\cos t\right)}^2+{\left(3\sin t\right)}^2\\ =9{\cos }^{2}t+9{\sin }^{2}t\\ =9\left(1\right)\\ =9\end{array}$

Thus, after eliminating $t,$ obtained equation is $x^2+y^2=9$ . It is a circle centered at origin with radius 3 units.

Graph of the circle $x^2+y^2=9$:

  

Verification:

Reproduce above graph using parametric equations: $x=3\cos t,$

  $y=3\sin t$

The points $\left(x,y\right)$ for different values of $t$ given below:

$t$	$0$	$\pi /2$	$\pi$	$3\pi /2$
$x=3\cos t$	$3$	$0$	$-3$	$0$
$y=3\sin t$	$0$	$3$	$0$	$-3$

All the points $\left(3,0\right),$ $\left(0,3\right),$ $\left(-3,0\right),$ and $\left(0,-3\right)$ lie on the circle $x^2+y^2=9$ .