Chapter 9, Problem 4CRT

(a)

To determine

To graph:For the given parametric equations.

Explanation of Solution

Given information:

The given parametric equationsis.

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

Graph:

Create a table to plot the graph.

$x$	$y$	$t$
$2$	$1$	$0$
$1$	$0$	$\frac{\pi }{2}$
$2$	$-1$	$\pi$
$\frac{3}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\pi }{4}$
$1$	$0$	$\frac{3\pi }{2}$
$2$	$1$	$2\pi$

The graph for the parametric equations using above table is shown in figure (1).

    figure (1)

Interpretation:Graph for the parametric equations $g\left(x\right)=\frac{1}{x+5}$ is shown in figure (1).

(b)

To determine

To find:The equation for the curve in rectangular coordinates by eliminating parameter and type of the curve.

Answer to Problem 4CRT

The equation for the curve in rectangular coordinates by eliminating parameter is $x=y^2+1$ and type of the curve is parabola.

Explanation of Solution

Given information:

The given parametric equationsis.

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

Calculation:

Eliminate the parameters by solving ${\sin }^{2}t$ .

  $2-x={\sin }^{2}t$

Squaring on both side of the second parametric equation.

  $y^2={\cos }^{2}t$

Using the formula ${\sin }^{2}t+{\cos }^{2}t=1$ .

  $\begin{array}{c}2-x+y^2=1\\ x=y^2+1\end{array}$

The type of graph for the above equation is parabola.

Therefore, the equation for the curve in rectangular coordinates by eliminating parameter is $x=y^2+1$ and type of the curve is parabola.