Chapter 6.3, Problem 65E

Solution

a)

To find: graph of given parametric equations for different values of $a$ .

The graph of given parametric equations for different values of $a$ is shown in Figure (1).

Given information:

Given parametric equations: $x=a\cos t$ , $y=a\sin t$ , $0\le t\le 2\pi$ .

Calculation:

Graph of the parametric equations for $a=1$:

$t$	$0°$	$90°$	$180°$	$270°$
$x=a\cos t$	$1$	$0$	$-1$	$0$
$y=a\sin t$	$0$	$1$	$0$	$-1$

By joining the points $\left(1,0\right),$

  $\left(0,1\right),$

  $\left(-1,0\right),$

  $\left(0,-1\right)$ , and $\left(1,0\right),$ get a circle of radius 1.

Graph of the parametric equations for $a=2$:

$t$	$0°$	$90°$	$180°$	$270°$
$x=a\cos t$	$2$	$0$	$-2$	$0$
$y=a\sin t$	$0$	$2$	$0$	$-2$

By joining the points $\left(2,0\right),$

  $\left(0,2\right),$

  $\left(-2,0\right),$

  $\left(0,-2\right)$ , and $\left(2,0\right),$ get a circle of radius 2.

Graph of the parametric equations for $a=3$:

$t$	$0°$	$90°$	$180°$	$270°$
$x=a\cos t$	$3$	$0$	$-3$	$0$
$y=a\sin t$	$0$	$3$	$0$	$-3$

By joining the points $\left(3,0\right),$

  $\left(0,3\right),$

  $\left(-3,0\right),$

  $\left(0,-3\right)$ , and $\left(3,0\right),$ get a circle of radius 3.

Graph of the parametric equations for $a=4$:

$t$	$0°$	$90°$	$180°$	$270°$
$x=a\cos t$	$4$	$0$	$-4$	$0$
$y=a\sin t$	$0$	$4$	$0$	$-4$

By joining the points $\left(4,0\right),$

  $\left(0,4\right),$

  $\left(-4,0\right),$

  $\left(0,-4\right)$ , and $\left(4,0\right),$ get a circle of radius 4.

Plot all circles as shown in Figure (1).

  

Figure (1)

b)

To find: rectangular equation after eliminating $t$ .

After eliminating $t$ , obtained equation is $x^2+y^2=a^2$ . It is a circle of radius $a$ with center as origin.

Given information:

Given parametric equations: $x=a\cos t$ , $y=a\sin t$ , $0\le t\le 2\pi$ .

Formula used:

  ${\cos }^2\theta +{\sin }^2\theta =1$

Calculation:

Elimination of $t$:

Compute $x^2+y^2$:

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

Thus, after eliminating $t$ , obtained equation is $x^2+y^2=a^2$ , which is a circle of radius $a$ .

c)

To find: graph of given parametric equations for different values of $h$ , and $k$ .

The graphs of given parametric equation for different values of $h$ , and $k$ are shown in figures (2), (3), (4), and (5).

Given information:

The parametric equations: $x=h+a\cos t$ , $y=k+a\sin t$ , $0\le t\le 2\pi$ .

Also given that $a=1$ and the pair of values for $h$ and $k$ are given.

$h$	2	$-2$	$-4$	$3$
$k$	3	3	$-2$	$-3$

Calculation:

Substitute the values 2 for $h$ , 3 for $k$ and $a=1$ in the parametric equations:

  $x=h+a\cos t$ , $y=k+a\sin t$

The following table shows that the different points of $\left(x,y\right)$ for different values of $t$ .

$t$	$0°$	$90°$	$180°$	$270°$
$x=h+a\cos t$	$3$	$2$	$1$	$2$
$y=k+a\sin t$	$3$	$4$	$3$	$2$

By joining the points $\left(3,3\right),$

  $\left(2,4\right),$

  $\left(1,3\right),$

  $\left(2,2\right)$ , and $\left(3,3\right)$ , get a circle of radius 1 centers at $\left(2,3\right)$ . It is shown in figure (2).

  

Figure (2)

Substitute the values $-2$ for $h$ , 3 for $k$ and $a=1$

in the parametric equations:

  $x=h+a\cos t$ , $y=k+a\sin t$

The following table shows that the different points of $\left(x,y\right)$ for different values of $t$ .

$t$	$0°$	$90°$	$180°$	$270°$
$x=h+a\cos t$	$-1$	$-2$	$-3$	$-2$
$y=k+a\sin t$	$3$	$4$	$3$	$2$

By joining the points $\left(-1,3\right),$

  $\left(-2,4\right),$

  $\left(-1,3\right),$

  $\left(-2,2\right)$ , and $\left(-1,3\right),$ get a circle of radius 1 centered at $\left(-2,3\right)$ . It is shown in figure (3).

  

Figure (3)

Substitute the values $-4$ for $h$ , $-2$ for $k$ and $a=1$

in the parametric equations:

  $x=h+a\cos t$ , $y=k+a\sin t$

The following table shows that the different points of $\left(x,y\right)$ for different values of $t$ .

$t$	$0°$	$90°$	$180°$	$270°$
$x=h+a\cos t$	$-3$	$-4$	$-5$	$-4$
$y=k+a\sin t$	$-2$	$-1$	$-2$	$-3$

By joining the points $\left(-3,-2\right),$

  $\left(-2,-1\right),$

  $\left(-5,-2\right),$

  $\left(-4,-3\right)$ , and $\left(-3,-2\right),$ get a circle of radius 1 centered at $\left(-4,-2\right)$ . It is shown in figure (4).

  

Figure (4)

Substitute the values $-3$ for $h$ , 3 for $k$ and $a=1$

in the parametric equations:

  $x=h+a\cos t$ , $y=k+a\sin t$

The following table shows that the different points of $\left(x,y\right)$ for different values of $t$ .

$t$	$0°$	$90°$	$180°$	$270°$
$x=h+a\cos t$	$-2$	$-3$	$-4$	$-3$
$y=k+a\sin t$	$3$	$4$	$3$	$2$

By joining the points $\left(-2,3\right),$

  $\left(-3,4\right),$

  $\left(-4,3\right),$

  $\left(-3,2\right)$ , and $\left(-2,3\right),$ , get a circle of radius 1 centers at $\left(-3,3\right)$ . It is shown in figure (5).

  

Figure (5)

d)

To find: rectangular equation after eliminating $t$ .

After eliminating $t$ , obtained equation is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$ , which is a circle centered $\left(h,k\right)$ of radius $a$ .

Given information:

Given parametric equations: $x=h+a\cos t$ , $y=k+a\sin t$ , $0\le t\le 2\pi$ .

Formula used:

  ${\cos }^2\theta +{\sin }^2\theta =1$

Calculation:

Elimination of $t$:

Compute ${\left(x-h\right)}^2+{\left(y-k\right)}^2$:

  $\begin{array}{c}{\left(x-h\right)}^2+{\left(y-k\right)}^2={\left(a\cos t\right)}^2+{\left(a\sin t\right)}^2\\ =a^2\left({\cos }^{2}t+{\sin }^{2}t\right)\\ =a^2\end{array}$

Thus, after eliminating $t$ , obtained equation is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$ , which is a circle centered $\left(h,k\right)$ of radius $a$ .

e)

To find: the parametric equation of the circle with center $\left(-1,4\right)$ and radius 3.

The parametric equation of the circle with center $\left(-1,4\right)$ and radius 3 is

  $x=-1+3\cos t$ , $y=4+3\sin t$ , $0\le t\le 2\pi$ .

Given information:

Center of the circle is $\left(-1,4\right)$ and radius is 3.

Formula used:

Parametric equation of the circle with center $\left(h,k\right)$ having the radius $r$ is given as $x=h+a\cos t$ , $y=k+a\sin t$ , $0\le t\le 2\pi$ .

Calculation:

Substitute $-1$ for $h$ , 4 for $k$ , and 3 for $r$ in the equations: $x=h+a\cos t$ , $y=k+a\sin t$ , $0\le t\le 2\pi$ .

  $\begin{array}{c}x=h+a\cos t\\ =-1+r\cos t\end{array}$

And

  $\begin{array}{c}y=k+a\sin t\\ =4+3\sin t\end{array}$

Thus, parametric equation of the circle having center $\left(-1,4\right)$ with radius 3 is $x=-1+3\cos t$ , $y=4+3\sin t$ , $0\le t\le 2\pi$ .