Chapter 6.3, Problem 42E

Solution

a)

To give: an argument that the graphs of the pairs of parametric equations are the same.

As both the parametric equations represent the same cartesian equation of a circle and when the co-ordinates of the first circle is swapped, we get a point on the second circle. Hence it is proved that both the pairs of parametric equations are the same.

Given information:

  $x=3\cos t$ , $y=3\sin t$ and $x=3\sin t$ , $y=3\cos t$ for $0\le t\le 2\pi$ .

Formula used:

  $x=r\cos t$

  $y=r\sin t$

  $x^2+y^2=r^2$

Calculation:

Compare the general equation of the circle in parameters, $x=r\cos t$ , $y=r\sin t$ with the given equations $x=3\cos t$ and $y=3\sin t$ .

It is evident that $r=3$ .

Now, substitute $3$ for $r$ in $x^2+y^2=r^2$ .

  $\begin{array}{l}{x}^2+y^2=3^2\\ x^2+y^2=9\end{array}$

Compare the general equation of the circle in parameters, $x=r\sin t$ , $y=r\cos t$ with the given equations $x=3\sin t$ and $y=3\cos t$ .

It is evident that $r=3$ .

Now, substitute $3$ for $r$ in $x^2+y^2=r^2$ .

  $\begin{array}{l}{x}^2+y^2=3^2\\ x^2+y^2=9\end{array}$

So, it is clear that both the parametric equations represent the same the cartesian equation of the circle $x^2+y^2=9$ .

Also,

Substitute $2$ for $x$ in $x^2+y^2=9$ .

  $\begin{array}{l}{2}^2+y^2=9\\ 4+y^2=9\\ y^2=9-4\\ y^2=5\\ y=\sqrt{5}\\ \approx 2.24\end{array}$ . .

Therefore, when $x=2,y=2.24$ .

Similarly,

Substitute $2$ for $y$ in $x^2+y^2=9$ .

  $\begin{array}{l}{x}^2+2^2=9\\ x^2+4=9\\ x^2=9-4\\ x^2=5\\ x=\sqrt{5}\\ \approx 2.24\end{array}$ .

Therefore, when $y=2,x=2.24$ .

Hence, it is evident that when the co-ordinates of the point $\left(2,2.4\right)$ is swapped, we get $\left(2.4,2\right)$ which is a point on the same circle.

b)

To explain: how the parametrizations are different.

The first set of equations trace the curve in clockwise direction and the second set of equations trace the curve in anti-clockwise direction.

Given information:

  $x=3\cos t$ , $y=3\sin t$ and $x=3\sin t$ , $y=3\cos t$ for $0\le t\le 2\pi$ .

Formula used:

  $x=r\cos t$

  $y=r\sin t$

  $x^2+y^2=r^2$

Calculation:

Substitute $0$ for $t$ and $3$ for $r$ in $x=r\cos t$ .

  $\begin{array}{l}x=3\cos 0°\\ =3\left(1\right)\\ =3\end{array}$

Substitute $0$ for $t$ and $3$ for $r$ in $x=r\sin t$ .

  $\begin{array}{l}y=3\sin 0°\\ =3\left(0\right)\\ =0\end{array}$

So, the circle is formed from the point $\left(0,3\right)$ and it traces the curve in clockwise direction.

Similarly,

Substitute $0$ for $t$ and $3$ for $r$ in $x=r\sin t$ .

  $\begin{array}{l}x=3\sin 0°\\ =3\left(0\right)\\ =0\end{array}$

Similarly,

Substitute $0$ for $t$ and $3$ for $r$ in $y=r\cos t$ .

  $\begin{array}{l}y=3\cos 0°\\ =3\left(1\right)\\ =3\end{array}$

So, the circle is formed from the point $\left(3,0\right)$ and it traces the curve in anti-clockwise direction.