Find rectangular equations for the plane curves represented by each of the parametric equations. Graph each curve and indicate its orientation.

(a) $x( t) =R\, \cos t\quad y( t) = R\, \sin t\quad 0\leq t\leq \pi$ and $R> 0$

(b) $x( t) =R\, \sin t\quad y( t) = R\, \cos t\quad 0\leq t\leq \pi$ and $R> 0$

Solution (a) We eliminate the parameter $t$ using a Pythagorean Identity. $$\begin{eqnarray*} \cos^{2}t+\sin ^{2}t &=&1 \\ \left( \dfrac{x}{R}\right) ^{2}+\left( \dfrac{y}{R}\right) ^{2} &=&1 \qquad \color{#0066A7}{\cos t=\dfrac{x}{R},\quad \sin t=\dfrac{y}{R}} \\ x^{2}+y^{2} &=&R^{2} \tag{1} \end{eqnarray*}$$

Figure 4 $y=\sqrt{R^{2}-x^{2}}$, $-R\leq x \leq R$.

The rectangular equation represents a circle with radius $R$ and center at the origin. In the parametric equations, $0\leq t\leq \pi$, so the curve begins when $t=0$ at the point $( R, 0)$, passes through the point $( 0, R)$ when $t=\dfrac{\pi }{2}$, and ends when $t=\pi$ at the point $( -R,0)$. The curve is an upper semicircle of radius $R$ with counterclockwise orientation, as shown in Figure 4. If we solve equation (1) for $y$, we obtain the rectangular equation of the semicircle $$y=\sqrt{R^{2}-x^{2}}\qquad\hbox{where }-R\leq x\leq R$$

(b) We eliminate the parameter $t$ as we did in (a), and again we obtain $$\begin{eqnarray*} x^{2}+y^{2}=R^{2}\tag{2} \end{eqnarray*}$$

Figure 5 $x=\sqrt{R^{2}-y^{2}}$, $-R\leq y \leq R$.

The rectangular equation represents a circle with radius $R$ and center at $( 0,0)$. But in the parametric equations, $0\leq t\leq \pi$, so now the curve begins when $t=0$ at the point $( 0, R)$, passes through the point $( R, 0)$ when $t=\dfrac{\pi }{2}$, and ends at the point $\left( 0,-R\right)$ when $t=\pi$. The curve is a right semicircle of radius $R$ with a clockwise orientation, as shown in Figure 5. If we solve equation (2) for $x$, we obtain the rectangular equation of the semicircle $$x=\sqrt{R^{2}-y^{2}}\qquad \hbox{where }-R\leq y\leq R$$