Answer
(a) The corresponding points in (B) are denoted by lowercase letters. They are listed in the table below:
$\begin{array}{l}
\begin{array}{*{20}{c}}
{Figure21\left( {\rm{A}} \right)}\\
{Polar:\left( {r,\theta } \right)}\\
{A = \left( {0,0} \right)}\\
{B = \left( {1,\frac{\pi }{4}} \right)}\\
{C = \left( {0,\frac{\pi }{2}} \right)}\\
{D = \left( { - 1,\frac{{3\pi }}{4}} \right)}
\end{array}\begin{array}{*{20}{c}}
{Figure21\left( {\rm{B}} \right)}\\
{Rectangular:\left( {x,y} \right)}\\
{a = \left( {0,0} \right)}\\
{b = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)}\\
{c = \left( {0,0} \right)}\\
{d = \left( {\frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)}
\end{array}\\
\begin{array}{*{20}{c}}
{E = \left( {0,\pi } \right)}\\
{F = \left( {1,\frac{{5\pi }}{4}} \right)}\\
{G = \left( {0,\frac{{3\pi }}{2}} \right)}\\
{H = \left( { - 1,\frac{{7\pi }}{4}} \right)}\\
{I = \left( {0,2\pi } \right)}
\end{array}\begin{array}{*{20}{c}}
{e = \left( {0,0} \right)}\\
{f = \left( { - \frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)}\\
{g = \left( {0,0} \right)}\\
{h = \left( { - \frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)}\\
{i = \left( {0,0} \right)}
\end{array}
\end{array}$
(b)
The interval: $\left[ {0,\frac{\pi }{2}} \right]$
The part of the curve in this interval is in the first quadrant.
The interval: $\left[ {\frac{\pi }{2},\pi } \right]$
The part of the curve in this interval is in the fourth quadrant.
The interval: $\left[ {\pi ,\frac{{3\pi }}{2}} \right]$
The part of the curve in this interval is in the third quadrant.
The interval: $\left[ {\frac{{3\pi }}{2},2\pi } \right]$
The part of the curve in this interval is in the second quadrant.
Work Step by Step
From Figure 21 (A) and using $r=\sin 2\theta$, the points $A$ - $I$ in polar coordinates are
$A = \left( {0,0} \right)$, $B = \left( {1,\frac{\pi }{4}} \right)$, $C = \left( {0,\frac{\pi }{2}} \right)$, $D = \left( { - 1,\frac{{3\pi }}{4}} \right)$, $E = \left( {0,\pi } \right)$, $F = \left( {1,\frac{{5\pi }}{4}} \right)$, $G = \left( {0,\frac{{3\pi }}{2}} \right)$, $H = \left( { - 1,\frac{{7\pi }}{4}} \right)$, $I = \left( {0,2\pi } \right)$.
(a) To find the points in Figure 21 (B) corresponding to points $A$ - $I$ in Figure 21 (A), we use the conversion formula from polar coordinates to rectangular coordinates given by
$x=r \cos \theta$, ${\ \ \ }$ $y=r \sin \theta$.
Denote the corresponding points in (B) by lowercase letters. The results are listed in the table below:
$\begin{array}{l}
\begin{array}{*{20}{c}}
{Figure21\left( {\rm{A}} \right)}\\
{Polar:\left( {r,\theta } \right)}\\
{A = \left( {0,0} \right)}\\
{B = \left( {1,\frac{\pi }{4}} \right)}\\
{C = \left( {0,\frac{\pi }{2}} \right)}\\
{D = \left( { - 1,\frac{{3\pi }}{4}} \right)}
\end{array}\begin{array}{*{20}{c}}
{Figure21\left( {\rm{B}} \right)}\\
{Rectangular:\left( {x,y} \right)}\\
{a = \left( {0,0} \right)}\\
{b = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)}\\
{c = \left( {0,0} \right)}\\
{d = \left( {\frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)}
\end{array}\\
\begin{array}{*{20}{c}}
{E = \left( {0,\pi } \right)}\\
{F = \left( {1,\frac{{5\pi }}{4}} \right)}\\
{G = \left( {0,\frac{{3\pi }}{2}} \right)}\\
{H = \left( { - 1,\frac{{7\pi }}{4}} \right)}\\
{I = \left( {0,2\pi } \right)}
\end{array}\begin{array}{*{20}{c}}
{e = \left( {0,0} \right)}\\
{f = \left( { - \frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)}\\
{g = \left( {0,0} \right)}\\
{h = \left( { - \frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)}\\
{i = \left( {0,0} \right)}
\end{array}
\end{array}$
(b)
The interval: $\left[ {0,\frac{\pi }{2}} \right]$
Since the points $a$, $b$, $c$ are in this interval, the part of the curve in this interval is in the first quadrant.
The interval: $\left[ {\frac{\pi }{2},\pi } \right]$
Since the points $c$, $d$, $e$ are in this interval, the part of the curve in this interval is in the fourth quadrant.
The interval: $\left[ {\pi ,\frac{{3\pi }}{2}} \right]$
Since the points $e$, $f$, $g$ are in this interval, the part of the curve in this interval is in the third quadrant.
The interval: $\left[ {\frac{{3\pi }}{2},2\pi } \right]$
Since the points $g$, $h$, $i$ are in this interval, the part of the curve in this interval is in the second quadrant.
