## Calculus (3rd Edition)

(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.
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. 