\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
9.6  p. 755:   1,  3,  5,  7, 11, 13, 23, 29, 31, 37, 55.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np
a) and e) both label the Cartesian point $(3,0)$.

b) and g) both label the Cartesian point $(-3,0)$.

c) and h) both label the Cartesian point $(-1,\sqrt3)$.

d)  and f) both label the Cartesian point $(1,\sqrt3)$.

Clearly there are no further identities.

\np \vskip-\baselineskip\itemitem{a)} $(2,\frac\pi2+2\pi k)$ and
$(-2,\frac{3\pi}2+2\pi k)$, $k$ any integer are all the polar coordinates
for this point, $(\sqrt2,\sqrt2)$ in Cartesian corrdinates.

\itemitem{b)}$(2,2\pi k)$ and $(-2,\pi+2\pi k)$, $k$ any integer are all the polar coordinates
for this point, $(2,0)$ in Cartesian corrdinates.

\itemitem{c)} $(-2,\frac\pi2+2\pi k)$ and $(2,\frac{3\pi}2+2\pi k)$, $k$ any integer are all the polar coordinates
for this point, $(-\sqrt2,-\sqrt2)$ in Cartesian corrdinates.

\itemitem{d)} $(-2,2\pi k)$ and $(2,\pi+2\pi k)$, $k$ any integer are all the polar coordinates
for this point, $(-2,0)$ in Cartesian corrdinates.


\np
The answers are written in \#1 above.


\np
$r=2$ is a circle centered at the pole with radius $2$.

\np
This is a wedge with its tip at the pole.

\np
This is a piece of the line through the origin with slope
$\tan\bigl(\frac\pi3\bigr)=\sqrt3$.
The left-most point on the line has Cartesian coordinates
$(\frac{-\sqrt3}{2},\frac{-1}{2})$ and
the right-most point has Cartesian coordinates
$(\frac{3\sqrt3}{2},\frac{3}{2})$.

\np$r\cos\theta=2$ or $x=2$: a vertical line whose $x$ coordinate
is $2$.

\np$r\cos\theta+r\sin\theta=1$, or $x+y=1$.
The line with slope $-1$ through the Cartesian point $(1,0)$.

\np$r^2=1$, or $x^2+y^2=1$ which is the circle centered at the origin 
with radius $1$.

\np$r=\csc\theta\ e^{r\cos\theta}$, or
$r\sin\theta=e^{r\cos\theta}$, so
$y=e^x$ which is the standard graph of the exponential function.

\np$\Frac{x^2}{9}+\Frac{y^2}{4}=1$, or
$\Frac{r^2\cos^2\theta}{9}+\Frac{r^2\sin^2\theta}{4}=1$.
One can now write this in many forms.
Multiply both sides by $36$ to get the answer in the back of the book.

\end