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\begin{center}
{\bf Coordinate Geometry}
\end{center}

In this section we shall discuss various co-ordinate systems in 2
and 3 dimensions and equations for a variety of curves.

${}$

Cartesian coordinates are formed by two sets of parallel lines,
usually orthogonal, although not necessarily so.  A point is
specified by an ordered pair $(a,b)$, and can be thought of as the
intersection of the lines $x=a$ and $y=b$.

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You will be familiar with many equations and the corresponding
curves in rectangular cartesian co-ordinates.  You should note
that the shape of a curve connected with an equation depends on
the co-ordinate system.

e.g. $\ds x^2+y^2=1$ is the equation of a circle in rectangular
cartesian co-ordinates, but in oblique cartesian co-ordinates it
represents an ellipse.

\begin{center}
{\bf Polar Co-ordinates}
\end{center}

This system is set up with an origin $O$ and a direction fixed
through $O$.  The co-ordinates of a point $P$ are specified as the
distance $OP$ and the angle $OP$ makes with the fixed direction.
Thus we have $P(r,\theta)$.  Each pair of numbers specifies a
point, but without any restrictions on $r$ and $\theta$ a point
may be assigned more than one set of co-ordinates.

e.g $(1,0)~(1,2\pi)$

If we interpret negative $r$ as reflected through the origin then
$(1,0)~(-1,\pi)$.

In order to achieve uniqueness we make the restriction $r>0$, and
we restrict $\theta$ to belong to some interval of length $2\pi$.
e.g $-\pi<\theta\leq\pi$ is the most common.  In describing some
curves with more than one point in a given direction like spirals
it is however convenient to relax this restriction on $\theta$. We
shall always take $r\geq0$ however.  There is still a slight
problem about the origin $O$.  Clearly $r=0$, but what about
$\theta$?  We do not specify co-ordinates uniquely for $O$, but if
we obtain $r=0$ from an equation then that will correspond to $O$.

Just as the grid for Cartesian co-ordinates consists of lines
$x$=const, $y$=const so the grid for polar co-ordinates consists
of $r$=const (circles centre $O$), $\theta$=const (half-lines
starting at $O$).

We can transform from cartesian to polar co-ordinates as follows.

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If P has cartesian co-ordinates $(x,y)$ and polar co-ordinates
$(r,\theta)$ then

$\ds r=\sqrt{x^2+y^2}, \,\,\, \tan\theta=\frac{y}{x}, \,\,\,
x=r\cos\theta, \,\,\, y=r\sin\theta$.

Note that it is ambiguous to write $\ds
\theta=\tan^{-1}\frac{y}{x}$

$\begin{array}{llll} {\rm since\ if} & (x,y)=(1,1) & \frac{y}{x}=1
& {\rm and\ } \theta=\frac{\pi}{4}\\ {\rm and\ if} & (x,y)=(-1,-1)
& \frac{y}{x}=1 & {\rm but\ } \theta=-\frac{3\pi}{4}\end{array}$

We could write unambiguously
$\begin{array}{lccl}\theta=\tan^{-1}\frac{y}{x} & {\rm where} &
0<\theta<\pi & {\rm if\ }y>0\\ & {\rm and} & -\pi<\theta<\pi &
{\rm if\ }y<0 \end{array}$

(We still need the special cases $x=0, \,\,\, y=0$ for
completeness) but it is better to draw a diagram as well.

${}$

Notice that to say "the point $P$ has co-ordinates $(a,b)$" is
ambiguous out of context.  It depends what co-ordinate system we
are using.

${}$

$\begin{array}{ll} {\rm So\ rectangular\ cartesia} & P=(2,1)\\
60^\circ {\rm \ oblique,\ same\ }O {\rm \ and\ }x {\rm axis} &
P=(2-\frac{1}{\sqrt3})\\ {\rm polars,\ same\ }O {\rm \ and\ }x
{\rm axis} & P=(\sqrt5,0.464) {\rm \ \ (radians)}\end{array}$

\begin{center}
{\bf Some Equations in Polar Co-ordinates}
\end{center}

\begin{itemize}
\item[i)]
Straight line

Since the cartesian equation is $ax+by+c=0$

we use $x=r\cos\theta, \,\,\, y=r\sin\theta$ to obtain

$$ r(a\cos\theta+b\sin\theta)+c=0 \,\,\,\, (*)$$

alternatively we have $r\cos(\theta-\alpha)=p \,\,\,\,$ (*)

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You should try and relate the two forms of equation above.

Note that the equation $\theta=\frac{\pi}{4}$ does not represent
the whole line

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You should analyse what happens to the equations (*) when the line
passes through $O$.

\item[ii)]
circle

A circle centred at $O$ has equation $r=a$.

A circle having $O$ on the circumference and the initial line as
diameter.

DIAGRAM

$r=a\cos\theta$

DIAGRAM

$r=a\cos(\theta-\alpha)$

DIAGRAM

Applying the cosine formula to the triangle $POC$ gives

$$ a^2=r^2+d^2-2rd\cos(\theta-\alpha).$$

Again you should convert from cartesians to polars and try to
relate the equations.

We now consider slopes of tangents in polar co-ordinates.

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$\ds
\tan\phi\approx\frac{PN}{NP'}\approx\frac{r\delta\theta}{\delta
r}\approx r\frac{d\theta}{dr}$

$\ds PN=OP\sin PON=r\sin\delta\theta$

\begin{eqnarray*} P'N &=& OP'-ON=r+\delta
r-r\cos\delta\theta=\delta r+r(1-\cos\delta\theta)\\ &=& \delta
r+r*2\sin^2\frac{1}{2}\delta\theta\end{eqnarray*}

\begin{eqnarray*} \tan OP'P &=& \frac{PN}{P'N}=\frac{rsin\delta\theta}{\delta
r+r(2\sin^2\frac{1}{2}\delta\theta)}=
\frac{r*2\sin\frac{1}{2}\delta\theta\cos\frac{1}{2}\delta\theta}
{\delta r+r(2sin^2\frac{1}{2}\delta\theta)}\\ &=&
\frac{r\frac{\sin\frac{1}{2}\delta\theta}{\frac{1}{2}\delta\theta}\frac{\delta\theta}{\delta
r}\cos\frac{1}{2}\delta\theta}{1+r\frac{sin\frac{1}{2}\delta\theta}{\frac{1}{2}\delta\theta}
\frac{\delta\theta}{\delta r}\sin\frac{1}{2}\delta\theta}\\ &\to&
\frac{r*1*\frac{d\theta}{dr}*1}{1+r*1*\frac{d\theta}{dr}*0} =
r\frac{d\theta}{dr} {\rm \ \ \ as\ } P'\to P \end{eqnarray*}

$OP'P$ tends to the angle $\phi$ shown below

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$\tan\phi=r\frac{d\theta}{dr}, \,\,\,\,
\cot\phi=\frac{1}{r}\frac{dr}{d\theta}$

You should check through the above in the case when

$\delta r<0,$ or $OP'<ON$.

{\bf Example}

Consider the circle $r=a\cos\theta$

DIAGRAM

$\ds \frac{dr}{d\theta}=-a\sin\theta \hspace{0.3in}
\frac{1}{r}\frac{dr}{d\theta}=-\tan\theta=\cot\phi$

So
$-\tan\theta=\tan(\frac{\pi}{2}-\phi)=-\tan(\phi-\frac{\pi}{2})$

so $\theta=\phi-\frac{\pi}{2}$ \hspace{0.3in} or
$\phi=\frac{\pi}{2}+\theta$.

Consider the equation $r=a(1+\cos\theta)$.  Because
$\cos\theta=\cos(-\theta)$ this curve is symmetrical about the
line $\theta=0$.  As $\theta$ increases from 0 to $\pi, \,\,\, r$
decreases from $2a$ to 0.

$\ds \frac{dr}{d\theta}=-a\sin\theta$

so $\ds r\frac{d\theta}{dr}=-\frac{1+\cos\theta}{\sin\theta}=
-\cot\frac{1}{2}\theta=\tan\phi$

$\ds \tan\phi=-\cot\frac{1}{2}\theta=\tan\frac{1}{2}(\pi+\theta)$

so $\ds \phi=\frac{1}{2}(\pi+\theta)$

when $\theta=0, \,\,\,\, \phi=\frac{1}{2}\pi$ so the curve is at
right angles to the initial line.

DIAGRAM

when $\psi=0$, since $\psi=\phi+\theta$ we have
$\theta=-\frac{\pi}{3}.$

when $\theta=\frac{\pi}{3}, \,\,\,\, \psi=\pi$ so the highest
points are at $\theta=\frac{\pi}{3}$.

\end{itemize}

\end{document}