Polar Coordinates

Polar Coordinates Let P $(x,y)$ be a point on a plane. Then we represent P as a polar coordinate. Let the origin O be a pole. Then consider ray from the origin. In this case, $x$ -axis is called a polar axis and the angle $\theta$ formed by ray and polar axis is called argument. Let $\vert r\vert$ be the distance between the origin and a point P. Then the position of P can be expressed as a pair such as $[r, \theta]$ . Now we call this pair $[r, \theta]$ polar coordinate of P.

$[r, \theta]$ Note that $r$ can be negative.

NOTE Let a rectangular coordinate of P be $(x,y)$ and a polar coordinate of P be $[r, \theta]$ . Then we have

$\displaystyle x = r\cos{\theta}, y = r\sin{\theta}$

Thus,

$\displaystyle \tan{\theta} = \frac{y}{x} , r^2 = x^2 + y^2$

Then given $r$

and $\theta$ , P $(x,y)$

is fixed. On the other hand, even P $(x,y)$

is given, the value of $r$

and $\theta$ can not be determined uniquely.

Example 2..20 Find a polar coordinats of the point P that has rectangular coordinate $(1,\sqrt{3})$ .

**Figure 2.14:** Example2-20
$\includegraphics[width=3.5cm]{SOFTFIG-2/polarcoordinate.eps}$

SOLUTION Since $x = 1, y = \sqrt{3}$ , $\tan{\theta} = \sqrt{3}, r^2 = 1 + 3 = 4$ . Thus the point P is on the ray $\displaystyle{\theta = \frac{\pi}{3}}$ and the distance from the origin is 2. Thus the rectangular coordinate of P can be expressed as $\displaystyle{[2,\frac{\pi}{3}]}$ . Note that $\displaystyle{\theta = \frac{4\pi}{3}}$ and $r = -2$ also represents the point P. $\blacksquare$

Exercise 2..20 Find the rectangular coordinates of the point P with polar coordinates $[1,\frac{\pi}{3}]$ .

SOLUTION $r = 1, \theta = \frac{\pi}{3}$

$\displaystyle x$	$\displaystyle =$	$\displaystyle r\cos{\theta} = \cos{\frac{\pi}{3}} = \frac{1}{2}$
$\displaystyle y$	$\displaystyle =$	$\displaystyle r\sin{\theta} = \sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2}\ensuremath{ \blacksquare}$

Polar Equation Suppose the curve of a function is given by rectangular coordinates $g(x,y) = 0$ . Then the equation expressed by the polar coordinates ,

$\displaystyle g(r\cos{\theta}, r\sin{\theta}) = 0$

is called Polar Equation.

Example 2..21 Write the equation in polar coordinates

$\displaystyle x^2 + y^2 = 4$

Setting $x = r\cos{\theta}, y = r\sin{\theta}$ and eliminate $x,y$ .

SOLUTION Let $x = r\cos{\theta}, y = r\sin{\theta}$ . Then we have $r^2 = x^2 + y^2 = 4$ which implies $r = 2, -2$ $\blacksquare$

Exercise 2..21 Write the equation in polar coordinates

$\displaystyle (x-1)^2 + y^2 = 1$

SOLUTION Let $x = r\cos{\theta}, y = r\sin{\theta}$ . Then $(r\cos{\theta} - 1)^2 + (r\sin{\theta})^2 = 1$ $r^2 \cos^{2}{\theta} - 2r\cos{\theta} + 1 + r^2 \sin^{2}{\theta} = 1.$ Simplifying, $r^2 - 2r\cos{\theta} = r(r - 2\cos{\theta}) = 0.$ From this, $r = 2\cos{\theta}$ $\blacksquare$

Example 2..22 Sketch the following polar curve

$\displaystyle r(\theta) = 1 + \cos{\theta}$

**Figure 2.15:** Example2-22
$\includegraphics[width=3.5cm]{CALCFIG/FIg2-6-3.eps}$

If $r(-\theta) = r(\theta)$ , then for positive $\theta$ and negative $\theta$ give the same $r$ . Thus, the curve is symmetric with respect to the $x$ -axis.

SOLUTION 1. Since $\cos{\theta}$ is a even function, we have $r(-\theta) = r(\theta)$ . Thus it is symmetric with respect to the $x$ -axis. Thus to draw the curve of a function, we only need to check from $\theta = 0$ to $\theta = \pi$ .
2. Next we write a table for a polar coordinates of a curve.

$\begin{displaymath}\begin{array}{c\vert ccccccc} \theta & 0 & \frac{\pi}{6} & \... ... 1 & 1 - \frac{1}{2} & 1 - \frac{\sqrt{3}}{2} & 0 \end{array} \end{displaymath}$

From this, we have Cardioid $\blacksquare$

Exercise 2..22 Sketch the following polar curve

$\displaystyle r(\theta) = 1 + 2\cos{\theta}$

**Figure 2.16:** Exercise2-22
$\includegraphics[width=3.5cm]{CALCFIG/Fig2-6-4.eps}$

SOLUTION

$\cos{\theta}$ is even implies $r(-\theta) = r(\theta)$ . Thus it's curve is symmetric with respect ot the -axis.
Next we write a table for a polar coordinates of a curve.

$\begin{displaymath}\begin{array}{c\vert ccccccc} \theta & 0 & \frac{\pi}{6} & \... ... + \sqrt{3} & 1 + 1 & 1 & 1- 1 & 1- \sqrt{3} & -1 \end{array} \end{displaymath}$

The curve is called Limason $\blacksquare$

Exercise A

1.

Find the critical points of $f$

and the local extreme values. Describe the concavity of the graph of $f$

and find the points of inflection.

(a) $\displaystyle{f(x) = x^{3} - 3x + 2}$ (b) $\displaystyle{f(x) = x + \frac{1}{x}}$ (c) $\displaystyle{f(x) = x(x+1)(x+2)}$

(d) $\displaystyle{f(x) = \frac{x}{1+x^{2}}}$ (e) $\displaystyle{f(x) = \vert x-1\vert\vert x+2\vert}$

2.

Find the asymtote of the following functions

(a) $\displaystyle{f(x) = \frac{x}{3x-1}}$ (b) $\displaystyle{y = \frac{x^{2}}{x-2}}$ (c) $\displaystyle{y = \frac{2x}{x^{2}-9}}$

2.

Determine the following function has a vertical cuspWhen a function satisfies the condition $f'(c-0) = \pm \infty$ and $f'(c+0) = \mp \infty$ , we say $(c,f(c)$