Polar Coordinates

Polar Coordinates
Let P 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, -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.
NOTE Let a rectangular coordinate of P be 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 and $\theta$ , P is fixed. On the other hand, even P is given, the value of and $\theta$ can not be determined uniquely.

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

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

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

$\displaystyle r(\theta) = 1 + \cos{\theta}$
SOLUTION 1. Since $\cos{\theta}$ is a even function, we have $r(-\theta) = r(\theta)$ . Thus it is symmetric with respect to the -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

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

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 and the local extreme values. Describe the concavity of the graph of 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?D
(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 cusp?DWhen a function satisfies the condition $f'(c-0) = \pm \infty$ and $f'(c+0) = \mp \infty$ , we say is a vertical cusp?D
(a) $\displaystyle{f(x) = x^{1/3}}$ (b) $\displaystyle{f(x) = x^{2/3}}$ (c) $\displaystyle{f(x) = \sqrt{\vert x-2\vert}}$

Exercise B

4.

Find the local extreme value of and describe the concavity of the graph of ?D
(a) $\displaystyle{f(x) = x^{3} - 6x^2 + 9x + 3}$ (b) $\displaystyle{f(x) = x^{2}e^{-x}}$

1.

Sketch the graph?D
(a) $\displaystyle{y = \frac{1}{1 + e^{x}}}$ (b) $\displaystyle{y = \frac{1-x}{1 + x}}$

2.

Sketch the graph?D
(a) $\displaystyle{r = a\cos{\theta}, \ a > 0}$ (circle)
(b) $\displaystyle{r = 3\theta}$ , Archmedes' spiral
(c) $\displaystyle{r^2 = 4\cos{2\theta}}$ Bernoulli's lemniscate

Next: Integration Up: Extreme Value of Functions Previous: Curve Sketching Contents Index

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