Review of Polar Coordinates

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Objectives:

1.: Review polar coordinates.

Recall that the transformation to get from polar $(r,\theta )$ coordinates to Cartesian $(x,y)$ coordinates is $x =r\cos (\theta ),\quad y = r\sin (\theta ).$ The picture relating $(r,\theta )$ to $(x,y)$ is shown below:

It is useful to note that $r^2 = x^2+y^2$ .

The point $(r,\theta ) = (6,\pi /3)$ corresponds to the Cartesian point $(x,y) = (\answer {3},\answer {3\sqrt {3}}).$

Remember that polar coordinates was great for describing certain objects, such as circles.

Transform the equation $x^2+y^2 = 1$ into polar coordinates.

To transform the equation into polar, we plug in $x = r\cos (\theta )$ , $y = r\sin (\theta )$ into the equation to get $\answer {r^2} = 1,$ which means the equation is $r = \answer {1}$ .

One issue we will encounter with polar is the non-uniqueness of points. From now on, we will restrict ourselves to the situation where $r \geq 0$ (i.e. we will not look at negative $r$ -values). This will help us a little bit later.

What does $r=0$ look like in Cartesian coordinates?

The point $(x,y) = (0,0)$ The $x$ -axis The $y$ -axis The line $y=x$

Transform the line $y =x$ into polar coordinates.

To transform the equation, we plug in $x = r\cos (\theta )$ , $y = r\sin (\theta )$ to get $r\sin (\theta ) = r\cos (\theta ).$ Since $r=0$ is just a point (and will not describe the whole line), we can divide by $r$ to get $\sin (\theta ) = \cos (\theta ).$ On the interval $[0,2\pi ]$ , this happens when $\theta = \answer {\frac {\pi }{4}}, \frac {5\pi }{4}.$ Therefore, the line can be described as $\theta = \answer {\frac {\pi }{4}}$ and $\theta = \frac {5\pi }{4}$ . Note that the former gives the portion of the line with $x \geq 0$ and the latter gives the portion of the line with $x \leq 0$ .

The line $x=1$ in polar coordinates becomes $r = \answer {\frac {1}{\cos (\theta )}}.$

The circle $(x-1)^2+y^2 = 1$ in polar coordinates becomes $r = \answer {2\cos (\theta )}.$

Expand the circle to get $x^2-2x+1+y^2=1$ , or $x^2+y^2-2x=0$ . In polar, this is $r^2-2r\cos (\theta ) = 0$ . Now solve for $r$ .