# Question #54b36

Aug 4, 2015

$r = 6$

#### Explanation:

To convert rectangular to polar in general:

Plot the point $\left(x , y\right)$ on the coordinate system.

Draw the line segment fro the origin to $\left(x , y\right)$.

Call the length of that line segment $r$. (Note that $r$ and is also the distance between the point $\left(x , y\right)$ and the origin.)
Use the Pythagorean Theorem (or its offspring, the distance formula) to see that:

$r = \sqrt{{x}^{2} + {y}^{2}}$

Find the smallest positive angle from the positive $x$ axis to the line segment you drew. Call that angle $\theta$

Now from one of the definitions of the trigonometric functions:

If the point $\left(x , y\right)$ lies on the terminal side of $\theta$ and $r = \sqrt{{x}^{2} + {y}^{2}}$, then

$\sin \theta = \frac{y}{r} \text{ }$ this gives us the conversion: $y = r \sin \theta$

$\cos \theta = \frac{x}{r} \text{ }$ this gives us the conversion: $x = r \cos \theta$

$\tan \theta = \frac{y}{x} \text{ }$ this gives us the conversion to find $\theta$

In this question , we have:

${x}^{2} + {y}^{2} = 36$

The long way is to replace $x$ and $y$ by $r \cos \theta$ and $r \sin \theta$ and simplify:

${x}^{2} + {y}^{2} = 36$

${\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = 36$

${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = 36$

${r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = 36$

${r}^{2} \left(1\right) = 36$

${r}^{2} = 36$

With the usual definition $r$ is non-negative, so we get:

$r = 6$

The short way "Hey, look! The left is just the same as ${r}^{2}$"

We immediately get: ${r}^{2} = 36$ so $r = 6$.

The geometric way

The equation ${x}^{2} + {y}^{2} = 36$ has a graph that is the circle, centered at the origin with radius $6$.

So in terms of the polar coordinates, $\theta$ can be anything it wants to be and $r$ must always be $6$.

The equation is $r = 6$