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Section 9.6 : Polar Coordinates

3. The Cartesian coordinate of a point are \(\left( {2, - 6} \right)\). Determine a set of polar coordinates for the point.

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Let’s first determine \(r\). That’s always simple.

\[r = \sqrt {{x^2} + {y^2}} = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {40} = 2\sqrt {10} \] Show Step 2

Next let’s get \(\theta \). As we do this we need to remember that we actually have two possible values of which only one will work with the \(r\) we found in the first step.

Here are the two possible values of \(\theta \).

\[{\theta _{\,1}} = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = {\tan ^{ - 1}}\left( {\frac{{ - 6}}{2}} \right) = - 1.2490\hspace{0.25in}\hspace{0.25in}{\theta _{\,2}} = {\theta _{\,1}} + \pi = 1.8926\]

So, we can see that \( - \frac{\pi }{2} = - 1.57 < {\theta _{\,1}} = - 1.2490 < 0\) and so \({\theta _1}\) is in the fourth quadrant. Likewise, \(\frac{\pi }{2} = 1.57 < {\theta _2} = 1.8926 < \pi = 3.14\) and so \({\theta _2}\) is in the second quadrant.

We can also see from the Cartesian coordinates of the point that our point must be in the fourth quadrant and so, for this problem, \({\theta _1}\) is the correct value.

The polar coordinates of the point using the \(r\) from the first step and \(\theta \) from this step is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {2\sqrt {10} , - 1.2490} \right)}}\]

Note of course that there are many other sets of polar coordinates that are just as valid for this point. These are simply the set that we get from the formulas discussed in this section.