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Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°. I don't understand how to do this, I'm lost, can someone please explain this?

Guest Aug 12, 2017

Best Answer 

 #1
avatar+1828 
+1

Polar coordiantes can be represented as (r, θ) where r equals the radius and θ equals the angle in degrees or in radians.  To convert the cartesian coordinate (2,-2) to polar coordinate, first figure out what r is. To find out what r is, use the formula known as pythagoras theorem: \({r}^{2}={x}^{2}+{y}^{2}\) where x is the x-coordinate and y is the y-coordinate.

 

\({r}^{2}={x}^{2}+{y}^{2}\)

 

\({r}^{2}={2}^{2}+{(-2)}^{2}\)

 

\({r}^{2}=4+{(-2)}^{2}\)

 

\({r}^{2}=4+4\)

 

\({r}^{2}=8\)

 

\(\sqrt{{r}^{2}}=\sqrt{8}\)

 

\(r=\sqrt{8}\)

 

\(r=2\sqrt{2}\)

 

Now figure out what θ is.  To figure out what θ is, use the formula known as tangent function:

\(tan(\Theta)=\frac{x}{y}\).

 

\(tan(\Theta)=\frac{x}{y}\)

 

\(tan(\Theta)=\frac{2}{-2}\)

 

\(tan(\Theta)=-\frac{2}{2}\)

 

\(tan(\Theta)=-1\)

 

\({tan}^{-1}(tan(\Theta))={tan}^{-1}(-1)\)

 

\(\Theta ={tan}^{-1}(-1)\)

 

\(\Theta =-45°\)or \(\Theta =-\frac{\pi}{4}\)

 

Because the question asks to be within the 0° ≤ θ < 360° parameter, ignore the radian answer above.  Since the degree answer is not within the 0° ≤ θ < 360° parameter, you need to change the answer to an equilivent answer that fits the 0° ≤ θ < 360°.  To do that add 360° to the degree answer.

 

\(\Theta=-45°+360°\)

 

\(\Theta=315°\)

 

Now put r and θ in polar cordinate form.

 

\((r,\Theta)\)

 

\((2\sqrt{2}, 315°)\)

 

To find another coordinate in polar form that is the same as the polar coordinate above that fits the 0° ≤ θ < 360°, first subtract 180° from 315°.

 

\(\Theta=315°-180°\)

 

\(\Theta=135°\)

 

Second, change \(2\sqrt{2}\) to \(-2\sqrt{2}\).

 

Now put r and θ in polar cordinate form.

 

\((r,\Theta)\)

 

\((-2\sqrt{2},135°)\)

gibsonj338  Aug 12, 2017
edited by gibsonj338  Aug 12, 2017
Sort: 

4+0 Answers

 #1
avatar+1828 
+1
Best Answer

Polar coordiantes can be represented as (r, θ) where r equals the radius and θ equals the angle in degrees or in radians.  To convert the cartesian coordinate (2,-2) to polar coordinate, first figure out what r is. To find out what r is, use the formula known as pythagoras theorem: \({r}^{2}={x}^{2}+{y}^{2}\) where x is the x-coordinate and y is the y-coordinate.

 

\({r}^{2}={x}^{2}+{y}^{2}\)

 

\({r}^{2}={2}^{2}+{(-2)}^{2}\)

 

\({r}^{2}=4+{(-2)}^{2}\)

 

\({r}^{2}=4+4\)

 

\({r}^{2}=8\)

 

\(\sqrt{{r}^{2}}=\sqrt{8}\)

 

\(r=\sqrt{8}\)

 

\(r=2\sqrt{2}\)

 

Now figure out what θ is.  To figure out what θ is, use the formula known as tangent function:

\(tan(\Theta)=\frac{x}{y}\).

 

\(tan(\Theta)=\frac{x}{y}\)

 

\(tan(\Theta)=\frac{2}{-2}\)

 

\(tan(\Theta)=-\frac{2}{2}\)

 

\(tan(\Theta)=-1\)

 

\({tan}^{-1}(tan(\Theta))={tan}^{-1}(-1)\)

 

\(\Theta ={tan}^{-1}(-1)\)

 

\(\Theta =-45°\)or \(\Theta =-\frac{\pi}{4}\)

 

Because the question asks to be within the 0° ≤ θ < 360° parameter, ignore the radian answer above.  Since the degree answer is not within the 0° ≤ θ < 360° parameter, you need to change the answer to an equilivent answer that fits the 0° ≤ θ < 360°.  To do that add 360° to the degree answer.

 

\(\Theta=-45°+360°\)

 

\(\Theta=315°\)

 

Now put r and θ in polar cordinate form.

 

\((r,\Theta)\)

 

\((2\sqrt{2}, 315°)\)

 

To find another coordinate in polar form that is the same as the polar coordinate above that fits the 0° ≤ θ < 360°, first subtract 180° from 315°.

 

\(\Theta=315°-180°\)

 

\(\Theta=135°\)

 

Second, change \(2\sqrt{2}\) to \(-2\sqrt{2}\).

 

Now put r and θ in polar cordinate form.

 

\((r,\Theta)\)

 

\((-2\sqrt{2},135°)\)

gibsonj338  Aug 12, 2017
edited by gibsonj338  Aug 12, 2017
 #2
avatar+26102 
+3

The following image should help:

 

 

.

gibsonj338's other result requires a negative radial distance.  Difficult to visualise what this means!!

.

.

Alan  Aug 12, 2017
edited by Alan  Aug 12, 2017
edited by Alan  Aug 12, 2017
 #3
avatar+1828 
0

The negetave radical distance means to go the distance in the oppisite direction.  It is very easy to visualize what this means.

gibsonj338  Aug 12, 2017
 #4
avatar+26102 
0

Hmm!  

 

In polar notation the direction is given by \(\theta\) 

 

r gives the magnitude   It's the negative magnitude that I have difficulty visualizing!

.

Alan  Aug 12, 2017

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