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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

#1
+1802
+2

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:

#1
+1802
+2

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
+25975
+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
+1802
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
+25975
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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