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?
+1904 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.
\((r,\Theta)\)
\((-2\sqrt{2},135°)\)
+1904 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.
\((r,\Theta)\)
\((-2\sqrt{2},135°)\)
+33661 The following image should help:
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gibsonj338's other result requires a negative radial distance. Difficult to visualise what this means!!
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+1904 The negetave radical distance means to go the distance in the oppisite direction. It is very easy to visualize what this means.
