What is the polar form of $(- 200, 10)$ $(-200,10)$ ?

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What is the polar form of $(- 200, 10)$ $(-200,10)$ ?

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Answer 1 · 2017-07-15T19:22:21

$(10 \sqrt{401}, 3.09)$ $(10sqrt401, 3.09)$

Explanation:

To convert this rectangular coordinate $(x, y)$ $(x,y)$ to a polar coordinate $(r, θ)$ $(r, theta)$ , use the following formulas:

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$tan θ = \frac{y}{x}$ $tan theta=y/x$

$r^{2} = {(- 200)}^{2} + {(10)}^{2}$ $r^2=(-200)^2+(10)^2$
$r^{2} = 40100$ $r^2=40100$
$r = \sqrt{40100}$ $r=sqrt40100$
$r = 10 \sqrt{401}$ $r=10sqrt401$

$tan θ = \frac{y}{x}$ $tan theta=y/x$

$tan θ = \frac{10}{- 200}$ $tan theta=10/-200$
$θ = {tan}^{- 1} (\frac{10}{- 200})$ $theta=tan^-1(10/-200)$
$θ \approx - 0.05$ $theta~~-0.05$

The angle $- 0.05$ $-0.05$ radians is in Quadrant $I V$ $IV$ , while the coordinate $(- 200, 10)$ $(-200, 10)$ is in Quadrant $I I$ $II$ . The angle is wrong because we used the $arctan$ $arctan$ function, which only has a range of $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2, pi/2]$ . To find the correct angle, add $π$ $pi$ to $θ$ $theta$ .

$- 0.05 + π = 3.09$ $-0.05 + pi = 3.09$

So, the polar coordinate is $(10 \sqrt{401}, 3.09)$ $(10sqrt401, 3.09)$ or $(200.25, 3.09)$ $(200.25, 3.09)$ .

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What is the polar form of $(- 200, 10)$ $(-200,10)$ ?

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Explanation:

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