How do you find the polar coordinates given the rectangular coordinates (3,8)?

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How do you find the polar coordinates given the rectangular coordinates (3,8)?

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Answer 1 · 2016-11-18T00:11:48

Please see the explanation.

Explanation:

The polar coordinate system is an ordered pair $(r, θ)$ $(r,theta)$

To convert from Cartesian, $(x, y)$ $(x, y)$ to r:

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

To convert from Cartesian, $(x, y)$ $(x, y)$ to $θ$ $theta$ :

If $x > 0 and y \geq 0$ $x > 0 and y >= 0$ , then use: $θ = {tan}^{- 1} (\frac{y}{x}) [1]$ $theta = tan^-1(y/x)" [1]"$
If $x = 0 and y > 0$ $x = 0 and y > 0$ , then use: $θ = \frac{π}{2} [2]$ $theta = pi/2" [2]"$
If $x = 0 and y < 0$ $x = 0 and y < 0$ , then use: $θ = \frac{3 π}{3} [3]$ $theta = (3pi)/3" [3]"$
If $x < 0$ $x < 0$ , then use: $θ = π + {tan}^{- 1} (\frac{y}{x}) [4]$ $theta = pi + tan^-1(y/x)" [4]"$
If $x > 0 and y < 0$ $x > 0 and y < 0$ , then use: $θ = 2 π + {tan}^{- 1} (\frac{y}{x}) [5]$ $theta = 2pi + tan^-1(y/x)" [5]"$

For your point, $(3, 8)$ $(3, 8)$ :

$r = \sqrt{3^{2} + 8^{2}} = \sqrt{73} length units$ $r = sqrt(3^2 + 8^2) = sqrt(73) " length units"$

Use equation [1]:

$θ = {tan}^{- 1} (\frac{8}{3})$ $theta = tan^-1(8/3)$

$θ \approx 1.249 radians$ $theta ~~ 1.249 " radians"$

The polar point is $(\sqrt{73} length units, 1.249 radians)$ $(sqrt(73)" length units", 1.249" radians")$

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How do you find the polar coordinates given the rectangular coordinates (3,8)?

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Explanation:

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