Precalculus Examples

$(13, 8)$

Convert from rectangular coordinates $(x, y)$ to polar coordinates $(r, θ)$ using the conversion formulas.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Replace $x$ and $y$ with the actual values.

$r = \sqrt{{(13)}^{2} + {(8)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Find the magnitude of the polar coordinate.

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Raise $13$ to the power of $2$ .

$r = \sqrt{169 + {(8)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $8$ to the power of $2$ .

$r = \sqrt{169 + 64}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $169$ and $64$ .

$r = \sqrt{233}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{233}$

$θ = t a n^{- 1} (\frac{y}{x})$

Replace $x$ and $y$ with the actual values.

$r = \sqrt{233}$

$θ = t a n^{- 1} (\frac{8}{13})$

The inverse tangent of $\frac{8}{13}$ is $θ = 31.60750224 °$ .

$r = \sqrt{233}$

$θ = 31.60750224 °$

This is the result of the conversion to polar coordinates in $(r, θ)$ form.

$(\sqrt{233}, 31.60750224 °)$

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