Precalculus Examples

$(4, 80)$

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.

$r = \sqrt{{(4)}^{2} + {(80)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Find the magnitude of the polar coordinate.

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Remove parentheses around $4$ .

$r = \sqrt{4^{2} + {(80)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $4$ to the power of $2$ .

$r = \sqrt{16 + {(80)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Remove parentheses around $80$ .

$r = \sqrt{16 + 80^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $80$ to the power of $2$ .

$r = \sqrt{16 + 6400}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $16$ and $6400$ .

$r = \sqrt{6416}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $6416$ as $4^{2} \cdot 401$ .

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Factor $16$ out of $6416$ .

$r = \sqrt{16 (401)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $16$ as $4^{2}$ .

$r = \sqrt{4^{2} \cdot 401}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{4^{2} \cdot 401}$

$θ = t a n^{- 1} (\frac{y}{x})$

Pull terms out from under the radical.

$r = 4 \sqrt{401}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 4 \sqrt{401}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = 4 \sqrt{401}$

$θ = t a n^{- 1} (\frac{80}{4})$

The inverse tangent of $20$ is $θ = 87.13759477 °$ .

$r = 4 \sqrt{401}$

$θ = 87.13759477 °$

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

$(4 \sqrt{401}, 87.13759477 °)$

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