Precalculus Examples

$(23, 65)$

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.

$r = \sqrt{{(23)}^{2} + {(65)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Find the magnitude of the polar coordinate.

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Remove parentheses around $23$ .

$r = \sqrt{23^{2} + {(65)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $23$ to the power of $2$ to get $529$ .

$r = \sqrt{529 + {(65)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Remove parentheses around $65$ .

$r = \sqrt{529 + 65^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $65$ to the power of $2$ to get $4225$ .

$r = \sqrt{529 + 4225}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $529$ and $4225$ to get $4754$ .

$r = \sqrt{4754}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{4754}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \sqrt{4754}$

$θ = t a n^{- 1} (\frac{65}{23})$

The inverse tangent of $\frac{65}{23}$ is $θ = 70.513870427534 °$ .

$r = \sqrt{4754}$

$θ = 70.513870427534 °$

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

$(\sqrt{4754}, 70.513870427534 °)$

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