Precalculus Examples

$(- 5, - 1)$

Step 1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

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

$r = \sqrt{{(- 5)}^{2} + {(- 1)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $- 5$ to the power of $2$ .

$r = \sqrt{25 + {(- 1)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $- 1$ to the power of $2$ .

$r = \sqrt{25 + 1}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $25$ and $1$ .

$r = \sqrt{26}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{26}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \sqrt{26}$

$θ = t a n^{- 1} (\frac{- 1}{- 5})$

Step 5

The inverse tangent of $\frac{1}{5}$ is $θ = 191.30993247 °$ .

$r = \sqrt{26}$

$θ = 191.30993247 °$

Step 6

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

$(\sqrt{26}, 191.30993247 °)$