Precalculus Examples

$(11, - 4)$

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.

$r = \sqrt{{(11)}^{2} + {(- 4)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $11$ to the power of $2$ .

$r = \sqrt{121 + {(- 4)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{121 + 16}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $121$ and $16$ .

$r = \sqrt{137}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{137}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \sqrt{137}$

$θ = t a n^{- 1} (\frac{- 4}{11})$

Step 5

The inverse tangent of $- \frac{4}{11}$ is $θ = 340.01689347 °$ .

$r = \sqrt{137}$

$θ = 340.01689347 °$

Step 6

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

$(\sqrt{137}, 340.01689347 °)$