Precalculus Examples

$(- 4, 6)$

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.

$r = \sqrt{{(- 4)}^{2} + {(6)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

$r = \sqrt{16 + {(6)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $6$ to the power of $2$ .

$r = \sqrt{16 + 36}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $16$ and $36$ .

$r = \sqrt{52}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite $52$ as $2^{2} \cdot 13$ .

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Step 3.4.1

Factor $4$ out of $52$ .

$r = \sqrt{4 (13)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

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

$r = \sqrt{2^{2} \cdot 13}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{2^{2} \cdot 13}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

$r = 2 \sqrt{13}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 2 \sqrt{13}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 2 \sqrt{13}$

$θ = t a n^{- 1} (\frac{6}{- 4})$

Step 5

The inverse tangent of $- \frac{3}{2}$ is $θ = 123.69006752 °$ .

$r = 2 \sqrt{13}$

$θ = 123.69006752 °$

Step 6

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

$(2 \sqrt{13}, 123.69006752 °)$