Precalculus Examples

$(2, - 6)$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{4 + 36}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $4$ and $36$ .

$r = \sqrt{40}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 3.4.1

Factor $4$ out of $40$ .

$r = \sqrt{4 (10)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

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

$r = \sqrt{2^{2} \cdot 10}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{2^{2} \cdot 10}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

$r = 2 \sqrt{10}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 2 \sqrt{10}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 2 \sqrt{10}$

$θ = t a n^{- 1} (\frac{- 6}{2})$

Step 5

The inverse tangent of $- 3$ is $θ = 288.43494882 °$ .

$r = 2 \sqrt{10}$

$θ = 288.43494882 °$

Step 6

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

$(2 \sqrt{10}, 288.43494882 °)$

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