Precalculus Examples

$(3, 120 °)$

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.

$r = \sqrt{{(3)}^{2} + {(120 °)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $3$ to the power of $2$ .

$r = \sqrt{9 + {(120 °)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $120 °$ to the power of $2$ .

$r = \sqrt{9 + 14400}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $9$ and $14400$ .

$r = \sqrt{14409}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite $14409$ as $3^{2} \cdot 1601$ .

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Step 3.4.1

Factor $9$ out of $14409$ .

$r = \sqrt{9 (1601)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Rewrite $9$ as $3^{2}$ .

$r = \sqrt{3^{2} \cdot 1601}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3^{2} \cdot 1601}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

$r = 3 \sqrt{1601}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 3 \sqrt{1601}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 3 \sqrt{1601}$

$θ = t a n^{- 1} (\frac{120 °}{3})$

Step 5

The inverse tangent of $40$ is $θ = 5074.56708896 °$ .

$r = 3 \sqrt{1601}$

$θ = 5074.56708896 °$

Step 6

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

$(3 \sqrt{1601}, 5074.56708896 °)$