Precalculus Examples

$(- 3 \sqrt{3}, - 3)$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Simplify the expression.

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Step 3.1.1

Apply the product rule to $- 3 \sqrt{3}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

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

$r = \sqrt{9 {\sqrt{3}}^{2} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 {\sqrt{3}}^{2} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$r = \sqrt{9 {(3^{\frac{1}{2}})}^{2} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$r = \sqrt{9 \cdot 3^{\frac{1}{2} \cdot 2} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

Combine $\frac{1}{2}$ and $2$ .

$r = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4

Cancel the common factor of $2$ .

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Step 3.2.4.1

Cancel the common factor.

$r = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4.2

Rewrite the expression.

$r = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.5

Evaluate the exponent.

$r = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 3 + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $9$ by $3$ .

$r = \sqrt{27 + {(- 3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.2

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

$r = \sqrt{27 + 9}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.3

Add $27$ and $9$ .

$r = \sqrt{36}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.4

Rewrite $36$ as $6^{2}$ .

$r = \sqrt{6^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.5

Pull terms out from under the radical, assuming positive real numbers.

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 6$

$θ = t a n^{- 1} (\frac{- 3}{- 3 \sqrt{3}})$

Step 5

The inverse tangent of $\frac{\sqrt{3}}{3}$ is $θ = 210 °$ .

$r = 6$

$θ = 210 °$

Step 6

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

$(6, 210 °)$