Precalculus Examples

$(\sqrt{3}, \frac{π}{6})$

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.

$r = \sqrt{{(\sqrt{3})}^{2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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Step 3.1.1

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

$r = \sqrt{{(3^{\frac{1}{2}})}^{2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

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

$r = \sqrt{3^{\frac{1}{2} \cdot 2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.3

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

$r = \sqrt{3^{\frac{2}{2}} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.4

Cancel the common factor of $2$ .

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Step 3.1.4.1

Cancel the common factor.

$r = \sqrt{3^{\frac{2}{2}} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.4.2

Rewrite the expression.

$r = \sqrt{3 + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3 + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.5

Evaluate the exponent.

$r = \sqrt{3 + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3 + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Simplify the expression.

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Step 3.2.1

Apply the product rule to $\frac{π}{6}$ .

$r = \sqrt{3 + \frac{π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

$r = \sqrt{3 + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3 + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

To write $3$ as a fraction with a common denominator, multiply by $\frac{36}{36}$ .

$r = \sqrt{3 \cdot \frac{36}{36} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Combine $3$ and $\frac{36}{36}$ .

$r = \sqrt{\frac{3 \cdot 36}{36} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Simplify the expression.

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Step 3.5.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{3 \cdot 36 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5.2

Multiply $3$ by $36$ .

$r = \sqrt{\frac{108 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{108 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Rewrite $\sqrt{\frac{108 + π^{2}}{36}}$ as $\frac{\sqrt{108 + π^{2}}}{\sqrt{36}}$ .

$r = \frac{\sqrt{108 + π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Simplify the denominator.

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Step 3.7.1

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

$r = \frac{\sqrt{108 + π^{2}}}{\sqrt{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.2

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

$r = \frac{\sqrt{108 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{108 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{108 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{108 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{\frac{π}{6}}{\sqrt{3}})$

Step 5

The inverse tangent of $\frac{π \sqrt{3}}{18}$ is $θ = 16.82006135 °$ .

$r = \frac{\sqrt{108 + π^{2}}}{6}$

$θ = 16.82006135 °$

Step 6

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

$(\frac{\sqrt{108 + π^{2}}}{6}, 16.82006135 °)$