Precalculus Examples

$p = (1, - \frac{π}{6})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

One to any power is one.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.2.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $- 1$ to the power of $2$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{π^{2}}{6^{2}}$ by $1$ .

$r = \sqrt{1 + \frac{π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{1 + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Write $1$ as a fraction with a common denominator.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Combine the numerators over the common denominator.

$r = \sqrt{\frac{36 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the denominator.

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Step 3.9.1

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

$r = \frac{\sqrt{36 + π^{2}}}{\sqrt{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

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

$r = \frac{\sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{36 + π^{2}}}{6}$

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

Step 5

The inverse tangent of $- \frac{π}{6}$ is $θ = 332.36350066 °$ .

$r = \frac{\sqrt{36 + π^{2}}}{6}$

$θ = 332.36350066 °$

Step 6

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

$(\frac{\sqrt{36 + π^{2}}}{6}, 332.36350066 °)$