Precalculus Examples

$(\frac{- 5}{2}, \frac{π}{6})$

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.

$r = \sqrt{{(\frac{- 5}{2})}^{2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Move the negative in front of the fraction.

$r = \sqrt{{(- \frac{5}{2})}^{2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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 $- \frac{5}{2}$ .

$r = \sqrt{{(- 1)}^{2} {(\frac{5}{2})}^{2} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $\frac{5}{2}$ .

$r = \sqrt{{(- 1)}^{2} (\frac{5^{2}}{2^{2}}) + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{{(- 1)}^{2} (\frac{5^{2}}{2^{2}}) + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{1 (\frac{5^{2}}{2^{2}}) + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$r = \sqrt{\frac{5^{2}}{2^{2}} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Raise $5$ to the power of $2$ .

$r = \sqrt{\frac{25}{2^{2}} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

$r = \sqrt{\frac{25}{4} + {(\frac{π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

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

$r = \sqrt{\frac{25}{4} + \frac{π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

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

$r = \sqrt{\frac{25}{4} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

To write $\frac{25}{4}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$r = \sqrt{\frac{25}{4} \cdot \frac{9}{9} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.10.1

Multiply $\frac{25}{4}$ by $\frac{9}{9}$ .

$r = \sqrt{\frac{25 \cdot 9}{4 \cdot 9} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.2

Multiply $4$ by $9$ .

$r = \sqrt{\frac{25 \cdot 9}{36} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{25 \cdot 9}{36} + \frac{π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Combine the numerators over the common denominator.

$r = \sqrt{\frac{25 \cdot 9 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12

Multiply $25$ by $9$ .

$r = \sqrt{\frac{225 + π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.13

Rewrite $\sqrt{\frac{225 + π^{2}}{36}}$ as $\frac{\sqrt{225 + π^{2}}}{\sqrt{36}}$ .

$r = \frac{\sqrt{225 + π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.14

Simplify the denominator.

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Step 3.14.1

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

$r = \frac{\sqrt{225 + π^{2}}}{\sqrt{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.14.2

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

$r = \frac{\sqrt{225 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{225 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{225 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{225 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{\frac{π}{6}}{\frac{- 5}{2}})$

Step 5

The inverse tangent of $- \frac{π}{15}$ is $θ = 168.17098164 °$ .

$r = \frac{\sqrt{225 + π^{2}}}{6}$

$θ = 168.17098164 °$

Step 6

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

$(\frac{\sqrt{225 + π^{2}}}{6}, 168.17098164 °)$