Precalculus Examples

$(\frac{- 14}{2}, \frac{- 13}{2})$

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.

$r = \sqrt{{(\frac{- 14}{2})}^{2} + {(\frac{- 13}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Divide $- 14$ by $2$ .

$r = \sqrt{{(- 7)}^{2} + {(\frac{- 13}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $- 7$ to the power of $2$ .

$r = \sqrt{49 + {(\frac{- 13}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Move the negative in front of the fraction.

$r = \sqrt{49 + {(- \frac{13}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

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Step 3.4.1

Apply the product rule to $- \frac{13}{2}$ .

$r = \sqrt{49 + {(- 1)}^{2} {(\frac{13}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Apply the product rule to $\frac{13}{2}$ .

$r = \sqrt{49 + {(- 1)}^{2} (\frac{13^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{49 + {(- 1)}^{2} (\frac{13^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{49 + 1 (\frac{13^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Multiply $\frac{13^{2}}{2^{2}}$ by $1$ .

$r = \sqrt{49 + \frac{13^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Raise $13$ to the power of $2$ .

$r = \sqrt{49 + \frac{169}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

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

$r = \sqrt{49 + \frac{169}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

To write $49$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$r = \sqrt{49 \cdot \frac{4}{4} + \frac{169}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Combine $49$ and $\frac{4}{4}$ .

$r = \sqrt{\frac{49 \cdot 4}{4} + \frac{169}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Combine the numerators over the common denominator.

$r = \sqrt{\frac{49 \cdot 4 + 169}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12

Simplify the numerator.

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Step 3.12.1

Multiply $49$ by $4$ .

$r = \sqrt{\frac{196 + 169}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12.2

Add $196$ and $169$ .

$r = \sqrt{\frac{365}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{365}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.13

Rewrite $\sqrt{\frac{365}{4}}$ as $\frac{\sqrt{365}}{\sqrt{4}}$ .

$r = \frac{\sqrt{365}}{\sqrt{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.14

Simplify the denominator.

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Step 3.14.1

Rewrite $4$ as $2^{2}$ .

$r = \frac{\sqrt{365}}{\sqrt{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.14.2

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

$r = \frac{\sqrt{365}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{365}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{365}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{365}}{2}$

$θ = t a n^{- 1} (\frac{\frac{- 13}{2}}{\frac{- 14}{2}})$

Step 5

The inverse tangent of $\frac{13}{14}$ is $θ = 42.8789036 °$ .

$r = \frac{\sqrt{365}}{2}$

$θ = 42.8789036 °$

Step 6

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

$(\frac{\sqrt{365}}{2}, 42.8789036 °)$