Precalculus Examples

$(- 3, - \frac{2 π}{3})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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 $- \frac{2 π}{3}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $\frac{2 π}{3}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

Apply the product rule to $2 π$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{2^{2} π^{2}}{3^{2}}$ by $1$ .

$r = \sqrt{9 + \frac{2^{2} π^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{9 + \frac{4 π^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

$r = \sqrt{9 + \frac{4 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Combine $9$ and $\frac{9}{9}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the expression.

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Step 3.9.1

Combine the numerators over the common denominator.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

Multiply $9$ by $9$ .

$r = \sqrt{\frac{81 + 4 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{81 + 4 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Rewrite $\sqrt{\frac{81 + 4 π^{2}}{9}}$ as $\frac{\sqrt{81 + 4 π^{2}}}{\sqrt{9}}$ .

$r = \frac{\sqrt{81 + 4 π^{2}}}{\sqrt{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Simplify the denominator.

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Step 3.11.1

Rewrite $9$ as $3^{2}$ .

$r = \frac{\sqrt{81 + 4 π^{2}}}{\sqrt{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.2

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

$r = \frac{\sqrt{81 + 4 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{81 + 4 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{81 + 4 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{81 + 4 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{- \frac{2 π}{3}}{- 3})$

Step 5

The inverse tangent of $\frac{2 π}{9}$ is $θ = 214.92011439 °$ .

$r = \frac{\sqrt{81 + 4 π^{2}}}{3}$

$θ = 214.92011439 °$

Step 6

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

$(\frac{\sqrt{81 + 4 π^{2}}}{3}, 214.92011439 °)$