Precalculus Examples

$(9, \frac{π}{5})$

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.

$r = \sqrt{{(9)}^{2} + {(\frac{π}{5})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $9$ to the power of $2$ .

$r = \sqrt{81 + {(\frac{π}{5})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{81 + \frac{π^{2}}{5^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{81 + \frac{π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

To write $81$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$r = \sqrt{81 \cdot \frac{25}{25} + \frac{π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Combine $81$ and $\frac{25}{25}$ .

$r = \sqrt{\frac{81 \cdot 25}{25} + \frac{π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Simplify the expression.

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Step 3.6.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{81 \cdot 25 + π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.2

Multiply $81$ by $25$ .

$r = \sqrt{\frac{2025 + π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{2025 + π^{2}}{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Rewrite $\sqrt{\frac{2025 + π^{2}}{25}}$ as $\frac{\sqrt{2025 + π^{2}}}{\sqrt{25}}$ .

$r = \frac{\sqrt{2025 + π^{2}}}{\sqrt{25}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the denominator.

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Step 3.8.1

Rewrite $25$ as $5^{2}$ .

$r = \frac{\sqrt{2025 + π^{2}}}{\sqrt{5^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

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

$r = \frac{\sqrt{2025 + π^{2}}}{5}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{2025 + π^{2}}}{5}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{2025 + π^{2}}}{5}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{2025 + π^{2}}}{5}$

$θ = t a n^{- 1} (\frac{\frac{π}{5}}{9})$

Step 5

The inverse tangent of $\frac{π}{45}$ is $θ = 3.99352043 °$ .

$r = \frac{\sqrt{2025 + π^{2}}}{5}$

$θ = 3.99352043 °$

Step 6

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

$(\frac{\sqrt{2025 + π^{2}}}{5}, 3.99352043 °)$