Precalculus Examples

$(3, \frac{5 π}{6})$

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.

$r = \sqrt{{(3)}^{2} + {(\frac{5 π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Tap for more steps...

Step 3.1

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

$r = \sqrt{9 + {(\frac{5 π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

Tap for more steps...

Step 3.2.1

Apply the product rule to $\frac{5 π}{6}$ .

$r = \sqrt{9 + \frac{{(5 π)}^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $5 π$ .

$r = \sqrt{9 + \frac{5^{2} π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 + \frac{5^{2} π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{9 + \frac{25 π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

$r = \sqrt{9 + \frac{25 π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

To write $9$ as a fraction with a common denominator, multiply by $\frac{36}{36}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Combine $9$ and $\frac{36}{36}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Simplify the expression.

Tap for more steps...

Step 3.7.1

Combine the numerators over the common denominator.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.2

Multiply $9$ by $36$ .

$r = \sqrt{\frac{324 + 25 π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{324 + 25 π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Rewrite $\sqrt{\frac{324 + 25 π^{2}}{36}}$ as $\frac{\sqrt{324 + 25 π^{2}}}{\sqrt{36}}$ .

$r = \frac{\sqrt{324 + 25 π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the denominator.

Tap for more steps...

Step 3.9.1

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

$r = \frac{\sqrt{324 + 25 π^{2}}}{\sqrt{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

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

$r = \frac{\sqrt{324 + 25 π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{324 + 25 π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{324 + 25 π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{324 + 25 π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{\frac{5 π}{6}}{3})$

Step 5

The inverse tangent of $\frac{5 π}{18}$ is $θ = 41.11005286 °$ .

$r = \frac{\sqrt{324 + 25 π^{2}}}{6}$

$θ = 41.11005286 °$

Step 6

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

$(\frac{\sqrt{324 + 25 π^{2}}}{6}, 41.11005286 °)$