Convert to Polar Coordinates (5,(5pi)/6)

$(5, \frac{5 π}{6})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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

$r = \sqrt{25 + {(\frac{5 π}{6})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

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

$r = \sqrt{25 + \frac{{(5 π)}^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the product rule to $5 π$ .

$r = \sqrt{25 + \frac{5^{2} π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 + \frac{5^{2} π^{2}}{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \sqrt{25 + \frac{25 π^{2}}{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

To write $25$ as a fraction with a common denominator, multiply by $\frac{36}{36}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Combine $25$ and $\frac{36}{36}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Combine the numerators over the common denominator.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Factor $25$ out of $25 \cdot 36 + 25 π^{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the numerator.

Rewrite $25 (36 + π^{2})$ as $5^{2} (6^{2} + π^{2})$ .

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

$r = \frac{\sqrt{5^{2} (36 + π^{2})}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \frac{\sqrt{5^{2} (6^{2} + π^{2})}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{5^{2} (6^{2} + π^{2})}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Pull terms out from under the radical.

$r = \frac{5 \sqrt{6^{2} + π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \frac{5 \sqrt{36 + π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{5 \sqrt{36 + π^{2}}}{\sqrt{36}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the denominator.

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

$r = \frac{5 \sqrt{36 + π^{2}}}{\sqrt{6^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \frac{5 \sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{5 \sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{5 \sqrt{36 + π^{2}}}{6}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{5 \sqrt{36 + π^{2}}}{6}$

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

Step 5

The inverse tangent of $\frac{π}{6}$ is $θ = 27.63649933 °$ .

$r = \frac{5 \sqrt{36 + π^{2}}}{6}$

$θ = 27.63649933 °$

Step 6

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

$(\frac{5 \sqrt{36 + π^{2}}}{6}, 27.63649933 °)$

Convert to Polar Coordinates (5,(5pi)/6)

Related Questions

Support LCMGCF.com by disabling your adblocker.