Precalculus Examples

$(5, π)$

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.

$r = \sqrt{{(5)}^{2} + {(π)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

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

$r = \sqrt{25 + π^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \sqrt{25 + π^{2}}$

$θ = t a n^{- 1} (\frac{π}{5})$

Step 5

The inverse tangent of $\frac{π}{5}$ is $θ = 32.14190763 °$ .

$r = \sqrt{25 + π^{2}}$

$θ = 32.14190763 °$

Step 6

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

$(\sqrt{25 + π^{2}}, 32.14190763 °)$