Calculus Examples

$(2, 3)$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Tap for more steps...

Step 3.1

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

$r = \sqrt{4 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{4 + 9}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $4$ and $9$ .

$r = \sqrt{13}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{13}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \sqrt{13}$

$θ = t a n^{- 1} (\frac{3}{2})$

Step 5

The inverse tangent of $\frac{3}{2}$ is $θ = 56.30993247 °$ .

$r = \sqrt{13}$

$θ = 56.30993247 °$

Step 6

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

$(\sqrt{13}, 56.30993247 °)$