Calculus Examples

$(3, - 5)$

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.

$r = \sqrt{{(3)}^{2} + {(- 5)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{9 + 25}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $9$ and $25$ .

$r = \sqrt{34}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{34}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \sqrt{34}$

$θ = t a n^{- 1} (\frac{- 5}{3})$

Step 5

The inverse tangent of $- \frac{5}{3}$ is $θ = 300.96375653 °$ .

$r = \sqrt{34}$

$θ = 300.96375653 °$

Step 6

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

$(\sqrt{34}, 300.96375653 °)$