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Calculus Examples

(1, 3)

$(1, 3)$

Step 1

Convert from rectangular coordinates

(x, y)

$(x, y)$ to polar coordinates

(r, θ)

$(r, θ)$ using the conversion formulas.

r = \sqrt{x^{2} + y^{2}}

$r = \sqrt{x^{2} + y^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

Replace

x

$x$ and

y

$y$ with the actual values.

r = \sqrt{{(1)}^{2} + {(3)}^{2}}

$r = \sqrt{{(1)}^{2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

One to any power is one.

r = \sqrt{1 + {(3)}^{2}}

$r = \sqrt{1 + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise

3

$3$ to the power of

2

$2$ .

r = \sqrt{1 + 9}

$r = \sqrt{1 + 9}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add

1

$1$ and

9

$9$ .

r = \sqrt{10}

$r = \sqrt{10}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{10}

$r = \sqrt{10}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace

x

$x$ and

y

$y$ with the actual values.

r = \sqrt{10}

$r = \sqrt{10}$

θ = t a n^{- 1} (\frac{3}{1})

$θ = t a n^{- 1} (\frac{3}{1})$

Step 5

The inverse tangent of

3

$3$ is

θ = 71.56505117 °

$θ = 71.56505117 °$ .

r = \sqrt{10}

$r = \sqrt{10}$

θ = 71.56505117 °

$θ = 71.56505117 °$

Step 6

This is the result of the conversion to polar coordinates in

(r, θ)

$(r, θ)$ form.

(\sqrt{10}, 71.56505117 °)

$(\sqrt{10}, 71.56505117 °)$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$