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Calculus Examples

(4, 11)

$(4, 11)$

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{{(4)}^{2} + {(11)}^{2}}

$r = \sqrt{{(4)}^{2} + {(11)}^{2}}$

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

Raise

4

$4$ to the power of

2

$2$ .

r = \sqrt{16 + {(11)}^{2}}

$r = \sqrt{16 + {(11)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise

11

$11$ to the power of

2

$2$ .

r = \sqrt{16 + 121}

$r = \sqrt{16 + 121}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add

16

$16$ and

121

$121$ .

r = \sqrt{137}

$r = \sqrt{137}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{137}

$r = \sqrt{137}$

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{137}

$r = \sqrt{137}$

θ = t a n^{- 1} (\frac{11}{4})

$θ = t a n^{- 1} (\frac{11}{4})$

Step 5

The inverse tangent of

\frac{11}{4}

$\frac{11}{4}$ is

θ = 70.01689347 °

$θ = 70.01689347 °$ .

r = \sqrt{137}

$r = \sqrt{137}$

θ = 70.01689347 °

$θ = 70.01689347 °$

Step 6

This is the result of the conversion to polar coordinates in

(r, θ)

$(r, θ)$ form.

(\sqrt{137}, 70.01689347 °)

$(\sqrt{137}, 70.01689347 °)$

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

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