Precalculus Examples

(5, \frac{π}{2})

$(5, \frac{π}{2})$

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{{(5)}^{2} + {(\frac{π}{2})}^{2}}

$r = \sqrt{{(5)}^{2} + {(\frac{π}{2})}^{2}}$

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

5

$5$ to the power of

2

$2$ .

r = \sqrt{25 + {(\frac{π}{2})}^{2}}

$r = \sqrt{25 + {(\frac{π}{2})}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Apply the product rule to

\frac{π}{2}

$\frac{π}{2}$ .

r = \sqrt{25 + \frac{π^{2}}{2^{2}}}

$r = \sqrt{25 + \frac{π^{2}}{2^{2}}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise

2

$2$ to the power of

2

$2$ .

r = \sqrt{25 + \frac{π^{2}}{4}}

$r = \sqrt{25 + \frac{π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

To write

25

$25$ as a fraction with a common denominator, multiply by

\frac{4}{4}

$\frac{4}{4}$ .

r = \sqrt{25 \cdot \frac{4}{4} + \frac{π^{2}}{4}}

$r = \sqrt{25 \cdot \frac{4}{4} + \frac{π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Combine

25

$25$ and

\frac{4}{4}

$\frac{4}{4}$ .

r = \sqrt{\frac{25 \cdot 4}{4} + \frac{π^{2}}{4}}

$r = \sqrt{\frac{25 \cdot 4}{4} + \frac{π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Simplify the expression.

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Step 3.6.1

Combine the numerators over the common denominator.

r = \sqrt{\frac{25 \cdot 4 + π^{2}}{4}}

$r = \sqrt{\frac{25 \cdot 4 + π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.2

Multiply

25

$25$ by

4

$4$ .

r = \sqrt{\frac{100 + π^{2}}{4}}

$r = \sqrt{\frac{100 + π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{\frac{100 + π^{2}}{4}}

$r = \sqrt{\frac{100 + π^{2}}{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Rewrite

\sqrt{\frac{100 + π^{2}}{4}}

$\sqrt{\frac{100 + π^{2}}{4}}$ as

\frac{\sqrt{100 + π^{2}}}{\sqrt{4}}

$\frac{\sqrt{100 + π^{2}}}{\sqrt{4}}$ .

r = \frac{\sqrt{100 + π^{2}}}{\sqrt{4}}

$r = \frac{\sqrt{100 + π^{2}}}{\sqrt{4}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the denominator.

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Step 3.8.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

r = \frac{\sqrt{100 + π^{2}}}{\sqrt{2^{2}}}

$r = \frac{\sqrt{100 + π^{2}}}{\sqrt{2^{2}}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

Pull terms out from under the radical, assuming positive real numbers.

r = \frac{\sqrt{100 + π^{2}}}{2}

$r = \frac{\sqrt{100 + π^{2}}}{2}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \frac{\sqrt{100 + π^{2}}}{2}

$r = \frac{\sqrt{100 + π^{2}}}{2}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \frac{\sqrt{100 + π^{2}}}{2}

$r = \frac{\sqrt{100 + π^{2}}}{2}$

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 = \frac{\sqrt{100 + π^{2}}}{2}

$r = \frac{\sqrt{100 + π^{2}}}{2}$

θ = t a n^{- 1} (\frac{\frac{π}{2}}{5})

$θ = t a n^{- 1} (\frac{\frac{π}{2}}{5})$

Step 5

The inverse tangent of

\frac{π}{10}

$\frac{π}{10}$ is

θ = 17.44059449 °

$θ = 17.44059449 °$ .

r = \frac{\sqrt{100 + π^{2}}}{2}

$r = \frac{\sqrt{100 + π^{2}}}{2}$

$θ = 17.44059449 °$

Step 6

This is the result of the conversion to polar coordinates in

$(r, θ)$ form.

$(\frac{\sqrt{100 + π^{2}}}{2}, 17.44059449 °)$