Precalculus Examples

$(1, \frac{π}{4})$

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.

$r = \sqrt{{(1)}^{2} + {(\frac{π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

One to any power is one.

$r = \sqrt{1 + {(\frac{π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Apply the product rule to $\frac{π}{4}$ .

$r = \sqrt{1 + \frac{π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $4$ to the power of $2$ .

$r = \sqrt{1 + \frac{π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Write $1$ as a fraction with a common denominator.

$r = \sqrt{\frac{16}{16} + \frac{π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Combine the numerators over the common denominator.

$r = \sqrt{\frac{16 + π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Rewrite $\sqrt{\frac{16 + π^{2}}{16}}$ as $\frac{\sqrt{16 + π^{2}}}{\sqrt{16}}$ .

$r = \frac{\sqrt{16 + π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Simplify the denominator.

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Step 3.7.1

Rewrite $16$ as $4^{2}$ .

$r = \frac{\sqrt{16 + π^{2}}}{\sqrt{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.2

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

$r = \frac{\sqrt{16 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{16 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{16 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{16 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{\frac{π}{4}}{1})$

Step 5

The inverse tangent of $\frac{π}{4}$ is $θ = 38.14602598 °$ .

$r = \frac{\sqrt{16 + π^{2}}}{4}$

$θ = 38.14602598 °$

Step 6

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

$(\frac{\sqrt{16 + π^{2}}}{4}, 38.14602598 °)$