Precalculus Examples

$(0, \frac{1}{7})$

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.

$r = \sqrt{{(0)}^{2} + {(\frac{1}{7})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raising $0$ to any positive power yields $0$ .

$r = \sqrt{0 + {(\frac{1}{7})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Apply the product rule to $\frac{1}{7}$ .

$r = \sqrt{0 + \frac{1^{2}}{7^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

One to any power is one.

$r = \sqrt{0 + \frac{1}{7^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Raise $7$ to the power of $2$ .

$r = \sqrt{0 + \frac{1}{49}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Add $0$ and $\frac{1}{49}$ .

$r = \sqrt{\frac{1}{49}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Rewrite $\sqrt{\frac{1}{49}}$ as $\frac{\sqrt{1}}{\sqrt{49}}$ .

$r = \frac{\sqrt{1}}{\sqrt{49}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Any root of $1$ is $1$ .

$r = \frac{1}{\sqrt{49}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the denominator.

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Step 3.8.1

Rewrite $49$ as $7^{2}$ .

$r = \frac{1}{\sqrt{7^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

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

$r = \frac{1}{7}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{1}{7}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{1}{7}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{1}{7}$

$θ = t a n^{- 1} (\frac{\frac{1}{7}}{0})$

Step 5

The inverse tangent of Undefined is $θ = 90 °$ .

$r = \frac{1}{7}$

$θ = 90 °$

Step 6

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

$(\frac{1}{7}, 90 °)$