Precalculus Examples

$(0, 3 \sqrt{2})$

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.

$r = \sqrt{{(0)}^{2} + {(3 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Simplify the expression.

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

$r = \sqrt{0 + {(3 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the product rule to $3 \sqrt{2}$ .

$r = \sqrt{0 + 3^{2} {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $3$ to the power of $2$ .

$r = \sqrt{0 + 9 {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{0 + 9 {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$r = \sqrt{0 + 9 {(2^{\frac{1}{2}})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$r = \sqrt{0 + 9 \cdot 2^{\frac{1}{2} \cdot 2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Combine $\frac{1}{2}$ and $2$ .

$r = \sqrt{0 + 9 \cdot 2^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Cancel the common factor of $2$ .

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Cancel the common factor.

$r = \sqrt{0 + 9 \cdot 2^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite the expression.

$r = \sqrt{0 + 9 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{0 + 9 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Evaluate the exponent.

$r = \sqrt{0 + 9 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{0 + 9 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the expression.

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Multiply $9$ by $2$ .

$r = \sqrt{0 + 18}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $0$ and $18$ .

$r = \sqrt{18}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{18}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $18$ as $3^{2} \cdot 2$ .

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Factor $9$ out of $18$ .

$r = \sqrt{9 (2)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $9$ as $3^{2}$ .

$r = \sqrt{3^{2} \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3^{2} \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Pull terms out from under the radical.

$r = 3 \sqrt{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 3 \sqrt{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 3 \sqrt{2}$

$θ = t a n^{- 1} (\frac{3 \sqrt{2}}{0})$

Step 5

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

$r = 3 \sqrt{2}$

$θ = 90 °$

Step 6

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

$(3 \sqrt{2}, 90 °)$