Precalculus Examples

$(- 5 \sqrt{2}, - 5 \sqrt{2})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Simplify the expression.

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Step 3.1.1

Apply the product rule to $- 5 \sqrt{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

Raise $- 5$ to the power of $2$ .

$r = \sqrt{25 {\sqrt{2}}^{2} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 {\sqrt{2}}^{2} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

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Step 3.2.1

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

$r = \sqrt{25 {(2^{\frac{1}{2}})}^{2} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

$r = \sqrt{25 \cdot 2^{\frac{1}{2} \cdot 2} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

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

$r = \sqrt{25 \cdot 2^{\frac{2}{2}} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4

Cancel the common factor of $2$ .

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Step 3.2.4.1

Cancel the common factor.

$r = \sqrt{25 \cdot 2^{\frac{2}{2}} + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4.2

Rewrite the expression.

$r = \sqrt{25 \cdot 2 + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 \cdot 2 + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.5

Evaluate the exponent.

$r = \sqrt{25 \cdot 2 + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{25 \cdot 2 + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $25$ by $2$ .

$r = \sqrt{50 + {(- 5 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.2

Apply the product rule to $- 5 \sqrt{2}$ .

$r = \sqrt{50 + {(- 5)}^{2} {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.3

Raise $- 5$ to the power of $2$ .

$r = \sqrt{50 + 25 {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{50 + 25 {\sqrt{2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

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Step 3.4.1

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

$r = \sqrt{50 + 25 {(2^{\frac{1}{2}})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

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

$r = \sqrt{50 + 25 \cdot 2^{\frac{1}{2} \cdot 2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.3

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

$r = \sqrt{50 + 25 \cdot 2^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.4

Cancel the common factor of $2$ .

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Step 3.4.4.1

Cancel the common factor.

$r = \sqrt{50 + 25 \cdot 2^{\frac{2}{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.4.2

Rewrite the expression.

$r = \sqrt{50 + 25 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{50 + 25 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.5

Evaluate the exponent.

$r = \sqrt{50 + 25 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{50 + 25 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Simplify the expression.

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Step 3.5.1

Multiply $25$ by $2$ .

$r = \sqrt{50 + 50}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5.2

Add $50$ and $50$ .

$r = \sqrt{100}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5.3

Rewrite $100$ as $10^{2}$ .

$r = \sqrt{10^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5.4

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

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 10$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 10$

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

Step 5

The inverse tangent of $1$ is $θ = 225 °$ .

$r = 10$

$θ = 225 °$

Step 6

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

$(10, 225 °)$