Precalculus Examples

$(3 \sqrt{2}, \frac{3 π}{4})$

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.

$r = \sqrt{{(3 \sqrt{2})}^{2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Simplify the expression.

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Step 3.1.1

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

$r = \sqrt{3^{2} {\sqrt{2}}^{2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

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

$r = \sqrt{9 {\sqrt{2}}^{2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 {\sqrt{2}}^{2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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 $2^{\frac{1}{2}}$ .

$r = \sqrt{9 {(2^{\frac{1}{2}})}^{2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

$r = \sqrt{9 \cdot 2^{\frac{1}{2} \cdot 2} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

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

$r = \sqrt{9 \cdot 2^{\frac{2}{2}} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4

Cancel the common factor of $2$ .

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Step 3.2.4.1

Cancel the common factor.

$r = \sqrt{9 \cdot 2^{\frac{2}{2}} + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4.2

Rewrite the expression.

$r = \sqrt{9 \cdot 2 + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 2 + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.5

Evaluate the exponent.

$r = \sqrt{9 \cdot 2 + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 2 + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Multiply $9$ by $2$ .

$r = \sqrt{18 + {(\frac{3 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.4.1

Apply the product rule to $\frac{3 π}{4}$ .

$r = \sqrt{18 + \frac{{(3 π)}^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Apply the product rule to $3 π$ .

$r = \sqrt{18 + \frac{3^{2} π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{18 + \frac{3^{2} π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{18 + \frac{9 π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

$r = \sqrt{18 + \frac{9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

To write $18$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$r = \sqrt{18 \cdot \frac{16}{16} + \frac{9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Combine $18$ and $\frac{16}{16}$ .

$r = \sqrt{\frac{18 \cdot 16}{16} + \frac{9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the expression.

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Step 3.9.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{18 \cdot 16 + 9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

Multiply $18$ by $16$ .

$r = \sqrt{\frac{288 + 9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{288 + 9 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Rewrite $\sqrt{\frac{288 + 9 π^{2}}{16}}$ as $\frac{\sqrt{288 + 9 π^{2}}}{\sqrt{16}}$ .

$r = \frac{\sqrt{288 + 9 π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Simplify the numerator.

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Step 3.11.1

Rewrite $288 + 9 π^{2}$ as $3^{2} (32 + π^{2})$ .

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Step 3.11.1.1

Factor $9$ out of $288$ .

$r = \frac{\sqrt{9 (32) + 9 π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.1.2

Factor $9$ out of $9 π^{2}$ .

$r = \frac{\sqrt{9 (32) + 9 (π^{2})}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.1.3

Factor $9$ out of $9 (32) + 9 (π^{2})$ .

$r = \frac{\sqrt{9 (32 + π^{2})}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.1.4

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

$r = \frac{\sqrt{3^{2} (32 + π^{2})}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{3^{2} (32 + π^{2})}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.2

Pull terms out from under the radical.

$r = \frac{3 \sqrt{32 + π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{3 \sqrt{32 + π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12

Simplify the denominator.

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Step 3.12.1

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

$r = \frac{3 \sqrt{32 + π^{2}}}{\sqrt{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12.2

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

$r = \frac{3 \sqrt{32 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{3 \sqrt{32 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{3 \sqrt{32 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{3 \sqrt{32 + π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{\frac{3 π}{4}}{3 \sqrt{2}})$

Step 5

The inverse tangent of $\frac{π \sqrt{2}}{8}$ is $θ = 29.04605753 °$ .

$r = \frac{3 \sqrt{32 + π^{2}}}{4}$

$θ = 29.04605753 °$

Step 6

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

$(\frac{3 \sqrt{32 + π^{2}}}{4}, 29.04605753 °)$