Precalculus Examples

$(\sqrt{32}, \sqrt{32})$

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.

$r = \sqrt{{(\sqrt{32})}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Tap for more steps...

Step 3.1

Rewrite $32$ as $4^{2} \cdot 2$ .

Tap for more steps...

Step 3.1.1

Factor $16$ out of $32$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

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

$r = \sqrt{{\sqrt{4^{2} \cdot 2}}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{{\sqrt{4^{2} \cdot 2}}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Pull terms out from under the radical.

$r = \sqrt{{(4 \sqrt{2})}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Simplify the expression.

Tap for more steps...

Step 3.3.1

Apply the product rule to $4 \sqrt{2}$ .

$r = \sqrt{4^{2} {\sqrt{2}}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.2

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

Tap for more steps...

Step 3.4.1

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

$r = \sqrt{16 {(2^{\frac{1}{2}})}^{2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

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

$r = \sqrt{16 \cdot 2^{\frac{1}{2} \cdot 2} + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.3

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.4.4.1

Cancel the common factor.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.4.2

Rewrite the expression.

$r = \sqrt{16 \cdot 2 + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{16 \cdot 2 + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.5

Evaluate the exponent.

$r = \sqrt{16 \cdot 2 + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{16 \cdot 2 + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Multiply $16$ by $2$ .

$r = \sqrt{32 + {(\sqrt{32})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Rewrite $32$ as $4^{2} \cdot 2$ .

Tap for more steps...

Step 3.6.1

Factor $16$ out of $32$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.2

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

$r = \sqrt{32 + {\sqrt{4^{2} \cdot 2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{32 + {\sqrt{4^{2} \cdot 2}}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Pull terms out from under the radical.

$r = \sqrt{32 + {(4 \sqrt{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the expression.

Tap for more steps...

Step 3.8.1

Apply the product rule to $4 \sqrt{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

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

Tap for more steps...

Step 3.9.1

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

$r = \sqrt{32 + 16 {(2^{\frac{1}{2}})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

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

$r = \sqrt{32 + 16 \cdot 2^{\frac{1}{2} \cdot 2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.3

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.9.4.1

Cancel the common factor.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.4.2

Rewrite the expression.

$r = \sqrt{32 + 16 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{32 + 16 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.5

Evaluate the exponent.

$r = \sqrt{32 + 16 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{32 + 16 \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Simplify the expression.

Tap for more steps...

Step 3.10.1

Multiply $16$ by $2$ .

$r = \sqrt{32 + 32}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.2

Add $32$ and $32$ .

$r = \sqrt{64}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.3

Rewrite $64$ as $8^{2}$ .

$r = \sqrt{8^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.4

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

$r = 8$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 8$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 8$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 8$

$θ = t a n^{- 1} (\frac{\sqrt{32}}{\sqrt{32}})$

Step 5

The inverse tangent of $1$ is $θ = 45 °$ .

$r = 8$

$θ = 45 °$

Step 6

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

$(8, 45 °)$