Precalculus Examples

$(\frac{3}{8}, - \frac{\sqrt{55}}{8})$

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.

$r = \sqrt{{(\frac{3}{8})}^{2} + {(- \frac{\sqrt{55}}{8})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Tap for more steps...

Step 3.1

Apply the product rule to $\frac{3}{8}$ .

$r = \sqrt{\frac{3^{2}}{8^{2}} + {(- \frac{\sqrt{55}}{8})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{\frac{9}{8^{2}} + {(- \frac{\sqrt{55}}{8})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $8$ to the power of $2$ .

$r = \sqrt{\frac{9}{64} + {(- \frac{\sqrt{55}}{8})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

Tap for more steps...

Step 3.4.1

Apply the product rule to $- \frac{\sqrt{55}}{8}$ .

$r = \sqrt{\frac{9}{64} + {(- 1)}^{2} {(\frac{\sqrt{55}}{8})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Apply the product rule to $\frac{\sqrt{55}}{8}$ .

$r = \sqrt{\frac{9}{64} + {(- 1)}^{2} (\frac{{\sqrt{55}}^{2}}{8^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{9}{64} + {(- 1)}^{2} (\frac{{\sqrt{55}}^{2}}{8^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Simplify the expression.

Tap for more steps...

Step 3.5.1

Raise $- 1$ to the power of $2$ .

$r = \sqrt{\frac{9}{64} + 1 (\frac{{\sqrt{55}}^{2}}{8^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5.2

Multiply $\frac{{\sqrt{55}}^{2}}{8^{2}}$ by $1$ .

$r = \sqrt{\frac{9}{64} + \frac{{\sqrt{55}}^{2}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{9}{64} + \frac{{\sqrt{55}}^{2}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Rewrite ${\sqrt{55}}^{2}$ as $55$ .

Tap for more steps...

Step 3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{55}$ as $55^{\frac{1}{2}}$ .

$r = \sqrt{\frac{9}{64} + \frac{{(55^{\frac{1}{2}})}^{2}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.2

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

$r = \sqrt{\frac{9}{64} + \frac{55^{\frac{1}{2} \cdot 2}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.3

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

$r = \sqrt{\frac{9}{64} + \frac{55^{\frac{2}{2}}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.6.4.1

Cancel the common factor.

$r = \sqrt{\frac{9}{64} + \frac{55^{\frac{2}{2}}}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.4.2

Rewrite the expression.

$r = \sqrt{\frac{9}{64} + \frac{55}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{9}{64} + \frac{55}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6.5

Evaluate the exponent.

$r = \sqrt{\frac{9}{64} + \frac{55}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{9}{64} + \frac{55}{8^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Simplify the expression.

Tap for more steps...

Step 3.7.1

Raise $8$ to the power of $2$ .

$r = \sqrt{\frac{9}{64} + \frac{55}{64}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.2

Combine the numerators over the common denominator.

$r = \sqrt{\frac{9 + 55}{64}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.3

Add $9$ and $55$ .

$r = \sqrt{\frac{64}{64}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.4

Divide $64$ by $64$ .

$r = \sqrt{1}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.5

Any root of $1$ is $1$ .

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 1$

$θ = t a n^{- 1} (\frac{- \frac{\sqrt{55}}{8}}{\frac{3}{8}})$

Step 5

The inverse tangent of $- \frac{\sqrt{55}}{3}$ is $θ = 292.02431283 °$ .

$r = 1$

$θ = 292.02431283 °$

Step 6

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

$(1, 292.02431283 °)$