Precalculus Examples

$(- 4, - \frac{p}{4})$

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.

$r = \sqrt{{(- 4)}^{2} + {(- \frac{p}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

Tap for more steps...

Step 3.1

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

$r = \sqrt{16 + {(- \frac{p}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

Tap for more steps...

Step 3.2.1

Apply the product rule to $- \frac{p}{4}$ .

$r = \sqrt{16 + {(- 1)}^{2} {(\frac{p}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

$r = \sqrt{16 + {(- 1)}^{2} (\frac{p^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{16 + {(- 1)}^{2} (\frac{p^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{16 + 1 (\frac{p^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{p^{2}}{4^{2}}$ by $1$ .

$r = \sqrt{16 + \frac{p^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Combine the numerators over the common denominator.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Multiply $16$ by $16$ .

$r = \sqrt{\frac{256 + p^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Rewrite $\frac{256 + p^{2}}{16}$ as ${(\frac{1}{4})}^{2} (256 + p^{2})$ .

Tap for more steps...

Step 3.10.1

Factor the perfect power $1^{2}$ out of $256 + p^{2}$ .

$r = \sqrt{\frac{1^{2} (256 + p^{2})}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.2

Factor the perfect power $4^{2}$ out of $16$ .

$r = \sqrt{\frac{1^{2} (256 + p^{2})}{4^{2} \cdot 1}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.3

Rearrange the fraction $\frac{1^{2} (256 + p^{2})}{4^{2} \cdot 1}$ .

$r = \sqrt{{(\frac{1}{4})}^{2} (256 + p^{2})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{{(\frac{1}{4})}^{2} (256 + p^{2})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Pull terms out from under the radical.

$r = \frac{1}{4} \cdot \sqrt{256 + p^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.12

Combine $\frac{1}{4}$ and $\sqrt{256 + p^{2}}$ .

$r = \frac{\sqrt{256 + p^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{256 + p^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{256 + p^{2}}}{4}$

$θ = t a n^{- 1} (\frac{- \frac{p}{4}}{- 4})$

Step 5

The inverse tangent of $\frac{p}{16}$ is $θ = 0 °$ .

$r = \frac{\sqrt{256 + p^{2}}}{4}$

$θ = 0 °$

Step 6

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

$(\frac{\sqrt{256 + p^{2}}}{4}, 0 °)$