Precalculus Examples

$(3, - \frac{7 π}{4})$

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.

$r = \sqrt{{(3)}^{2} + {(- \frac{7 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

$r = \sqrt{9 + {(- \frac{7 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

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Step 3.2.1

Apply the product rule to $- \frac{7 π}{4}$ .

$r = \sqrt{9 + {(- 1)}^{2} {(\frac{7 π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $\frac{7 π}{4}$ .

$r = \sqrt{9 + {(- 1)}^{2} (\frac{{(7 π)}^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

Apply the product rule to $7 π$ .

$r = \sqrt{9 + {(- 1)}^{2} (\frac{7^{2} π^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 + {(- 1)}^{2} (\frac{7^{2} π^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{9 + 1 (\frac{7^{2} π^{2}}{4^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{7^{2} π^{2}}{4^{2}}$ by $1$ .

$r = \sqrt{9 + \frac{7^{2} π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Raise $7$ to the power of $2$ .

$r = \sqrt{9 + \frac{49 π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

$r = \sqrt{9 + \frac{49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

To write $9$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$r = \sqrt{9 \cdot \frac{16}{16} + \frac{49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Combine $9$ and $\frac{16}{16}$ .

$r = \sqrt{\frac{9 \cdot 16}{16} + \frac{49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the expression.

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Step 3.9.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{9 \cdot 16 + 49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

Multiply $9$ by $16$ .

$r = \sqrt{\frac{144 + 49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{144 + 49 π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Rewrite $\sqrt{\frac{144 + 49 π^{2}}{16}}$ as $\frac{\sqrt{144 + 49 π^{2}}}{\sqrt{16}}$ .

$r = \frac{\sqrt{144 + 49 π^{2}}}{\sqrt{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11

Simplify the denominator.

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Step 3.11.1

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

$r = \frac{\sqrt{144 + 49 π^{2}}}{\sqrt{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.11.2

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

$r = \frac{\sqrt{144 + 49 π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{144 + 49 π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{144 + 49 π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{144 + 49 π^{2}}}{4}$

$θ = t a n^{- 1} (\frac{- \frac{7 π}{4}}{3})$

Step 5

The inverse tangent of $- \frac{7 π}{12}$ is $θ = 298.62015342 °$ .

$r = \frac{\sqrt{144 + 49 π^{2}}}{4}$

$θ = 298.62015342 °$

Step 6

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

$(\frac{\sqrt{144 + 49 π^{2}}}{4}, 298.62015342 °)$