Precalculus Examples

$(\frac{7}{25}, \frac{24}{25})$

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.

$r = \sqrt{{(\frac{7}{25})}^{2} + {(\frac{24}{25})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Apply the product rule to $\frac{7}{25}$ .

$r = \sqrt{\frac{7^{2}}{25^{2}} + {(\frac{24}{25})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

$r = \sqrt{\frac{49}{25^{2}} + {(\frac{24}{25})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $25$ to the power of $2$ .

$r = \sqrt{\frac{49}{625} + {(\frac{24}{25})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Apply the product rule to $\frac{24}{25}$ .

$r = \sqrt{\frac{49}{625} + \frac{24^{2}}{25^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Raise $24$ to the power of $2$ .

$r = \sqrt{\frac{49}{625} + \frac{576}{25^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Raise $25$ to the power of $2$ .

$r = \sqrt{\frac{49}{625} + \frac{576}{625}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Combine the numerators over the common denominator.

$r = \sqrt{\frac{49 + 576}{625}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Add $49$ and $576$ .

$r = \sqrt{\frac{625}{625}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Divide $625$ by $625$ .

$r = \sqrt{1}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Any root of $1$ is $1$ .

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 1$

$θ = t a n^{- 1} (\frac{\frac{24}{25}}{\frac{7}{25}})$

Step 5

The inverse tangent of $\frac{24}{7}$ is $θ = 73.73979529 °$ .

$r = 1$

$θ = 73.73979529 °$

Step 6

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

$(1, 73.73979529 °)$