Calculus Examples

$(1, \frac{5 π}{3})$

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.

$r = \sqrt{{(1)}^{2} + {(\frac{5 π}{3})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

One to any power is one.

$r = \sqrt{1 + {(\frac{5 π}{3})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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 $\frac{5 π}{3}$ .

$r = \sqrt{1 + \frac{{(5 π)}^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $5 π$ .

$r = \sqrt{1 + \frac{5^{2} π^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{1 + \frac{5^{2} π^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $5$ to the power of $2$ .

$r = \sqrt{1 + \frac{25 π^{2}}{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

$r = \sqrt{1 + \frac{25 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Write $1$ as a fraction with a common denominator.

$r = \sqrt{\frac{9}{9} + \frac{25 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

Combine the numerators over the common denominator.

$r = \sqrt{\frac{9 + 25 π^{2}}{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Rewrite $\sqrt{\frac{9 + 25 π^{2}}{9}}$ as $\frac{\sqrt{9 + 25 π^{2}}}{\sqrt{9}}$ .

$r = \frac{\sqrt{9 + 25 π^{2}}}{\sqrt{9}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the denominator.

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Step 3.8.1

Rewrite $9$ as $3^{2}$ .

$r = \frac{\sqrt{9 + 25 π^{2}}}{\sqrt{3^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

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

$r = \frac{\sqrt{9 + 25 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{9 + 25 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{9 + 25 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{9 + 25 π^{2}}}{3}$

$θ = t a n^{- 1} (\frac{\frac{5 π}{3}}{1})$

Step 5

The inverse tangent of $\frac{5 π}{3}$ is $θ = 79.18752128 °$ .

$r = \frac{\sqrt{9 + 25 π^{2}}}{3}$

$θ = 79.18752128 °$

Step 6

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

$(\frac{\sqrt{9 + 25 π^{2}}}{3}, 79.18752128 °)$