Precalculus Examples

$(1.5, \frac{π}{4})$

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.

$r = \sqrt{{(1.5)}^{2} + {(\frac{π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Simplify the expression.

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Step 3.1.1

Raise $1.5$ to the power of $2$ .

$r = \sqrt{2.25 + {(\frac{π}{4})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

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

$r = \sqrt{2.25 + \frac{π^{2}}{4^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.3

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

$r = \sqrt{2.25 + \frac{π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{2.25 + \frac{π^{2}}{16}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Add $2.25$ and $\frac{π^{2}}{16}$ .

$r = \sqrt{2.86685027}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Evaluate the root.

$r = 1.69317756$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1.69317756$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 1.69317756$

$θ = t a n^{- 1} (\frac{\frac{π}{4}}{1.5})$

Step 5

The inverse tangent of $0.52359877$ is $θ = 27.63649933 °$ .

$r = 1.69317756$

$θ = 27.63649933 °$

Step 6

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

$(1.69317756, 27.63649933 °)$