How do you change (4, -1) from rectangular to cylindrical coordinates between [0, 2π)?

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How do you change (4, -1) from rectangular to cylindrical coordinates between [0, 2π)?

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Answer 1 · 2015-03-04T23:18:45

I will assume you meant "polar coordinates" since cylindrical coordinates are 3-dimensional and you have only supplied 2-dimensional Cartesian coordinates.

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From the graph:
$r = \sqrt{4^{2} + {(- 1)}^{2}} = \sqrt{17}$ $r = sqrt(4^2 + (-1)^2) = sqrt(17)$
and
$θ = arctan (\frac{- 1}{4})$ $theta = arctan((-1)/4)$

Note that standard arctan functions will return a value which will need to be adjusted to compensate for the point being in the IV quadrant so $θ$ $theta$ will fall in the $[0, 2 π)$ $[0,2pi)$ range.

Depending upon the version used you will typically get
$arctan (\frac{- 1}{4}) = - 0.245$ $arctan((-1)/4) = -0.245$ (radians)
or
$arctan (\frac{- 1}{4}) = - 14.040$ $arctan((-1)/4) = -14.040$ (degrees)

Add $2 π$ $2pi$ or $360^{o}$ $360^o$ respectively to get the "correct" answer.

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How do you change (4, -1) from rectangular to cylindrical coordinates between [0, 2π)?

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