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Full Version: Polar and Rectangular conversion
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rectangular_coordinates(-(sqrt(2)),(π/4)) ==> [-1,-1]

polar_coordinates(-1,-1) ==> [sqrt(2),(3/4)*π]
polar_coordinates([-1,-1]) ==> [sqrt(2),(-3/4)*π]

polar_coordinates() expects a point, and returns a vector.
rectangular_coordinates() expects a vector and returns a vector.

This is confusing for infrequent users of these commands. The polar_coordinates() command can lead to an incorrect result, if a vector is supplied as the argument, such as might be the case if the result of rectangular_coordinates() was being re-used there.

NOTE: I was using the emulator, and I have been noticing some odd behaviors (including the above). After performing a reset, I am now getting the same result from polar_coordinates() whether or not a point or vector is used. I apologize for this diversion!

-Dale-
(07-26-2018 07:01 PM)DrD Wrote: [ -> ]rectangular_coordinates(-(sqrt(2)),(π/4)) ==> [-1,-1]

polar_coordinates(-1,-1) ==> [sqrt(2),(3/4)*π]
polar_coordinates([-1,-1]) ==> [sqrt(2),(-3/4)*π]

polar_coordinates() expects a point, and returns a vector.
rectangular_coordinates() expects a vector and returns a vector.

This is confusing for infrequent users of these commands. The polar_coordinates() command can lead to an incorrect result, if a vector is supplied as the argument, such as might be the case if the result of rectangular_coordinates() was being re-used there.

polar_coordinates([-1,-1]) ==> [sqrt(2),(-3/4)*π] is correct.
The original polar coordinate -(sqrt(2)),(π/4) has a rotation of π added due to the negative sign of the magnitude.
The angle sum (π+π/4) can be written +5/4*π or -3/4*π which is the result given by polar_coordinates([-1,-1])
Using polar_coordinates without the vector is omitting the sign of the angle.
(07-26-2018 07:01 PM)DrD Wrote: [ -> ]polar_coordinates(-1,-1) ==> [sqrt(2),(3/4)*π]
polar_coordinates([-1,-1]) ==> [sqrt(2),(-3/4)*π]

I was confused why those gave different results, but it's a typo. They give the same (expected) result: $$\begin{bmatrix} \sqrt 2& -\frac{3}{4}\ast\pi \end{bmatrix}$$.
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