Polar and Rectangular conversion

07262018, 07:01 PM
(This post was last modified: 07272018 11:28 AM by DrD.)
Post: #1




Polar and Rectangular conversion
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 reused 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 

07272018, 04:47 AM
Post: #2




RE: Polar and Rectangular conversion
(07262018 07:01 PM)DrD Wrote: rectangular_coordinates((sqrt(2)),(π/4)) ==> [1,1] 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. 

07272018, 10:54 AM
Post: #3




RE: Polar and Rectangular conversion
(07262018 07:01 PM)DrD Wrote: 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: \( \begin{bmatrix} \sqrt 2& \frac{3}{4}\ast\pi \end{bmatrix} \). — Ian Abbott 

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