Given any Cartesian coordinates, (x,y), there are polar coordinates (R,THETA) wi

Question

Given any Cartesian coordinates, (x,y), there are polar coordinates (R,THETA) with -pi/2<0<0

Explanation / Answer

Find polar coordinates with -p/2 = ? = p/2 for the following Cartesian coordinates: NOTE: Because -p/2 = ? = p/2 (this is a semicircle) then r > 0 when x >0 and r < 0 when x < 0 to allow for the other half of the circle in which points lie (a) If (x, y) = (19, 3) then (r, theta) = ( , ), i got r as 19.24 and my theta was wrong and i said it was 8.972627. r = v(x² + y²) = v(19² + 3²) = v(370) ˜ 19.24 ? = arctan (y/x) = arctan (3/19) ˜ 0.157 radians So if (x, y) = (19, 3) then (r, ?)= (19.24, 0.157) (approx.) (b) If (x,y) = (10, 6) then (r, theta) = ( , ), I got r as 11.661903 and again my theta was wrong and i said it was 30.963757. r = v(x² + y²) = v(10² + 6²) = v(136) ˜ 11.66 ? = arctan (y/x) = arctan (6/10) ˜ 0.540 radians So if (x, y) = (10, 6) then (r, ?) = (11.66, 0.540) (approx.) (c) If (x,y) = (-9, -8) then (r, theta) = ( , ), I got both r and theta wrong and no clue why r = v(x² + y²) = v((-9)² + (-8)²) = - v(145) (- because x < 0 it is in 2nd quadrant or 3rd quadrants) ˜ -12.04 ? = arctan (y/x) = arctan (-8/-9) ˜ 0.844 radians So if (x, y) = (-9, -8) then (r, ?) = (-12.04, 0.844) (approx.) (d) If (x,y) = (16, 1) then (r, theta) = ( , ), I got r as 16.03122 and it was right and again my theta was wrong r = v(x² + y²) = v((16)² + (1)²) = v(257) ˜ 16.03 ? = arctan (y/x) = arctan (1/16) ˜ 1.508 radians So if (x, y) = (1, 16) then (r, ?) = (16.03, 1.508) (approx.) (e) If (x,y) = (-4, 6) then (r, theta) = ( , ), both r and theta were wrong r = v(x² + y²) = v((-4)² + (6)²) = -v(52) (- because x < 0 it is in 2nd quadrant or 3rd quadrants) ˜ -7.211 ? = arctan (y/x) = arctan (6/-4) ˜ -0.983 radians So if (x, y) = (-4, 6) then (r, ?) = (-7.211, -0.983) (approx.) (f) If (x,y) = (0, -6) then (r, theta) = ( , ). both r and theta were wrong r = v(x² + y²) = v((0)² + (-6)²) = v(36) = 6 ? = arctan (y/x) = arctan (-6/0) = arctan (-8) = - p/2 radians So if (x, y) = (0, -6) then (r, ?) = (6, - p/2 ) Actually this last one was obvious ... on the negative Y - axis and 6 units away means it is on a circle 6 units in radius and 90 degrees (= p/2 radians) clockwise from X-axis and so at - p/2 radians.

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