okay i have no clue why these coordinates are not working?

Question

Given any Cartesian coordinates, (x,y), there are polar coordinates (r,(theta)) with -{pi}/{2} < theta <(and equal) {pi}/{2} .
Find polar coordinates with -{pi}{2} (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.
(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.
(c) If (x,y) = (-9, -8) then (r, theta) = ( , ), I got both r and theta wrong and no clue why
(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
(e) If (x,y) = (-4, 6) then (r, theta) = ( , ), both r and theta were wrong
(f) If (x,y) = (0, -6) then (r, theta) = ( , ). both r and theta were wrong

i know you are suppose to use r^2=x^2+y^2 to find r
and then for theta use arctan of (y/x) . why am i getting these wrong?

okay i got a,b, and d finally okay so i plugged in for question c and i got r= 12.04 and its not right....what am i doing wrong.

Answer 1

Find polar coordinates with -π/2 ≤ θ ≤ π/2 for the following Cartesian coordinates:

NOTE: Because -π/2 ≤ θ ≤ π/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 = √(x² + y²)
= √(19² + 3²)
= √(370)
≈ 19.24

θ = arctan (y/x)
= arctan (3/19)
≈ 0.157 radians

So if (x, y) ≡ (19, 3) then (r, θ)= (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 = √(x² + y²)
= √(10² + 6²)
= √(136)
≈ 11.66

θ = arctan (y/x)
= arctan (6/10)
≈ 0.540 radians

So if (x, y) ≡ (10, 6) then (r, θ) ≡ (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 = √(x² + y²)
= √((-9)² + (-8)²)
= - √(145) (- because x < 0 it is in 2nd quadrant or 3rd quadrants)
≈ -12.04

θ = arctan (y/x)
= arctan (-8/-9)
≈ 0.844 radians

So if (x, y) ≡ (-9, -8) then (r, θ) ≡ (-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 = √(x² + y²)
= √((16)² + (1)²)
= √(257)
≈ 16.03

θ = arctan (y/x)
= arctan (1/16)
≈ 1.508 radians

So if (x, y) ≡ (1, 16) then (r, θ) ≡ (16.03, 1.508) (approx.)

(e) If (x,y) = (-4, 6) then (r, theta) = ( , ), both r and theta were wrong

r = √(x² + y²)
= √((-4)² + (6)²)
= -√(52) (- because x < 0 it is in 2nd quadrant or 3rd quadrants)
≈ -7.211

θ = arctan (y/x)
= arctan (6/-4)
≈ -0.983 radians

So if (x, y) ≡ (-4, 6) then (r, θ) ≡ (-7.211, -0.983) (approx.)

(f) If (x,y) = (0, -6) then (r, theta) = ( , ). both r and theta were wrong

r = √(x² + y²)
= √((0)² + (-6)²)
= √(36)
= 6

θ = arctan (y/x)
= arctan (-6/0)
= arctan (-∞)
= - π/2 radians

So if (x, y) ≡ (0, -6) then (r, θ) ≡ (6, - π/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 (= π/2 radians) clockwise from X-axis and so at - π/2 radians.

Answer 2

Well, the calculator being in radians/degrees would not effect the r calculation, but it will gum up the theta.

r=sqrt(x**2 + y**2)

I notice you tend to have r wrong when one or the other of x or y is negative. Since x and y are squared, the negative signs always go away. Both x**2 and y**2 are always positive.

It looks like all your thetas are wrong.

Consider the test point (3,4) The radius should be 5.

Consider the test point (1,2) Theta should be pi/6, or 60 degrees.

Answer 3

I suppose you checked to make sure your calculator was in the proper form of radians or degrees depending on which the problem called for. Other than that I couldnt tell ya