CONVERT RECTANGULAR COORDINATES INTO POLAR FORM
Let P be the rectangular coordinate in the form (x, y), we should convert it into the form of (r, θ).
Then,
r2 = x2 + y2
θ = tan-1 (y/x)
To find the general solution,
θ = tan-1 (y/x) + nπ
Example 1 :
Convert to polar coordinates on the interval 0 < θ < 2π
(a) (-1, 1) (b) (1, √3)
Solution :
(a) (-1, 1)
x = -1 and y = 1
|
r2 = x2 + y2 r2 = 12 + 12 r2 = 2 r = ± √2 |
θ = tan-1 (y/x) θ = tan-1 (-1/1) θ = tan-1 (-1) θ = -π/4 |
If r = √2 then θ = -π/4 ==> (√2, -π/4)
If r = -√2 then θ = -π/4 + π ==> (√2, 3π/4)
If r = √2 then θ = -π/4 + 2π ==> (√2, 7π/4)
(b) (1, √3)
x = 1 and y = √3
|
r2 = x2 + y2 r2 = 12 + √32 r2 = 1 + 3 r2 = 4 r = ±2 |
θ = tan-1 (y/x) θ = tan-1 (1/1) θ = tan-1 (1) θ = π/4 |
If r = 2 then θ = π/4 ==> (2, π/4)
If r = -2 then θ = π/4 + π ==> (-2, 5π/4)
If r = 2 then θ = π/4 + 2π ==> (2, 9π/4)
If we have any restriction like the angle must between 0 to 360, then we have to ignore the last option.
Example 2 :
Convert the given rectangular coordinate to polar coordinate between 0 ≤ θ ≤ 2π
(1, -√3)
x = 1 and y = -√3
|
r2 = x2 + y2 r2 = 12 + (-√3)2 r2 = 1 + 3 r2 = 4 r = ±2 |
θ = tan-1 (y/x) θ = tan-1 (-√3/1) θ = tan-1 (-√3) θ = -π/3 |
If r = 2 then θ = -π/3 ==> (2, -π/3)
If r = -2 then θ = -π/3 + π ==> (-2, 2π/3)
If r = 2 then θ = -π/3 + 2π ==> (2, 5π/3)
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