CONVERTING POLAR COORDINATES TO RECTANGULAR COORDINATES

Let P be the point have the polar coordinates (r,  θ) and its rectangular coordinates will be (x, y).

Then,

x = rcosθ and y = rsinθ

r2 = x2 + y2

tanθ = y/x

Example 1 :

Convert the given polar coordinates to rectangular coordinates.

(a)  (-4, 2π/3)    (b)  (√3, π/6)

Solution :

From the point (-4, 2π/3), r is -4 and θ is 2π/3

x  =  r cos θ

x  =  -4 cos 2π/3

x  =  -4(-1/2)

x  =  2

y  =  r sin θ

y  =  -4 sin 2π/3

y  =  -4(√3/2)

y  =  -2√3

So, the required rectangular co ordinate is (2, -2√3).

(b)  (√3, π/6)

From the point (√3, π/6), r is √3 and θ is π/6

x  =  r cos θ

x  =  √3 cos π/6

x  =  √3 (√3/2)

x  =  3/2

y  =  r sin θ

y  =  √3 sin π/6

y  =  √3 (1/2)

y  =  √3/2

So, the required rectangular co ordinate is (3/2√3/2).

Example 2 :

Convert the given polar coordinates to rectangular coordinates.

(a)  (2, π/4)       (b)  (-3, 5π/6)  

(c)  (5, 10π/3)    (d)  (47, 17π/2)

Solution :

(a)  (2, π/4) 

From the point (2, π/4), r is 2 and θ is π/4

x  =  r cos θ

x  =  2 cos π/4

x  =  2(1/2)

x  =  2/2

y  =  r sin θ

y  =  2 sin π/4

y  =  2(1/2)

y  =  2/√2

So, the required rectangular co ordinate is (2/2, 2/2).

(b)  (-3, 5π/6)

From the point (-3, 5π/6), r is -3 and θ is 5π/6

x  =  r cos θ

x  =  -3 cos 5π/6

x  =  -3(√3/2)

x  =  -3√3/2

y  =  r sin θ

y  =  -3 cos 5π/6

y  =  -3(1/2)

y  =  -3/2

So, the required rectangular co ordinate is

(-3√3/2, -3/2).

(c)  (5, 10π/3)

From the point (5, 10π/3), r is 5 and θ is 10π/3

x  =  r cos θ

x  =  5 cos 10π/3

x  =  5(-√3/2)

x  =  -5√3/2

y  =  r sin θ

y  =  5 cos 10π/3

y  =  5(-1/2)

y  =  -5/2

So, the required rectangular co ordinate is

(-5√3/2, -5/2).

 (d)  (47, 17π/2)

From the point (47, 17π/2), r is 47 and θ is 17π/2

x  =  r cos θ

x  =  47 cos 17π/2

x  =  47(0)

x  =  0

y  =  r sin θ

x  =  47 sin 17π/2

x  =  47(1)

x  =  47

So, the required rectangular co ordinate is (0, 47).

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