The Cartesian coordinates of a point are (-2,3).?

Question

(b) The Cartesian coordinates of a point are (-2,3). 

(i) Find polar coordinates (r,\theta) of the point, where r>0 and 0 \le \theta < 2\pi. 

r = 

\theta = 

(ii) Find polar coordinates (r,\theta) of the point, where r<0 and 0 \le \theta < 2\pi. 

r = 

\theta =

 

 

Answer 1

(b)

(i)

The Cartesian coordinates of a point (x,y) = (-2,3).

x = -2 , y = 3

The Polar coordinates of a point (r,Θ) = ?

r = √ x² + y² )

r = √[(-2)² + 3²]

r = √(4+9)

r = √13

r = 3.62 > 0

Θ = 180 + tan^-1(y/x)  when x<0

Θ = 180 + tan^-1(3/(-2))

Θ = 180 - 56.3099

Θ = 123.69°

0 < Θ < 360

Solution:

r = 3.62

Θ = 123.69°

Answer 2

(b)

(ii)

The Cartesian coordinates of a point (x,y) = (-2,3).

x = -2 , y = 3

The Polar coordinates of a point (r,Θ) = ?

r = √ x² + y² )

r = √[(-2)² + 3²]

r = √(4+9)

r = √13

r = ± 3.62

But the condition is r < 0.

r = - 3.62

This is shown in graph.

Θ = 180 + tan^-1(y/x)  when x<0

Θ = 180 + tan^-1(3/(-2))

Θ = 180 - 56.3099

Θ = 123.69°

As r is negative , angle slides 180° in anticlock wise direction.

Θ = 180° + 123.69°

Θ = 303.69°

0 < Θ < 360

Solution:

r = -3.62

Θ = 303.69°