Question

Convert the Cartesian coordinate (6,-3) to polar coordinates, 0≤θ<2π, r>0

Answer 1

Answer:

The polar coordinates is (3√5 , 333.4°) OR (3√5 , 5.82 rad)

Step-by-step explanation:

The polar form of the Cartesian coordinates (x , y) is (r , Ф), where

• $r=\sqrt{x^{2}+y^{2}}$
• Ф = $tan^{-1}(\frac{y}{x})$

The Cartesian coordinates is (6 , -3)

That means the point lies in the fourth quadrant because the x-coordinate is positive and the y-coordinate is negative, so Ф will be equal [2π -  $tan^{-1}(\frac{y}{x})$ ] (neglect the negative sign of y-coordinate)

∵ x = 6 and y = -3

∵ r > 0

∵ $r=\sqrt{x^{2}+y^{2}}$

- Substitute x and y in the rule of r

∴ $r=\sqrt{(6)^{2}+(-3)^{2}}$

∴ $r=\sqrt{36+9}$

∴ $r=\sqrt{45}$

∴ $r=3\sqrt{5}$

Now let us find Ф

∵ 0 ≤ Ф < 2π

∴ Ф = 2π -  $tan^{-1}(\frac{y}{x})$

- Neglect the negative sign of the y-coordinate

∴ Ф = 2π -  $tan^{-1}(\frac{3}{6})$

∴ Ф = 333.4°  OR  Ф = 5.82 radiant

∴ The polar coordinates is (3√5 , 333.4°) OR (3√5 , 5.82 rad)