Question

The polar coordinates of a certain point are (r = 2.75 cm, θ = 219°). (a) Find its Cartesian coordinates x and y. x = cm y = cm (b) Find the polar coordinates of the points with Cartesian coordinates (−x, y). r = cm θ = ° (c) Find the polar coordinates of the points with Cartesian coordinates (−2x, −2y). r = cm θ = ° (d) Find the polar coordinates of the points with Cartesian coordinates (3x, −3y). r = cm θ = °

Answer 1

Answer:

(a) x = 2.14 cm and y = 1.73 cm ⇒ the point is (-2.14 , -1.73)

(b) r = 2.75 cm, Ф = 321.05°

(c) r = 5.50 cm, Ф = 38.95

(d) r = 8.26 cm, Ф = 141.05°

Step-by-step explanation:

* Lets explain the relation between the polar coordinates and the

  Cartesian coordinates

- The polar coordinates are (r , Ф°)

- The Cartesian coordinates are (x , y)

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

- x = r cosФ

- y = r sinФ

* Lets solve the problem

- The polar form of the point is (r = 2.75 cm, θ = 219°)

(a)

∵ r = 2.75 cm and Ф = 219°

∵ x = r cosФ and y = r sinФ

∴ x = 2.75 (cos 219) = -2.14

∴ y = 2.75 (sin 219) = -1.73

- The point (x , y) lies on the 3rd quadrant

∴ x = 2.14 cm and y = 1.73 cm

- We neglect (-) in the answer only because no negative values for cm

(b)

- We want to find the polar coordinates of (-x , y)

- That means we want to find the polar coordinates of (2.14 , -1.73)

∵ This points lies on the 4th quadrant

∴ Ф will lies between 270° and 360°

∵ $r=\sqrt{x^{2}+y^{2}}$

∴ $r=\sqrt{(2.14)^{2}+(-1.73)^{2}}=2.75$

∴ r = 2.75

∵ α = $tan^{-1}\frac{1.73}{2.14}=38.95$, where α is acute angle

∵ Ф lies in the 4th quadrant, then Ф = 360° - α

∴ Ф = 360 - 38.95 = 321.05°

∴ r = 2.75 cm, Ф = 321.05°

(c)

- We want to find the polar coordinates of (-2x , -2y)

- That means we want to find the polar coordinates of (4.28 , 3.46)

∵ This points lies on the 1nd quadrant

∴ Ф will lies between 0° and 90°

∵ $r=\sqrt{x^{2}+y^{2}}$

∴ $r=\sqrt{(4.28)^{2}+(3.46)^{2}}=5.50$

∴ r = 5.50

∵ α = $tan^{-1}\frac{3.46}{4.28}=38.95$, where α is acute angle

∵ Ф lies in the 1st quadrant, then Ф = α

∴ Ф = 38.95

∴ r = 5.50 cm, Ф = 38.95

(d)

- We want to find the polar coordinates of (3x , -3y)

- That means we want to find the polar coordinates of (-6.42 , 5.19)

∵ This points lies on the 2nd quadrant

∴ Ф will lies between 90° and 180°

∵ $r=\sqrt{x^{2}+y^{2}}$

∴ $r=\sqrt{(-6.42)^{2}+(5.19)^{2}}=8.26$

∴ r = 8.26

∵ α = $tan^{-1}\frac{5.19}{6.42}=38.95$, where α is acute angle

∵ Ф lies in the 2nd quadrant, then Ф = 180° - α

∴ Ф = 180 - 38.95 = 141.05°

∴ r = 8.26 cm, Ф = 141.05°