Question

The Cartesian coordinates of a point are given. (a) (−3, 3) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = (ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) = (b) (4, 4 3 ) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ) = (ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ) =

Answer 1

Answer:

a) (-3, 3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (3√2, 0.75π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-3√2, 1.75π)

b) (4, 4√3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (8, 0.13π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-8, 1.13π)

Step-by-step explanation:

We know that polar coordinates are related to (x, y) coordinates through

x = r cos θ

y = r sin θ

And r = √[x² + y²]

a) For (-3, 3)

(i) x = -3, y = 3

r = √[x² + y²] = √[(-3)² + (3)²] = √18 = ±3√2

If r > 0, r = 3√2

x = r cos θ

-3 = 3√2 cos θ

cos θ = -3 ÷ 3√2 = -(1/√2)

y = r sin θ

3 = 3√2 sin θ

sin θ = 3 ÷ 3√2 = (1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 0.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (3√2, 0.75π)

(ii) When r < 0, r = -3√2

x = r cos θ

-3 = -3√2 cos θ

cos θ = -3 ÷ -3√2 = (1/√2)

y = r sin θ

3 = -3√2 sin θ

sin θ = 3 ÷ -3√2 = -(1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (-3√2, 1.75π)

b) For (4, 4√3)

(i) x = 4, y = 4√3

r = √[x² + y²] = √[(4)² + (4√3)²] = √64 = ±8

If r > 0, r = 8

x = r cos θ

4 = 8 cos θ

cos θ = 4 ÷ 8 = 0.50

y = r sin θ

4√3 = 8 sin θ

sin θ = 4√3 ÷ 8 = (√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (8, 0.13π)

(ii) When r < 0, r = -8

x = r cos θ

4 = -8 cos θ

cos θ = 4 ÷ -8 = -0.50

y = r sin θ

4√3 = -8 sin θ

sin θ = 4√3 ÷ -8 = -(√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 1.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (-8, 1.13π)

Hope this Helps!!!