Respuesta :
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!!!
The cartesian co-ordinates of point (-3, 3) when r > 0 are (3√2, 0.75π).
The cartesian co-ordinates of point (-3, 3) when r < 0 are (-3√2, 1.75π)
The cartesian co-ordinates of point when r > 0 (4, [tex]4\sqrt{3}[/tex]) are (8, 0.13π).
The cartesian co-ordinates of point when r < 0 (-4, [tex]4\sqrt{3}[/tex]) are (-8, 0.13π).
We have to determine, the Cartesian coordinates of a point are given below.
According to the question,
The polar coordinates are related to (x, y) coordinates through
x = rcosθ, and y = rsinθ
And r = √[x² + y²]
- The polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ).
Where x = -3 and y = 3
Then,
[tex]r^{2} = x^{2} + y^{2} \\\\r^{2} = (-3)^2 + (3)^2\\\\r^2 = 9+9 \\\\r^2 = 18\\\\r = \pm3\sqrt{2}\\\\When, \ r>0 \\r = 3\sqrt{2}[/tex]
Then,
[tex]X =r cos\theta \\\\-3 =3 {\sqrt{2}}cos\theta \\\\cos\theta = \dfrac{-1}{2}[/tex]
And,
[tex]y = rsin\theta\\\\3 = 3\sqrt{2} sin\theta\\\\sin\theta = \dfrac{1}{\sqrt{2}}[/tex]
Therefore,
[tex]Tan\theta = \dfrac{sin\theta}{cos\theta}\\\\Tan\theta = \dfrac{\dfrac{1}{\sqrt{2} } }{\dfrac{-1}{\sqrt{2} } }\\\\Tan\theta = -1\\\\\theta = 0.75\pi \ or \ 1.75\pi[/tex]
Here, θ = 0.75π and 1.75π satisfy the tanθ equation, only the 0.75π satisfies the sinθ and cosθ equations.
So, The cartesian co-ordinates of point (-3, 3) when r >0 are (3√2, 0.75π).
- To Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.
[tex]When, \ r < 0, \ r = -3\sqrt{2}\\\\X= r cos \theta \\\\-3 = -3\sqrt{2} cos\theta\\\\cos \theta = \dfrac{-3} { -3\sqrt{2}}= \dfrac{1}{\sqrt{2}}\\\\And,\\\\y = r sin\theta\\\\3 = -3\sqrt{2} sin \theta\\\\sin \theta= \dfrac{3 }{ -3\sqrt{2}} = \dfrac{-1}{ \sqrt{2}}\\\\Tan \theta = \dfrac{sin\theta}{cos\theta}\\\\Tan \theta = \dfrac{\dfrac{-1}{\sqrt{2}} } {\dfrac{1}{\sqrt{2}} } \ \\\\ Tan\theta = -1\\\\\theta = 0.75\pi \ or\ 1.75\pi[/tex]
Here, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.
So, The cartesian co-ordinates of point (-3, 3) when r < 0 are (-3√2, 1.75π)
- The polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ).
Where x = 4 and y = [tex]4\sqrt{3[/tex]
Then,
[tex]r^{2} = x^{2} + y^{2} \\\\r^{2} = (4)^2 + (4\sqrt{3} )^2 \\\\r^2 = 64\\\\r = \pm 8\\\ \\When \ \ r>0 \\\\r = 8[/tex]
Then,
[tex]X =r cos\theta \\\\4 =8cos\theta \\\\cos\theta = \dfrac{1}{2}[/tex]
And,
[tex]y = rsin\theta\\\\4\sqrt{3}= 8 sin\theta\\\\sin\theta = \dfrac{\sqrt{3} }{2}[/tex]
Therefore,
[tex]Tan\theta = \dfrac{sin\theta}{cos\theta}\\\\Tan\theta = \dfrac{\dfrac{\sqrt{3}}{\sqrt{2} } }{\dfrac{1}{\sqrt{2} } }\\\\Tan\theta = \sqrt{3}\\\\\theta = 0.13\pi \ or \ 1.13\pi[/tex]
Here, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies sinθ and cosθ equations.
So, The cartesian co-ordinates of point when r>0 (4, [tex]4\sqrt{3}[/tex]) are (8, 0.13π).
- To find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)
Where x = 4 and y = [tex]4\sqrt{3[/tex]
Then,
[tex]r^{2} = x^{2} + y^{2} \\\\r^{2} = (4)^2 + (4\sqrt{3} )^2 \\\\r^2 = 64\\\\r = \pm 8\\\ \\When \ \ r<0 \\\\r = -8[/tex]
Then,
[tex]X =r cos\theta \\\\4 =-8cos\theta \\\\cos\theta = \dfrac{-1}{2}[/tex]
And,
[tex]y = rsin\theta\\\\4\sqrt{3}=- 8 sin\theta\\\\sin\theta = \dfrac{-\sqrt{3} }{2}[/tex]
Therefore,
[tex]Tan\theta = \dfrac{sin\theta}{cos\theta}\\\\Tan\theta = \dfrac{\dfrac{-\sqrt{3}}{2} }{\dfrac{-1}{2} }\\\\Tan\theta = -{\sqrt{3}}\\\\\theta = 0.13\pi \ or \ 1.13\pi[/tex]
Here, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.
So, The cartesian co-ordinates of point when r <0 (-4, [tex]4\sqrt{3}[/tex]) are (-8, 0.13π).
To know more about Co-ordinates click the link given below.
https://brainly.com/question/14751805
