Answer to Question #273950 in Analytic Geometry for Aliyah

Question #273950
  1. Convert the polar coordinates (-8, 2π/3) into rectangular coordinates.
  2. Convert the rectangular coordinates (3, -3) into polar coordinates with r > 0 and 0 ≤ θ < 2π.
  3. Convert the rectangular equation x2 + y2 = 100 into a polar equation that expresses r in terms of θ.
  4. Convert the polar equation 4r cos θ + r sin θ = 8 into a rectangular equation that expresses y in terms of x.
1
Expert's answer
2021-12-02T07:20:48-0500

Let (r,θ)(r, \theta) - polar coordinates and (x,y)(x, y) - rectangular coordinates.


1) r1=8r_{1} = - 8

θ1=2π3\theta_{1} = \dfrac{2\pi}{3}

x1=r1cos(θ1)=(8)cos(2π3)=(8)(12)=4x_{1} = r_{1}cos(\theta_{1}) = (-8) * cos \left(\dfrac{2\pi}{3}\right) = (-8) * \left(-\dfrac{1}{2}\right) = 4

y1=r1sin(θ1)=(8)sin(2π3)=(8)(32)=43y_{1} = r_{1}sin(\theta_{1}) = (-8) * sin \left(\dfrac{2\pi}{3}\right) = (-8) * \left(\dfrac{\sqrt{3}}{2}\right) = -4\sqrt{3}

2) x2=3x_{2} = 3

y2=3y_{2} = - 3

r2>0r_{2} > 0

0θ2<2π0 \le \theta_{2} < 2\pi

r22=x22+y22=32+(3)2=18r2=32r_{2}^{2} = x_{2}^{2} + y_{2}^{2} = 3^{2} + (-3)^{2} = 18 \rArr r_{2} = 3\sqrt{2}

tan(θ2)=y2x2=1tan(\theta_{2}) = \dfrac{y_{2}}{x_{2}} = -1

x2>0,y2<0θ2=7π4x_{2} > 0, y_{2} < 0 \rArr \theta_{2} = \dfrac{7\pi}{4}


3)x2+y2=100x^{2} + y^{2} = 100

r2=100r=10r^{2} = 100 \rArr r = 10


4) 4rcosθ+rsinθ=84 r cos\theta + rsin \theta = 8

4x+y=84x + y = 8

y=4x+8y = - 4x + 8



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