PLEASE HELP!!!! 20 POINTS!!!!
2 answers:
These are two questions and two answers. Question 1) Which of the following polar equations is equivalent to the parametric equations below? <span>
x=t²
y=2t</span>Answer: option <span>A.) r = 4cot(theta)csc(theta) </span>Explanation: 1) Polar coordinates ⇒ x = r cosθ and y = r sinθ 2) replace x and y in the parametric equations: r cosθ = t² r sinθ = 2t 3) work r sinθ = 2t r sinθ/2 = t (r sinθ / 2)² = t² 4) equal both expressions for t² r cos θ = (r sin θ / 2 )² 5) simplify r cos θ = r² (sin θ)² / 4 4 = r (sinθ)² / cos θ r = 4 cosθ / (sinθ)² r = 4 cot θ csc θ ↔ which is the option A.Question 2) Which polar equation is equivalent to the parametric equations below? <span>
x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)</span>Answer: option B) r = sinθ + 1 Explanation: 1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ 2) replace x and y in the parametric equations: a) r cosθ = sin(θ)cos(θ)+cos(θ) <span>
b) r sinθ =sin²(θ)+sin(θ)</span> 3) work both equations a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1] ⇒ r = sinθ + 1 <span>
b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1 </span><span> </span><span> </span>Therefore, the answer is r = sinθ + 1 which is the option B.
*In order from the original test*
1. B. x=3cos^2 theta, y=3cos theta sin theta
2. A. r=4cot theta csc theta
3. B. r=sin theta+1
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