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e-lub [12.9K]
2 years ago
11

У (x, y) (?x, ?y) A' 3-4-3-2-10 1 2 B C B с

Mathematics
1 answer:
vovangra [49]2 years ago
8 0

Answer:c

Step-by-step explanation:

anser :c

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PLEASE HELP!!!! 20 POINTS!!!!
galben [10]
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.
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