Question

PLEASE HELP!!!! 20 POINTS!!!!

1.) Which of the following polar equations is equivalent to the parametric equations below?
x=t^2
y=2t
A.) r=4cot(theta)csc(theta)
B.) r=4tan(theta)sec(theta)
C.) r=tan(theta)sec(theta)/4
D.) r=16cot(theta)csc(theta)

2.) Which polar equation is equivalent to the parametric equations below?
x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)
A.) r=cos(theta)+1
B.) r=sin(theta)+1
C.) r=sin^2(theta)+1
D.) r=1-cos(theta)

Answer 1

These are two questions and two answers.

Question 1) Which of the following polar equations is equivalent to the parametric equations below?

x=t²
y=2t

Answer: option A.) r = 4cot(theta)csc(theta)


Step-by-step 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?

x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)

Answer: option B) r = sinθ + 1


Step-by-step 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)
b) r sinθ =sin²(θ)+sin(θ)

3) work both equations

a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1] ⇒ r = sinθ + 1


b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1


Therefore, the answer is r = sinθ + 1 which is the option B.

Answer 2

*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