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

2 answers:

6 0

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)


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

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.

Zarrin [17]2 years ago

4 0

*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

You might be interested in

In a circle a chord 10 inches long is parallel to a tangent and bisects the radius drawn to the point of tangency. How long is t

Lorico [155]

Check the picture below.

keeping in mind that the point of tangency for a radius line and a tangent is alway a right-angle, since the "red" chord is parallel to the "green" tangent line outside, then the chord is cutting the "green" radius there in two equal halves at a right-angle, as you see in the picture.

we know the chord is 10 units long, so 5 + 5, since is perpendicularity with the radius will also cut the chord in two equal halves.

anyhow, all that said, we end up with triangle you see on the right-hand-side, and then we can just use the pythagorean theorem.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
-------\\
c=2a\\
a=a\\
b=5
\end{cases}\implies (2a)^2=a^2+5^2$

$\bf (2^2a^2)=a^2+25\implies 4a^2=a^2+25\implies 3a^2=25
\\\\\\
a^2=\cfrac{25}{3}\implies a=\sqrt{\cfrac{25}{3}}\implies a=\cfrac{\sqrt{25}}{\sqrt{3}}\implies a=\cfrac{5}{\sqrt{3}}
\\\\\\
\textit{and then we can \underline{rationalize the denominator} }
\\\\\\
a=\cfrac{5}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies a=\cfrac{5\sqrt{3}}{\sqrt{3^2}}\implies a=\cfrac{5\sqrt{3}}{3}$