Question

Which is a polar form of the following parametric equations?
x equals 4 sine squared theta
y = 4 sin θ cos θ
(1 point)
Responses

lower r equals one-fourth sine theta cosine theta
Image with alt text: lower r equals one-fourth sine theta cosine theta

r = 2
r = 2

lower r equals 16 sine squared theta
Image with alt text: lower r equals 16 sine squared theta

r = 4 sin θ

Answer 1

To find the polar form of the given parametric equations \( x = 4 \sin^2 \theta \) and \( y = 4 \sin \theta \cos \theta \), we can express them in terms of \( r \) and \( \theta \).

1. Recall the relationships between Cartesian and polar coordinates:
\[
x = r \cos \theta, \quad y = r \sin \theta
\]

2. Rewrite \( x \) and \( y \) in terms of polar coordinates:
\[
x = 4 \sin^2 \theta
\]
and
\[
y = 4 \sin \theta \cos \theta
\]

3. We can also use the double angle identity for sine:
\[
y = 4 \sin \theta \cos \theta = 2 \sin(2\theta)
\]

4. Now, we can find \( r \) in terms of \( \theta \):
From the equation \( y = 2 \sin(2\theta) \), we can express \( r \sin \theta \):
\[
r \sin \theta = 2 \sin(2\theta) = 4 \sin \theta \cos \theta
\]
This gives us:
\[
r = 4 \cos \theta
\]

5. Next, for \( x = 4 \sin^2 \theta \):
We can express \( \sin^2 \theta \) using \( \sin \theta = \frac{y}{r} \):
\[
x = 4 \left( \frac{y}{r} \right)^2
\]

6. From these transformations, we derive \( r \) in terms of \( y \) or \( x \), but \( r = 4 \sin \theta \) simplifies to \( r = 4 \sin \theta \).

Thus, the polar form of the given equation is:
\[
r = 4 \sin \theta
\]

So the correct response is: **\( r = 4 \sin \theta \)**.