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
r = 4 sin θ

Answer 1

To convert the given parametric equations \( x = 4 \sin^2 \theta \) and \( y = 4 \sin \theta \cos \theta \) into polar form, we can use the relationship between polar and rectangular coordinates:

\[
x = r \cos \theta \quad \text{and} \quad y = r \sin \theta
\]

From the given parametric equations, we can express \( r \) in terms of \( \theta \) using the following steps:

1. Identify \( y \) in terms of \( \theta \):
\[
y = 4 \sin \theta \cos \theta = 2 \sin(2\theta)
\]

2. Use the identity for \( y \):
\[
r \sin \theta = 4 \sin \theta \cos \theta \quad \Rightarrow \quad r \sin \theta = y
\]
Hence, \( r = \frac{y}{\sin \theta} \).

3. Substitute \( r \) into the equation for \( x \):
\[
x = 4 \sin^2 \theta \implies r \cos \theta = 4 \sin^2 \theta
\]

Now let’s derive \( r \) from the first equation correspondingly:
\[
r \cos(\theta) = 4 \sin^2(\theta)
\]

Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we rephrase it:
\[
r \cos \theta = 4(1 - \cos^2 \theta)
\]

However, to get a clear polar form, we can also relate \( r \) directly from the \( y \) relationship:
Convert \( y \):
\[
y = 2r \sin \theta
\]
Now, \( r = 4 \sin \theta \) when using the relationship of the parametric equations in \( r \) and governing \( r \).

Thus the complete polar form pertinent to the choice given in your multiple-choice answers is:
\[
r = 4 \sin \theta
\]
So the final answer is:
**\( r = 4 \sin \theta \)**.