Question

Which polar equation is equivalent to the parametric equations below?

1st row x equals sine theta cosine theta plus cosine theta 2nd row y equals sine squared theta plus sine theta
(1 point)
Responses

r = cos θ + 1
r = cos θ + 1

r = sin θ + 1
r = sin θ + 1

r equals sine squared theta plus 1
Image with alt text: r equals sine squared theta plus 1

r equals 1 minus cosine theta

Answer 1

To find a polar equation equivalent to the given parametric equations \( x = \sin \theta \cos \theta + \cos \theta \) and \( y = \sin^2 \theta + \sin \theta \), we first need to express \( x \) and \( y \) in terms of \( r \) and \( \theta \).

In polar coordinates, we have:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)

1. **Analyzing the parametric equations:**
- The first equation is:
\[
x = \sin \theta \cos \theta + \cos \theta
\]
We can factor out \( \cos \theta \):
\[
x = \cos \theta (\sin \theta + 1)
\]

- The second equation is:
\[
y = \sin^2 \theta + \sin \theta
\]
This can be factored as:
\[
y = \sin \theta (\sin \theta + 1)
\]

2. **Relating to polar coordinates:**
From the polar angle definitions:
- \( r = \sqrt{x^2 + y^2} \)
- To express \( \sin \theta \) and \( \cos \theta \) in terms of \( r \) and \( \theta \), we use:
\[
\sin \theta = \frac{y}{r}
\]
\[
\cos \theta = \frac{x}{r}
\]

3. **Substituting and simplifying:**
Now we substitute \( \sin \theta \) and \( \cos \theta \) into the parametric equations.

Using the first parametric equation:
\[
x = \cos \theta (\sin \theta + 1) \Rightarrow x = \frac{x}{r}\left(\frac{y}{r} + 1\right)
\]

Multiply through by \( r \) (to eliminate the denominator):
\[
xr = x \left(\frac{y + r}{r}\right)
\]
This results in:
\[
xr = xy + x
\]

Next, rearranging gives:
\[
xr - xy - x = 0 \Rightarrow x(r - y - 1) = 0
\]

From the second parametric equation:
\[
y = \sin \theta (\sin \theta + 1) \Rightarrow y = \frac{y}{r}\left(\frac{y}{r} + 1\right)
\]

Multiply through by \( r^2 \) (collect terms and isolate):
The polar equation can be found through combinations. In polar angles, checking each order:

Going through the options, we substitute back to the original basis which lead to geometry similar understanding.

From original calculations, searching through potential values surfaces:
- Working through the options simplifies to assess arcs derived.

The equivalent polar equation that matches the original parameterizations simplifies back directly through testing:
\[
r = \sin\theta + 1
\]

Thus, the answer is:
- \( r = \sin \theta + 1 \) (this reflects the proper relocation alongside sine derivational effects especially reflected geometrically).

So the final equivalent polar equation is:

**Response: \( r = \sin \theta + 1 \)**.