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:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
-
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: \[ y = \sin \theta (\sin \theta + 1) \]
-
-
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: \[ \sin \theta = \frac{y}{r} \] \[ \cos \theta = \frac{x}{r} \]
- 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: \( r = \sin \theta + 1 \).