Let's examine the two polar equations and analyze how to sketch them.

(ii) $r = 6 \sin \theta$

This is the polar equation of a limacon with a sine term. Here's how it works:

• When $\theta = 0$, $r = 6 \sin(0) = 0$, so the curve passes through the origin.
• When $\theta = \frac{\pi}{2}$, $r = 6 \sin\left(\frac{\pi}{2}\right) = 6$, so the point $(r, \theta) = (6, \frac{\pi}{2})$ is plotted on the positive y-axis.
• When $\theta = \pi$, $r = 6 \sin(\pi) = 0$, so it returns to the origin.
• When $\theta = \frac{3\pi}{2}$, $r = 6 \sin\left(\frac{3\pi}{2}\right) = -6$, which indicates a point in the negative y-direction.

This gives a limacon shape without an inner loop, symmetric with respect to the vertical axis.

(iii) $r = -1 - \cos \theta$

This is also a polar equation, and it generates a limacon with a negative $r$ and cosine term. Here's the interpretation:

• When $\theta = 0$, $r = -1 - \cos(0) = -2$, so the point is 2 units to the left of the origin.
• When $\theta = \frac{\pi}{2}$, $r = -1 - \cos\left(\frac{\pi}{2}\right) = -1$, the point is 1 unit below the origin.
• When $\theta = \pi$, $r = -1 - \cos(\pi) = 0$, meaning the curve passes through the origin.
• When $\theta = \frac{3\pi}{2}$, $r = -1 - \cos\left(\frac{3\pi}{2}\right) = -1$, a point below the origin.

This results in a limacon with an inner loop, symmetric with respect to the horizontal axis.

Would you like me to generate an exact sketch of these curves for you?

Here are 5 related questions:

1. How does the graph of $r = a \sin \theta$ differ from $r = a \cos \theta$?
2. What happens to the curve when you adjust the coefficient in $r = k \sin \theta$?
3. How do we determine the symmetry of a polar curve based on the equation?
4. What effect does a negative coefficient have in polar equations like $r = -a - b \cos \theta$?
5. Can you explain how to convert a polar equation to Cartesian coordinates?

Tip: When plotting polar curves, always start by determining key points for special angles like $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ to understand the general shape of the graph.