Polar Plots and square roots

When I plot a polar plot of $r=\sin (3 \theta)$, and $r=\sqrt{\sin (3 \theta)}$ I get nearly identical graphs, both $3$ pedal rose type plots. In the case without the square root, it is easy to understand the plot. However, for the plot involving the square root of $\sin 3 \theta$, it is strange to me how the graph would handle thetas for which the $\sin$ of the $3 \theta$ is negative. It would seem that the negative values inside the square root should cause the 2 graphs to be dissimilar, yet it appears this is not the case.

Example: For $\theta = 65^\circ, r=\sin (3 \theta) = -0.259$; I would expect this it be an issue with the graph of $r= \sqrt{\sin (3 \theta)}$. Enclosed are images of the graphs of the 2 polar plots. URL for 2 graphs --> https://s3.amazonaws.com/grapher/exports/gtwzzdokst.png

Solution 1:

These images might give you a better sense of things. Very often, the nature of a parametric plot is best understood as a path drawn as the parameter changes, instead of just a static set of points that satisfy a relation.

You'll notice that the $r = \sin 3\theta$ curve draws smoothly, actually plotting "backwards" through the pole for values of $\theta$ (for instance, $90^\circ$) when $r$ is negative.

On the other hand, the $r = \sqrt{\sin 3\theta}$ curve "pauses" as $\theta$ sweeps through values that make the radicand negative. (The red flashing is Mathematica expressing its frustration at being asked to plot imaginary values. Normally, I'd suppress that; however, here, it's actually directly relevant to the discussion, so I left it.)

Solution 2:

Notice that if you plug in $r=|\sin\:(3θ)|$ instead of $r=\sin\:(3θ)$ you get a six-petal graph. The concept of a negative radius doesn't make much sense, so in reality the negative results of $r=\sin\:(3θ)$ are just not being drawn either.