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I'm having a bit of trouble figuring out what the polar plot of the following equation would look like: $$r^{2} = 16\sin 3\theta.$$

1.) One way to interpret the above is to rewrite it as $r = 4\sqrt{\sin 3\theta}$ and $r = -4\sqrt{\sin 3\theta}$, thus yielding a graph that is very similar to a 6-leaf rose.

2.) The other way to interpret this is to rewrite the graph in rectangular form, resulting in the equation $$(x^{2}+y^{2})^{3}=3x^{2}y - y^{3}$$, which yields a 3-leaf rose.

I was wondering which is correct, and why?

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    Isn't the rectangular form from $r^2r^3=r^3(16(3(\cos{\theta})^2\sin{\theta}-(\sin{\theta})^3))$ the 6-leaf rose: $$(x^2+y^2)^5=(16(3x^2y-y^3))^2.$$ – 2017-01-31

1 Answers 1

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The curve is defined by the equation

$$r^2=16\sin(3\theta).$$

when taking the square root, the OP writes:

$$r=\pm 4\sqrt{\sin(3\theta)}.$$

This seems to be correct since both of the signs give the same square. However, according to the definition of the polar coordinates

The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

So, $r$ being a distance, is always positive. Negative distances are meaningless...

Now, this is how I drew the curve:

By the definition of the trig functions we have

$$x(\theta)=4\sqrt{\sin(3\theta)}\cos(\theta)\ \text{ and } y(\theta)=4\sqrt{\sin(3\theta)}\sin(\theta).$$

This is a parametric description of the curve ($0\leq\theta\leq2\pi$). During our manipulations we did not do anithing wrong, anything that would have changed the shape.

And here is the shape:

enter image description here

So the true shape is of three leaves.

Taking into account the negative distances we would get three more leaves: the mirror images (through the origin) of the existing leaves.