Question

What does the graph `r = sqrt(sintheta)` look like in plane polar coordinates? How do you graph it?

Answer 1

It should look like this:

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In polar coordinates, you have a radius `r` that is a function of `theta`, and an angle `theta` from the righthand horizontal.

So, to plot the graph, measure the angle from the righthand horizontal, and acquire the radius at that angle; that's one point on the graph. This function is valid in `[0,180^@]` since `sintheta` is only positive for `sin0^@` through `sin180^@`.

You can get an idea for how the value of `r` changes with `theta` by calculating each value at, say, `45^@` increments, to find that it's something like a semicircle. If you use Excel in `1^@` increments, it gives:

assuming `r` is only vertical.

But that's not how the graph actually looks; in actual polar coordinates, `r` is not vertical, but radial.

So, take the above graph, which plots vertical `r`, and vary the angle of `r = r(theta)` so that `r` is radial. Also, take `90^@` on the above graph as your new origin of `(0,0)`.

In other words, take your finger and use the origin as an axis of rotation. Then, sweep through the first two quadrants (I,II) from the righthand horizontal to the lefthand horizontal, which traces `0^@ -> 180^@`, and follow the way `r` changes based on the above graph.

This distorts the above graph so that the endpoints are at `(0,0)`, and we move in a bit more of a circular fashion.

Here is this GIF to illustrate what is happening with `r` as a function of `theta`:

And so, the resultant graph looks like a squashed bouncing ball in slow motion: