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in looking up some specific info on polar coordinates, I came across a plot of the polar conversion of $y=sin(6x)+2$ on the wikipedia page. This equation converts to $r=sin(6\theta)+2$. I've been trying to prove this to myself, but can't figure out how to do it.

I begin by writing $r\cdot sin\theta =sin(6\cdot r \cdot cos\theta)$ and get stuck rather quickly. I've tried looking for trig identities that may help me simplify this, but I haven't been able to figure it out. I also tried converting the sines and cosines to complex exponetials using Euler's formula but I still was unable to get to the equation in polar form.

I'm looking for any guidance on how to make this conversion. Thank you.

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    $\begingroup$ Are you sure you have to convert the sine wave into polar so that it maintains the exact same shape. More likely, I think that your professor/teacher asked you to do something like this in Desmos $\endgroup$
    – Toby Mak
    Commented Oct 11, 2017 at 11:59
  • $\begingroup$ Not sure what asking/saying. On the wikipedia pave for polar coordinates, they show this mapping- I was just trying to prove the conversion to myself, but can’t figure it out..... no professor, just me curious with what i saw on en.wikipedia.org/wiki/Polar_coordinate_system $\endgroup$
    – jrive
    Commented Oct 11, 2017 at 12:03
  • $\begingroup$ You can't make this conversion because it isn't the same thing at all. Polar coordinates for $y=\sin (6x)$ should represent the very same graph but in terms of $r$ and $\theta$. What is shown in the wikipedia example is a completely different function, and they point out that this function can be mapped to the $(-\pi,\pi)$ sine. This is likely done to illustrate "the circular nature of the polar coordinate system". $\endgroup$
    – Cure
    Commented Oct 11, 2017 at 12:26
  • $\begingroup$ Thank you, @Cure....I don't completely understand --need to understand what is meant then to be "mapped"....i will look this up... $\endgroup$
    – jrive
    Commented Oct 11, 2017 at 12:29
  • $\begingroup$ r=(x^2+y^2)^0.5, teta= Arc tan (y/x); x=(Arc sin(y-2))/6 and y=Sin( 6x) +2, Is this useful? $\endgroup$
    – sirous
    Commented Oct 11, 2017 at 14:52

1 Answer 1

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Polar and Cartesian coordinates are two coordinate systems for describing the same points - so, for example, the point one unit above the origin is $(0,1)$ in Cartesian coordinates and $(\frac{\pi}{2},1)$ in polar (using the order $(\theta, r)$). Think of this as a different way of talking about the same point.

So when you "convert" an equation from Cartesian into polar, you shouldn't be changing the graph - you're just changing how you talk about the graph. For example, the equation $y = \sqrt{1 - x^2}$ in Cartesian coordinates can be rewritten as $r = 1$, $0 \leq \theta \leq \pi$ in polar. Notice that these equations look nothing alike - they produce the same graph, but the coordinate systems are so different that the equations have to be written very differently.

Likewise, $y = 6\sin{x} + 2$ can't be "converted" into polar by writing $r = 6\sin{\theta} + 2$ - a quick test with any graphing software you like should convince you that $r = 6\sin{\theta} + 2$ makes a big round shape, while $y = 6\sin{x} + 2$ makes a wiggly curve extending infinitely in each direction. Since those are different graph, this isn't a conversion.

But you can think of it as a transformation - the idea is that the act of "replacing" coordinates in one system with coordinates in the other system can be thought of as a function that turns points into other points. For example, the point $(0,1)$ in Cartesian coordinates could be transformed into the point $(0,1)$ in polar coordinates - which would mean $r = 1$ and $\theta = 0$, which would be the point with Cartesian coordinates $(1,0)$. Turning $y = 6\sin{x} + 2$ into $r = 6\sin{\theta} + 2$ is doing exactly this: you're transforming every point on the plane this way, and watching where the ones along your curve wound up.

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