Using polar coordinates to describe rose petals

[Worn_Out_Tools]

I encountered a question which asked me to describe the rose petal sketched below in polar coordinates. The complete answer is
R = {(r, θ): 0 ≤ r ≤ 6 cos(3θ), 0 ≤ θ ≤ π}. That makes sense to me for the right petal. What about the other two on the left?

 

[berkeman]

I encountered a question which asked me to describe the rose petal sketched below in polar coordinates. The complete answer is
R = {(r, θ): 0 ≤ r ≤ 6 cos(3θ), 0 ≤ θ ≤ π}. That makes sense to me for the right petal. What about the other two on the left?

View attachment 267067

Welcome to the PF.

What is different about the other two petals in polar coordinates?

 

[etotheipi]

$r(\theta + \frac{2\pi}{3}) = r(\theta)$ since $r$ is periodic at $\frac{2\pi}{3}$, so once you have drawn one petal you can insert the rest

 

[Worn_Out_Tools]

$r(\theta + \frac{2\pi}{3}) = r(\theta)$ since $r$ is periodic at $\frac{2\pi}{3}$, so once you have drawn one petal you can insert the rest

Wait, I don’t understand, how’d you determine that?

 

[berkeman]

When you phase shift a sine wave in rectangular coordinates, how do you adjust the arguments to the sine function?

https://mathbitsnotebook.com/Algebra2/TrigGraphs/phaseabc.jpg

So the analogous adjustment of the arguments in polar coordinates would look like what?

 

[etotheipi]

Wait, I don’t understand, how’d you determine that?

There are two ways you can think about it. Consider the function $r = 6\cos{(3\theta)}$. If we consider some angle $\theta_0$ where $r(\theta_0) = r_0$, then we see that$$r \left(\theta_0 + \frac{2\pi}{3} \right) = 6\cos{\left(3(\theta_0 + \frac{2\pi}{3})\right)} = 6\cos{(3\theta_0 + 2\pi)} = 6\cos{(3\theta_0)} = r(\theta_0) = r_0$$which means that both $\theta_0$ and $\theta_0 + \frac{2\pi}{3}$ correspond to equal radial coordinates. The other way to look at it is what I think @berkeman was suggesting, i.e. consider transforming the original function by a constant translation,$$r_2(\theta) := r(\theta + \frac{2\pi}{3})$$so $r_2$ is shifted left of $r$ by the amount $\Delta \theta = -\frac{2\pi}{3}$. But the periods of both $r$ and $r_2$ are equal to $\frac{2\pi}{3}$, so we've just shifted the waveform left an entire period. Since $r$ and $r_2$ are then equal everywhere, we have$$r(\theta) = r_2(\theta) = r(\theta + \frac{2\pi}{3})$$And finally, what amounts to a horizontal shift in $r$ vs $\theta$ on Cartesian axes corresponds to a rotation in polar coordinates!