calculuspolar coordinates
I'm learning about polar coordinates. What is the best way to understand $r=4\cos(6\theta)$ without a computer?
I already knew what the graph of $r=\cos(2\theta)$ looks like. Is there a way to get an idea about $r=4\cos(6\theta)$ using this "parent graph"?
What about the graph of $r=4\cos(5\theta)$. Does the fact that $5$ is odd while $6$ is even make a big difference as to the number of "leaves" that appear on the "rose"? Thanks
I'd say that it's convenient to include it in the set of rose curves—it can be useful to think about $r=a\sin n\theta$ for noninteger values of $n$ (even starting at $0$), rather than just integers greater than $1$:
Each one may be solved individually: Polar to cartesian form of $ r = \sin(2\theta)$
$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$ $$r^3 = 2(r\sin\theta)(r\cos\theta)$$
$$x = r\cos\theta$$ $$y = r\sin\theta$$
$$r^3 =2xy$$
$$r = (x^2 + y^2)^{\frac 12}$$
$$(x^2+y^2)^{\frac 32} =2xy$$ $$(x^2+y^2)^3=4x^2y^2$$
See Also: Polar to cartesian form of r=cos(2θ)
Best Answer
We have that $r=\cos ( \theta)$ makes a loop for any interval $\left[-\frac \pi 2 +k\pi ,\frac \pi 2 +k\pi\right]$ but since it leads to negative values for $k$ odd we have that only the loops with $k$ even can be plotted, which is a circle in this case (note that the same plot is obtained if we allow negative values for $r$ since in this case the plots coincide).
Therefore $r=\cos ( 2\theta)$ makes a loop for any interval $\left[-\frac \pi 4 +k\frac \pi 2 ,\frac \pi 4 +k\frac \pi 2\right]$ which correspond to $2$ loops if we restrict to positive values for $r$ and to $4$ loops if we allow negative values also.
So also for $\cos ( 6\theta)$ we have a loop for any interval $\left[-\frac \pi {12} +k\frac \pi 6 ,\frac \pi {12} +k\frac \pi 6\right]$ which correspond to $6$ loops if we restrict to positive values for $r$ and to $12$ loops if we allow negative values also.
For $r=\cos ( 5\theta)$ we have a loop for any interval $\left[-\frac \pi {10} +k\frac \pi {5} ,\frac \pi {10} +k\frac \pi {5}\right]$ which correspond in any case to $5$ loops since the loops for $k$ even ($r>0$) are the same we obtain for $k$ odd ($r<0$).