Finding the petal angles for r = 2cos(3θ)

[mateomy]

I don't remember ANYTHING from this section when I took Trig but we're finding the area of curves in the polar coords. Looking at the book they give us this equation

$$r=2cos3\theta$$

I can see, and I know how to figure out its a 3 leaf "rose" symmetrical about the theta= zero axis, but I can't figure out the next part which is the author stating "Finding the intervals we see that $\theta$=pi/6, 3pi/6, 5pi/6, 7pi/6, 9pi/6, and 11pi/6." Maybe (probably) its simple trig stuff that I am overlooking but how the H do they find those values?

Pointers, suggestions, and/or degrading comments will be greatly appreciated. Thanks.

 

[mege]

I don't remember ANYTHING from this section when I took Trig but we're finding the area of curves in the polar coords. Looking at the book they give us this equation

$$r=2cos\theta$$

I can see, and I know how to figure out its a 3 leaf "rose" symmetrical about the theta= zero axis, but I can't figure out the next part which is the author stating "Finding the intervals we see that $\theta$=pi/6, 3pi/6, 5pi/6, 7pi/6, 9pi/6, and 11pi/6." Maybe (probably) its simple trig stuff that I am overlooking but how the H do they find those values?

Pointers, suggestions, and/or degrading comments will be greatly appreciated. Thanks.

You may want to double check what the graph looks like. This isn't a 3-petal 'rose'.

Also, think back to your unit circle and how it relates to theta - all of the thetas they give should come out relatively nice. (Maybe think about them in their degree equivalents?)

 

[mateomy]

Did you see my edit on the LaTex? its actually 3theta not just theta.

 

[mege]

Did you see my edit on the LaTex? its actually 3theta not just theta.

Trickery!

I suggest graphing the 2sin3$\theta$ as an x-y plot. Using the graph you're more used to seeing, you should be able to identify how r varies a little better. For example: whenever 2sin3x = 0, 'drawing' your petal should be at the origin.

I think they're looking for 'intervals' where you have a continuous petal (since when $\theta = 0$, you're r = 2 and doesn't really 'close' a loop over the next $\frac{\pi}{3}$).


(Still learning latex myself, sorry :p)

 

[Ray Vickson]

For which values of t (t = theta) is cos(3*t) = 0? These would be the t-values at which the figure passes through the origin (if that is what you are after).

RGV

 

[mateomy]

For which values of t (t = theta) is cos(3*t) = 0? These would be the t-values at which the figure passes through the origin (if that is what you are after).

RGV

I know cos(theta) will equal zero and pi/2 and all of its multiples, but I can't figure out how theyre getting the pi/6. cos(3theta) will be zero at 3pi/2 right? I always hated trig graphing.

 

[mege]

I know cos(theta) will equal zero and pi/2 and all of its multiples, but I can't figure out how theyre getting the pi/6. cos(3theta) will be zero at 3pi/2 right? I always hated trig graphing.

What's 3(pi/6) ?