How Do You Determine the Key Angles for a Three-Leaf Rose in Polar Coordinates?

Click For Summary
SUMMARY

The discussion focuses on determining the key angles for a three-leaf rose represented by the polar equation r = 2cos(3θ). Participants clarify that the angles θ = π/6, 3π/6, 5π/6, 7π/6, 9π/6, and 11π/6 correspond to the points where the petals of the rose intersect the origin. The importance of understanding the unit circle and the behavior of the cosine function at these angles is emphasized, particularly in relation to the periodic nature of the function. Graphing the function is recommended for better visualization of the petals and their intervals.

PREREQUISITES
  • Understanding of polar coordinates and their graphical representation
  • Knowledge of trigonometric functions, specifically cosine
  • Familiarity with the unit circle and angle measurement in radians
  • Ability to graph polar equations and interpret their shapes
NEXT STEPS
  • Learn how to graph polar equations using software like Desmos or GeoGebra
  • Study the properties of the cosine function and its periodicity
  • Explore the derivation of polar coordinates from Cartesian coordinates
  • Investigate the area calculation of polar curves using integration techniques
USEFUL FOR

Students studying trigonometry, educators teaching polar coordinates, and anyone interested in graphing polar equations and understanding their properties.

mateomy
Messages
305
Reaction score
0
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

<br /> r=2cos3\theta<br />

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.
 
Last edited:
Physics news on Phys.org
mateomy said:
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

<br /> r=2cos\theta<br />

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?)
 
Did you see my edit on the LaTex? its actually 3theta not just theta.
 
mateomy said:
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)
 
Last edited by a moderator:
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
 
Ray Vickson said:
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.
 
mateomy said:
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) ?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
22
Views
4K