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

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Homework Help Overview

The discussion revolves around determining key angles for a three-leaf rose represented in polar coordinates, specifically using the equation r = 2cos(3θ). Participants are exploring how to find specific angle intervals related to the petals of the rose and the behavior of the function.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand how to derive the angles θ = π/6, 3π/6, 5π/6, 7π/6, 9π/6, and 11π/6 from the polar equation. Some suggest considering the unit circle and the properties of trigonometric functions to clarify the intervals.

Discussion Status

There is an ongoing exploration of the relationship between the angles and the graph of the polar function. Some participants have suggested graphing the function to visualize the petals better, while others are questioning the specific values and their derivation. No consensus has been reached, but various interpretations and methods are being discussed.

Contextual Notes

Participants have noted potential confusion regarding the correct interpretation of the equation and its implications for the graph. There are references to the need for clarity on the angles at which the function passes through the origin and how these relate to the overall shape of the rose.

mateomy
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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:
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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) ?
 

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