Graphing with polar coordinates Problem

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SUMMARY

The discussion focuses on graphing the polar equation r = 1/2 + cos(θ), which represents a Limacon. Key points include the maximum radius of 3/2 at θ = 0 and a minimum radius of -1/2 at θ = π. The curve exhibits symmetry about the x-axis, and specific radius values were calculated for angles from 0 to 180 degrees. The main challenge discussed is how to represent negative radius values, particularly at 120 degrees, by flipping them to positive values at corresponding angles.

PREREQUISITES
  • Understanding of polar coordinates and their graphical representation
  • Familiarity with Limacon curves and their properties
  • Basic knowledge of trigonometric functions, specifically cosine
  • Ability to interpret and plot mathematical equations
NEXT STEPS
  • Explore the properties of Limacon curves in detail
  • Learn about graphing techniques for polar equations
  • Study the effects of negative radius values in polar coordinates
  • Investigate symmetry in polar graphs and its implications
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Students studying mathematics, particularly those focused on polar coordinates and graphing techniques, as well as educators teaching these concepts in a classroom setting.

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Homework Statement



Draw the graph of r = 1/2 + cos(theta)


Homework Equations



The equation is itself given in the question. It is a Limacon.


The Attempt at a Solution



Step-1 ---> Max. value of r is 1/2 + 1 = 3/2 [ at cos (0) ]
Min. value of r is 1/2 - 1 = -1/2 [ at cos (Pi) ]

Step-2 ---> Symmetry The curve is symmetrical about x-axis.

Step-3 ---> I calculated all the values of "r" from 0 to 180 degrees. These are 1.5, 1.336, 1.207, 1, 0.5, 0, -0.207, -0.336, -0.5. These are in standard form like 0, 30, 45... and so on.

Step-4 ---> Now I draw the graph. Till 90 degrees its all right; but at 120 degrees the problem comes. How can I draw show 0 as well as 120 degrees? How can I show (-0.5, 180 degrees) ?
 
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Pretend you are drawing 0.5 at 180 degree, but since the radius is negative, flip it back to positive 0.5 at 0 degrees.
 
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Okay, thanks a lot.
 

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