Expert: Paul Klarreich - 2/22/2007

Question

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The text above is a follow-up to ...

-----Question-----
Hi again
Im now seeing the problem a little differently. Is it correct? it is a little different from the way you saw it.

Im seeing that the circle of equation:
r = acos@ is a circle with diameter equal to a that is a radius of a/2, centered at (a/2, 0). And the second circle of diameter a rolls on this circle.

So am i seeing what you are seeing differently? or are you seeing what i am seeing?
enlighten me please...
-----Answer-----
Hi, Vaulter,

Yes, my view was different.  I set things up with the 'inner' circle having radius a (yours has diameter a) and center at the origin (yours has center at (a/2,0) ).  

Then its polar equation is just r=a.  But I think the end result will be the same.  And you are certainly within your rights to object to the word 'cardioidish'; I don't love it, either.  

If the two circles are set up your way, the calculation should come out similarly, but with a different final polar equation.  The center of the cardioid would be at the origin (pole) instead of at (-a,0).

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follow up:

In your example: Did you take vector BP as Vetor BC plus vector CP?
What would be the vector CP?


Answer
Hi, Vaulter,

In your example: Did you take vector BP as Vetor BC plus vector CP?

>> Well, BP is equal to BC + CP, but that is not what I did.

What would be the vector CP?

In the diagram, OC is the line of centers, and CP is the radius from C, the center of the rolling circle, to point P.  Its length is a.  Now  OP, which is the vector we have to compute, is equal to OC + CP.

HOWEVER, I found the computation easier when I marked B as the perpendicular from P to OC, and then computed:

OP = OB + BP.

It is difficult to send a diagram through this site, I must agree.