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Is there a shorter way to get the answer, the polar equation of a cardioid, directly?

My solution involves some tedious work and doesn't give the polar equation directly:

The equation of the member curves (circles) is

##f(x, y, \phi)=(x-a\cos^2\phi)^2+(y-a\cos\phi\sin\phi)^2-a^2\cos^2\phi=0##. --- (1)

To find the envelope curve, we impose ##\frac{\partial f}{\partial\phi}=0##. Thus, we have

##2\sin\phi\cos\phi u-(cos^2\phi-sin^2\phi)v+\sin\phi\cos\phi a=0\,\,\,##, where ##u=x-a\cos^2\phi\,\,\,##and ##\,\,\,v=y-a\cos\phi\sin\phi##.

Making ##v## the subject and then substituting into equation (1), we have (after some tedious work)

##x=2a\cos^2\phi\cos2\phi##.

Making ##u## the subject and then substituting into equation (1), we have (after some tedious work)

##y=2a\cos^2\phi\sin2\phi##.

My solution involves some tedious work and doesn't give the polar equation directly:

The equation of the member curves (circles) is

##f(x, y, \phi)=(x-a\cos^2\phi)^2+(y-a\cos\phi\sin\phi)^2-a^2\cos^2\phi=0##. --- (1)

To find the envelope curve, we impose ##\frac{\partial f}{\partial\phi}=0##. Thus, we have

##2\sin\phi\cos\phi u-(cos^2\phi-sin^2\phi)v+\sin\phi\cos\phi a=0\,\,\,##, where ##u=x-a\cos^2\phi\,\,\,##and ##\,\,\,v=y-a\cos\phi\sin\phi##.

Making ##v## the subject and then substituting into equation (1), we have (after some tedious work)

##x=2a\cos^2\phi\cos2\phi##.

Making ##u## the subject and then substituting into equation (1), we have (after some tedious work)

##y=2a\cos^2\phi\sin2\phi##.

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