Expert: Paul Klarreich - 2/19/2007

Question
What will be the trajectory of a point on a circle of diameter   a
rolling on another circle defined by r=a*cos(theta). I guess its going to
be something liki heart... but how?????!!!!
it is something like r= a(1 plus cos(theta))  

Answer
Questioner:   Vaulter
Category:  Advanced Math
 
Subject:  Cycloids
Question:  What will be the trajectory of a point on a circle of diameter   a rolling on another circle defined by r=a*cos(theta). I guess its going to
be something like heart... but how?????!!!!
it is something like r= a(1 plus cos(theta))
.....................................
Hi, Vaulter,

Actually, it's not a cycloid, but an epicycloid.  I don't know where you got the idea that it's a cycloid.

Oh.  

You got it from me.  Sorry about that.

Anyway, this one does work out to be cardioidish.  There is a lot of work involved, so I will just outline the steps and indicate the result.

Construct the following figure:

Draw a circle with center at the origin, O, and radius a.
Draw a second circle with center at (2a,0) and radius a.  
This circle touches the first circle at A(a,0), which we also call P.  [I know, two names for the same point, but P moves and A remains fixed.]
This is the starting position for P.

Now construct a new diagram.  

In it, you let the second circle 'roll' around the first one by some small amount.  
Leave it in the first quadrant.
Mark the center of the second circle C, and draw the line of centers, OC, going through the contact point, R.
Note that the arc RA has been 'rolled' around.  Mark P on the second circle such that the arc RP is the same length as the arc RA.

Now the angle COA is to be labeled theta. [@]
Angle OCP is also theta. Some stuff about central angles and arcs.

From  P, draw the perpendicular to OC, meeting it at B.  Draw CP, forming right triangle CBP.

Obtain the coordinates of point B by considering the vector OB and its components.  

You will find that the length of BC = a cos @, and OB = 2a - BC = 2a - a cos @

So the coordinates of B are:

x1 = (2a - a cos @) cos @
y1 = (2a - a cos @) sin @

Now look at BP as a vector.  Its length is a sin @.  Draw a separate diagram to find its components.  You will get:

x2 = a sin @ cos (@ - pi/2)
y2 = a sin @ sin (@ - pi/2)

and

x2 =  a sin @ sin @
y2 = -a sin @ cos @

Now the coordinates of P (that's what we really want) are:

x = x1 + x2
 = 2a cos @ - a cos^2 @ + a sin^2 @
 = 2a cos @ - a cos(2@)

y = y1 + y2
 = 2a sin @ - a sin @ cos @ - a sin @ cos @
 = 2a sin @ - a sin(2@)

Now those are parametric equations.  But you want a polar equation?  Here's how you get it:

Compute  x^2 + y^2, which is r^2.  Do some good algebra, then use some trigonometric identities, and when the smoke clears, you have:

r^2 = a^2(5 - 4 cos @), and

r   = a sqrt(5 - 4 cos @)

Indeed, that is a cardioid.