Spring '99
 Math 50c

Generated using Matlab (Cartesian coordinates)

Description

        The Cardioid (the root cardi: Greek for heart) is the trace of a point on a circle rolling around the circumference of another circle with equal radius.


History:

        The cardioid was developed by the Danish astronomer Ole Christensen Roemer in 1674.  It was discovered during an investigation to find the best design for gear teeth.  The cardioid was given its name by de Castillon in the Philosophical Transactions of the Royal Society of 1741.  The arc length of the cardioid was discovered by La Hire in 1708.
        Since the cardioid is also an epicycloid and a special case of a Limacon of Pascal, it is believed that it could have originated from Etiene Pascal's studies (1588-1640).

Parametric Equation:
x = acos(t)(1+cos(t))
y = asin(t)(1+cos(t))

Polar Equation:   r = 2a(1 + cos(q))

Cartesian Equation:  (x² + y² - 2ax)² = 4a(x² + y²)

Graphed using Matlab (in Polar coordinate system)

Deriving the Parametric Equation:

Method One:

    Circles c1 and c2 have radius R.  The center of circle c1 is located at (R, 0).  Center c2 is located a distance of 2R from center c1.

Finding the the parametric equations:
    Using what we know so far,

x = L1(x) + L2(x) + L3(x)
y = L2(y) + L3(y)

    L1(x) is equal to the radius (R).  Now,  form a right triangle by sending a perpendicular line from the x-axis connecting to center c2.  The hypotenuse equals twice the radius (2R) and the angle of the triangle is a1.  The horizontal distance L2(x) is given by (2R)cos(a1), the vertical distance L2(y) is given by (2R)sin(a1).

x = R + ((2R)cos(a1)) + L3(x)
 y =  ((2R)sin(a1)) + L3(y)

   To find these last two measurements L3(x) and L3(y), another right triangle is formed with the adjacent leg parallel to the x-axis and the opposite leg parallel to the y-axis.  The angle measured for this triangle is the sum of the angles a1 and a2.  The hypotenuse is equal to the radius of the circle (R).  The horizontal distance L3(x) is given by (R)cos(a1 + a2), the vertical distance L3(y) is given by (R)sin(a1+a2).

x = R + ((2R)cos(a1)) + (R)cos(a1 + a2)
 y = ((2R)sin(a1)) + (R)sin(a1+a2)

    Because the arc length traveled by the two points is equal and the radius is equal, the angles a1 and a2 are equal. We will substitute a 't' in for a1 and a2.  So the new equations are:

x = R + (2R)cos(t) + (R)cos(2t)
 y = (2R)sin(t) + (R)sin(2t)
    First pull R out.  Then using the double angle formulas simplify the equations.
x = R(1 + 2cos(t) + cos(2t))
y = R(2sin(t)+sin(2t))
cos(2t) = 2cos²(t) - 1, sin(2t) = 2sin(t)cos(t)
x = R(1 + 2cos(t) + 2cos²(t) - 1)
y = R(2sin(t)+2sin(t)cos(t))
And in final parametric form:
x = 2Rcos(t)(1 + cos(t))
y = 2Rsin(t)(1 + cos(t))

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Method Two:  This is a method I discovered while playing with Geometer's Sketch Pad, and evidently I was able to prove that it is a viable process.

    The Center of circle c1 is located at (-R, 0) on the cartesian axes.  The radius of c1 is 2R.  The center of circle c2 is located at (2R, 0).  Circle c2 has a radius of R.  Point p1 is located at point (2Rcosq-R, 2Rsinq).  Circle c1 has double the radius of c2.  Because of this, point p2 is going to travel around circle c2 at twice the frequency of p1.  Therefore point p2 is located at (Rcos(2q)+2R, Rsin(2q)).  Now to find the coordinates of point M, which traces out the cardioid.  Point M is the midpoint of p1 and p2.

And now in Parametric form:

Which varies from the the other method by only a constant of 2.  So in other words, this cardioid is half the size of the cardiod from method, assuming R is the same value for both methods.

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Sources:

CRC Concise Encyclopedia Of Mathematics
http://www.astro.virginia.edu/%7Eeww6n/math/Cardioid.html

MacTutor Famous Curve Index
http://www-groups.dcs.st-andrews.ac.uk/%7Ehistory/Curves/Cardioid.html

Xah: Special Plane Curves
http://www.best.com/~xah/SpecialPlaneCurves_dir/Cardioid_dir/cardioid.html

Peter Gent: Cardioid
http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Sp98/PeterG/Cardiod.htm