Cardioid/Proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article Cardioid. This article is currently an experimental vehicle to see how well we can provide proofs and details for a math article without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for some current discussion. This article is "experimental" in the sense that it is a test of one way we may be able to incorporate more detailed proofs in Wikipedia.

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[edit] Theorem

The curve defined by the parametric equations

x(t) = 2 r \left( \cos t -{1 \over 2} \cos 2 t \right), \qquad \qquad (1)
y(t) = 2 r \left( \sin t - {1 \over 2} \sin 2 t \right) \qquad \qquad (2)

has the same shape as the curve defined in polar coordinates by the equation

\rho(\theta) = 2r(1 - \cos \theta). \

[edit] Proof

Starting from ρ(θ) = 2r(1 − cosθ), and using the polar to cartesian formulas

x = \rho(\theta) \cos(\theta) \,
y = \rho(\theta) \sin(\theta) \,

and double angle formulas we get the cartesian parametric equations:

x = 2r (1- \cos \theta) \cos \theta = 2r (\cos \theta - \cos^2 \theta) = 2r \left( \cos \theta - \frac{1+\cos 2\theta}{2} \right) = 2r \left( \cos \theta - \frac{1}{2} \cos 2 \theta \right) -r \,
y = 2r (1- \cos \theta) \sin \theta = 2r(\sin \theta - \sin \theta \cos \theta) = 2r \left( \sin \theta - \frac{1}{2} \sin 2\theta \right)\,

Simply replacing θ with t yields equations (1) and (2), with a shift to the left by r.

[edit] Another proof

Equations (1) and (2) define a cardioid whose cuspidal point is (r, 0). To convert to polar, the cusp should preferably be at the origin, so subtract r from the abscissa. Replacing t by θ yields

x(\theta) = 2r \left( -{1 \over 2} + \cos \theta - {1 \over 2} \cos 2 \theta \right) ,
y(\theta) = 2r \left( \sin \theta - {1 \over 2} \sin 2 \theta \right) .

The polar radius ρ(θ) is given by

\rho(\theta) = \sqrt{x^2(\theta) + y^2(\theta)}
= 2r \sqrt{\left( -{1 \over 2} + \cos \theta - {1 \over 2} \cos 2 \theta \right)^2 + \left( \sin \theta - {1 \over 2} \sin 2 \theta \right)^2 }.

Expanding this yields

\rho = 2r \sqrt{ {1 \over 4} + \cos^2 \theta + {1 \over 4} \cos^2 2 \theta - \cos \theta + {1 \over 2} \cos 2 \theta - \cos \theta \cos 2 \theta + \sin^2 \theta + {1 \over 4} \sin^2 2 \theta - \sin \theta \sin 2 \theta}.

We can simplify this by noticing that

\cos^2 \theta + \sin^2 \theta = 1, \qquad \qquad \mbox{(trigonometric identity)}
{1 \over 4} \cos^2 2 \theta + {1 \over 4} \sin^2 2 \theta = {1 \over 4}, \qquad \qquad \mbox{(variation of the above)}

and

\cos \theta \cos 2 \theta + \sin \theta \sin 2 \theta = \cos (\theta - 2 \theta) = \cos -\theta = \cos \theta. \

Thus,

\rho = 2r \sqrt{ {1 \over 4} + 1 + {1 \over 4} - 2 \cos \theta + {1 \over 2} \cos 2 \theta }
= 2r \sqrt{ {3 \over 2} - {4 \over 2} \cos \theta + {1 \over 2} \cos 2 \theta }
= 2r \sqrt{ {3 - 4 \cos \theta + \cos 2 \theta \over 2}}.

Then, since

\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1, \qquad \qquad \mbox{(trigonometric identity)}

it follows that

\rho = 2r \sqrt{ {3 - 4 \cos \theta + 2 \cos^2 \theta - 1 \over 2}} = 2r \sqrt{ {2 - 4 \cos \theta + 2 \cos^2 \theta \over 2}},
\rho = 2r \sqrt{ 1 - 2 \cos \theta + \cos^2 \theta} = 2r(1 - \cos \theta).

[edit] Area derivation

The objective is to integrate the area of the cardioid whose equation in polar coordinates is

r = 1 - \cos \theta . \,\!

The integral is

A = \iint dA = \int_0^{2\pi} \int_0^{(1 - \cos \theta)} r \, dr \, d\theta.

Integration with respect to dr yields

\begin{align} A &{}= \int_0^{2\pi} \left[ {1 \over 2} r^2 \right]_0^{(1-\cos\theta)} \, d\theta \\ &{}= \int_0^{2\pi} {1\over 2} (1 - \cos\theta)^2 \, d\theta \\ &{}= \int_0^{2\pi} {1 \over 2} (1 - 2 \cos\theta + \cos^2 \theta) \, d\theta. \end{align}

Distribute the integral among the three terms, and integrate the first two, to obtain

A = {1 \over 2} \left\{ [\theta]_0^{2\pi} - 2[\sin\theta]_0^{2\pi} + \int_0^{2\pi} \cos^2 \theta \, d\theta \right\}.

The second term vanishes, and integrating the third term yields

\begin{align} A &{}= {1 \over 2} \left\{ 2\pi + \left[ {1\over 2}\theta +{1\over 4}\sin 2\theta \right]_0^{2\pi} \right\} \\ &{}= \pi + {1\over 2} \left[\pi + {1\over 4} (\sin 4\pi - \sin 0)\right]. \end{align}

The last term within brackets vanishes, so that

A = \pi + {1 \over 2}\pi = {3 \over 2}\pi.

Cardioids of any size are all similar to each other, so increasing the cardioid's linear size by a factor of a increases the cardioid's areal size by a factor of a2,    Q.E.D.   (return to article)

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