Cardioid/Proofs

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Main article: Cardioid

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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)