Cardioid/Proofs

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

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

Proof

Starting from $rho(theta)\; =\; 2r(1\; -\; cos\; theta)$, 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:

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

$y\; =\; 2r\; (1-\; cos\; theta)\; sin\; theta\; =\; 2r(sin\; theta\; -\; sin\; theta\; cos\; theta)\; =\; 2r\; left(sin\; theta\; -\; frac\{1\}\{2\}\; sin\; 2theta\; right),$

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

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 $theta$ 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 $rho(theta)$ 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

$rho\; =\; 2r\; sqrt\{\; \{3\; -\; 4\; cos\; theta\; +\; 2\; cos^2\; theta\; -\; 1\; over\; 2\}\}\; =\; 2r\; sqrt\{\; \{2\; -\; 4\; cos\; theta\; +\; 2\; cos^2\; theta\; over\; 2\}\},$

$rho\; =\; 2r\; sqrt\{\; 1\; -\; 2\; cos\; theta\; +\; cos^2\; theta\}\; =\; 2r(1\; -\; cos\; theta).$

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

$A\; =\; iint\; dA\; =\; int\_0^\{2pi\}\; int\_0^\{(1\; -\; cos\; theta)\}\; r\; ,\; dr\; ,\; dtheta$.

Integration with respect to dr yields

$begin\{align\}$

A &{}= int_0^{2pi} left[{1 over 2} r^2 right]_0^{(1-costheta)} , dtheta &{}= int_0^{2pi} {1over 2} (1 - costheta)^2 , dtheta &{}= int_0^{2pi} {1 over 2} (1 - 2 costheta + cos^2 theta) , dtheta. end{align} Distribute the integral among the three terms, and integrate the first two, to obtain

$A\; =\; \{1\; over\; 2\}\; left\{\; [theta]\_0^\{2pi\}\; -\; 2[sintheta]\_0^\{2pi\}\; +\; int\_0^\{2pi\}\; cos^2\; theta\; ,\; dtheta\; right\}.$

The second term vanishes, and integrating the third term yields

$begin\{align\}$

A &{}= {1 over 2} left{ 2pi + left[{1over 2}theta +{1over 4}sin 2theta right]_0^{2pi} right} &{}= pi + {1over 2} left[pi + {1over 4} (sin 4pi - 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)