Kappa curve

Using the Cartesian coordinate system it can be expressed as

${\displaystyle x^2\left(x^2+y^2\right)=a^2{y}^2}$

or, using parametric equations,

${\displaystyle \begin{array}{cc}x& =a\sin t,\\ y& =a\sin t\tan t.\end{array}}$

In polar coordinates its equation is even simpler:

${\displaystyle r=a\tan \theta .}$

It has two vertical asymptotes at x = ±a, shown as dashed blue lines in the figure at right.

The kappa curve's curvature:

${\displaystyle \kappa (\theta )=\frac{8\left(3-{\sin }^2\theta \right){\sin }^4\theta }{a{\left({\sin }^2(2\theta )+4\right)}^{\frac{3}{2}}}.}$

Tangential angle:

${\displaystyle \varphi (\theta )=-\arctan \left(\frac{1}{2}\sin (2\theta )\right).}$

Tangents via infinitesimals

The tangent lines of the kappa curve can also be determined geometrically using differentials and the elementary rules of infinitesimal arithmetic. Suppose x and y are variables, while a is taken to be a constant. From the definition of the kappa curve,

${\displaystyle x^2\left(x^2+y^2\right)-a^2{y}^2=0}$

Now, an infinitesimal change in our location must also change the value of the left hand side, so

${\displaystyle d\left(x^2\left(x^2+y^2\right)-a^2{y}^2\right)=0}$

Distributing the differential and applying appropriate rules,

${\displaystyle \begin{array}{cc}d\left(x^2\left(x^2+y^2\right)\right)-d\left(a^2{y}^2\right)& =0\\ (2xdx)\left(x^2+y^2\right)+x^2(2xdx+2ydy)-a^{2}2ydy& =0\\ \left(4x^3+2x{y}^2\right)dx+\left(2y{x}^2-2a^{2}y\right)dy& =0\\ x\left(2x^2+y^2\right)dx+y\left(x^2-a^2\right)dy& =0\\ \frac{x\left(2x^2+y^2\right)}{y\left(a^2-x^2\right)}& =\frac{dy}{dx}\end{array}}$