Finding the curvature of a curve defined as the intersection of two specific surfaces Consider the following two surfaces:
$S_1: x^{11}+y^{10}+z^9+x^3y^2+x^2y^2+y^3+z^2+x^2+y^2+x+10y+2z=0$
$S_2: x^4-y^2z^2+x^3-y^2z+z=0$
How can I find the curvature at the point $(0,0,0)$ of the curve defined by the intersection of the two surfaces? If I will be able to find a parameterization of the curve then by Taylor expension it could be possible to find the curvature. Is there a way to find an explicit parameterization? Is there a simple way to find the curvature? 
 A: If all you need is the curvature at a single point, you can expand the parametrized intersection curve in a Taylor series about that point.  Here's how:
Assume that the intersection of the surfaces is a smooth curve passing through $(0,0,0)$, with parameterization $(x(t), y(t), z(t))$.  We can take $(x(0),y(0),z(0)) = (0,0,0)$.  We can expand each of these functions out in a power series:
$$
x(t) = x_1 t + x_2 t^2 + \mathcal{O}(t^3) \\
y(t) = y_1 t + y_2 t^2 + \mathcal{O}(t^3) \\
z(t) = z_1 t + z_2 t^2 + \mathcal{O}(t^3)
$$
Note that if we are just concerned with the curvature at $(0,0,0)$, we don't need to know the higher-order coefficients in these power series:  the curvature only depends on the first and second derivatives of $x, y$, and $z$, which are determined by the first two coefficients in these power series.
The constraint that these curves lie in both surfaces is then
$$
(x_1 t + x_2 t^2) + 10 (y_1 t + y_2 t^2) + 2 (z_1 t + z_2 t^2) + (x_1 t)^2 + (y_1 t)^2 + (z_1 t)^2 + \mathcal{O}(t^3) = 0 \\
(z_1 t + z_2 t^2) + \mathcal{O}(t^3) = 0
$$
Since these equations must be true for all $t$, it must be that the overall coefficients of $t$ & $t^2$ vanish in each case, implying that
$$
x_1 + 10 y_1 + 2 z_1 = 0 \\
x_2 + 10 y_2 + 2 z_2 + x_1^2 + y_1^2 + z_1^2 = 0 \\
z_1 = 0\\
z_2 = 0
$$
The general solution to this system can be written in terms of two parameters $\alpha$ and $\beta$:
$$
x_1 = - 10 \alpha\\ y_1 = \alpha \\ x_2 = -101 \alpha^2 - 10 \beta \\ y_2 = \beta.
$$ 
Finally, we can calculate the curvature of this curve by taking the quantity $\kappa = \| \gamma' \times \gamma'' \|/\|\gamma'\|^3$ at $t = 0$.  We have
$$ \gamma'(0) = (x_1, y_1, z_1) = (-10 \alpha, \alpha, 0) \\
\gamma''(0) = (2 x_2, 2 y_2, 2 z_2) = (-202 \alpha^2 - 20 \beta, 2 \beta, 0)
$$
Taking the cross & dot products, we have $\| \gamma'(0) \times \gamma''(0) \| = (0,0,202\alpha^3)$ and $\|\gamma'(0)\| = \sqrt{101} \alpha$;  thus, the curvature of the intersection curve is
$$
\boxed{\kappa = \frac{2}{\sqrt{101}}.}
$$
(There might be a nicer way to see that the $\beta$ terms cancel out "naturally" when you take the cross products, but I didn't see it immediately, so I left them in for completeness.)
