# Calculating the curvature of a surface

For a plane curve given parametrically in Cartesian coordinates as $f(t)=(x(t), y(t))$, the curvature is $$\kappa = \frac{|x'y''-y'x''|}{\left(x'^2+y'^2\right)^\frac32}$$

What is the general expression for a 3 or higher dimensional surface, e.g. the curvature of $f(t) = (x(t), y(t), z(t))$?

• That is the Gaussian curvature. In 'alternate formulas' on wikipedia you can see how to derive the Gaussian curvature of any surface given by a local parametrization: en.wikipedia.org/wiki/Gaussian_curvature – noctusraid Feb 6 '17 at 18:23
• Basically you need to compute the fundamental forms (first and second one) of your surface. Whenever you have given a local parametrization the former is easy the latter however requires you to find a Gauss map which can be troublesome. For 2-dimensional surfaces however you can use the cross product... – noctusraid Feb 6 '17 at 18:29
• What you wrote down at the end of your question is a curve in $3$-dimensional space, not a surface. I assume you actually want the curvature of the curve. Rather than writing an elaborate answer, I'll refer you to section 1.2 of my differential geometry text. If you truly want to know about curvature of surfaces, read the first few sections of chapter 2. :) – Ted Shifrin Feb 6 '17 at 19:53

For an arbitrary dimensional curve (that is, a mapping $f$ from $\mathbb{R}$ to $\mathbb{R}^n$), the curvature is the magnitude of the second derivative with respect to arclength: $|d^2f / ds^2|.$
At a point of a two dimensional surface embedded in three dimensional space, there are two principal curvatures, which are the curvatures of two curves on the surface in two orthogonal directions called the principal directions. The principal curvatures can be computed in various ways; one way is to compute the eigenvalues of the second fundamental form matrix. The second fundamental form matrix coefficient $b_{ij}$ is actually very easy to compute. It is just $S_{ij} \cdot n$, where $S_{ij}$ is the the mixed second partial derivative of the surface mapping $S$ (from $\mathbb{R}^2$ to $\mathbb{R}^3$) with respect to $u^i$ and $u^j$ (where $u^1=u$ and $u^2=v$), and $n$ is the unit surface normal. The principal curvatures are the basis for all types of curvature on a two-dimensional surface: Gauss curvature is the product of principal curvatures and mean curvature is the average of principal curvatures. The concept of principal curvatures also generalizes to higher dimensions.
Using the chain rule, it is possible to show that $$\frac{d^2 f}{ds^2} = \frac{f''|f'| - f'(f''\cdot f')/|f'|}{|f'|^3}$$ where $'$ denotes differentiation with respect to an arbitrary parameter $t.$ The denominator of this equation is $((x')^2 + (y')^2 + (z')^2 + \dots)^{3/2}.$ The squared norm of the numerator is $(f''\cdot f'')(f'\cdot f') - (f'\cdot f'')^2.$ For two dimensions, this is a perfect square, $(x''y' - x'y'')^2,$ but in three or higher dimensions it is no longer a perfect square.