If I have a surface defined as the graph of the function $z = f(x,y)$, is there a closed-form expression for the signed curvature of this surface in an arbitrary direction? That is, if $x(t) = x_0 + t\Delta x$ and $y(t) = y_0 + t\Delta y$, how can I compute the signed curvature of the intersection curve $\mathbf{C}(t) = \big(x(t), y(t), f(x(t), y(t))\big)$ at $t=0$?
Computing the signed curvature of a surface in an arbitrary direction
curvaturecurvesdifferential-geometrysurfaces
Related Solutions
Short answer: the curvature is a property of the second derviative, and it's perfectly possible for a $C^1$ function to have a badly behaved second derivative.
In one variable, think about the function $f(t) = |t|^{3/2}$; it is $C^1$ but $f''(0) = +\infty$.
To turn this into a surface with infinite curvature, try $$X(u,v) = (u, |u|^{3/2}, v)$$ i.e. just extending this curve in the $z$ direction, forming a sheet. You'll find the normal curvature at the origin in the direction normal to the $xy$-plane is infinite.
This is exercise $2.2.7$ in Pressley's Elementary Differential Geometry (which has solution sketches in the back of the text). I've expanded a bit on the solution provided.
Hints:
For signed curvature, what is the definition of signed curvature in terms of your signed normal and tangent? Remember that you're trying to find $\kappa_{s_{\epsilon}}$.
For the normal line part, what does it mean for a line to be tangent to $\epsilon$? What does a normal line at a specific point $s_0$ look like? We calculated the tangent to $\epsilon$ in the first part.
(More complete answer below)
$$\epsilon (s)=\gamma (s)+\frac{1}{\kappa_s(s)}n_s(s)$$
$\textbf{1.}$ We show that the arc length of $\epsilon$ is $\frac{-1}{\kappa_s(s)}$ up to a constant. First off, I'm going to forget about the $s$ notation-wise so assume everything is a function of $s$ unless stated otherwise.
Well, we take the derivative of $\epsilon$ with respect to $s$ to get the arc length: $$\dot{\epsilon} = \dot{\gamma} + \frac{1}{\kappa_s}(-\kappa_s \mathbf{t})-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}=-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}$$
(recall $\mathbf{\dot{t}} = \kappa_s \mathbf{n_s} \implies \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$ for a unit-speed curve since $\mathbf{n_s} \cdot \mathbf{t} = 0$ so $\mathbf{\dot{n_s}} \cdot \mathbf{t} + \mathbf{n_s} \cdot \mathbf{\dot{t}} = 0 \implies \mathbf{\dot{n_s}} \cdot \mathbf{t}= - \kappa_s(\mathbf{n_s \cdot n_s}) = -\kappa_s \implies \mathbf{\dot{n_s}}(\mathbf{t}\cdot \mathbf{t}) = \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$)
Now, the arc length is given by $$u=\int \| \dot{\epsilon} \| \,ds = \int \frac{\dot{\kappa_s}}{\kappa_s^2} \,ds = -\frac{1}{\kappa_s} + C$$ Note that the second equality holds since we assumed $\dot{\kappa_s} > 0$.
$\textbf{2.}$ We calculate the signed curvature $\kappa_{s_{\epsilon}}.$ Recall the signed curvature is the rate at which the tangent vector rotates. In particular, $$\mathbf{\dot{t}}_{\epsilon} = \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$ In this case, we take the tangent vector to be $\mathbf{t}_{\epsilon}=-\mathbf{n_s}$. Rotating the tangent vector counterclockwise by $-\pi/2$ gives us our signed unit normal. In particular, the signed normal is just $\mathbf{n_s}_{\epsilon}=\mathbf{t}$. Now, $$\frac{d (-\mathbf{n_s})}{\,du} = \kappa_s \mathbf{t} \frac{ds}{du} = \kappa_s \mathbf{t} \frac{\kappa_s^2}{\dot{\kappa_s}}= \frac{\kappa_s^3}{\dot{\kappa_s}}\mathbf{t}= \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$ In particular, since the derivative of the tangent vector is the signed curvature times the signed unit normal, dotting the derivative of the tangent vector with the signed unit normal gives the result. That is, take the dot product of the above expression with $\mathbf{t}$ to get the signed curvature of $\epsilon$: $$\frac{\kappa_s^3}{\dot{\kappa_s}}$$
$\textbf{3.}$ We show that all normal lines to $\gamma$ are tangent to $\epsilon$.
Well, let's look at a point on the normal line at $\gamma(s_0)$ for some arbitrary $s_0$. It looks like $\gamma(s_0) + C\mathbf{n_s}(s_0)$ for some $C$. Since $\epsilon(s_0) = \gamma(s_0) + \frac{1}{\kappa_s(s_0)}\mathbf{n_s}(s_0)$, the intersection occurs when $C=\frac{1}{\kappa_s(s_0)}$. Well, we calculated the tangent of $\epsilon$ at $s_0$: $$\dot{\epsilon}(s_0)=-\frac{\dot{\kappa_s(s_0)} \mathbf{n_s}(s_0)}{\kappa_s^2(s_0)}$$ so that the tangent there is parallel to $\mathbf{n_s}(s_0)$.
$\textbf{4.}$ Regarding the evolute of the cycloid, this is just a computation, with a lot of the steps highlighted above. Regarding the reparameterization, consider $\tilde{t} = t + \pi$.
Best Answer
Apply the second fundamental form of the surface (at the point $P=(x_0,y_0,f(x_0,y_0))$) to the unit tangent vector $\mathbf v$ of the curve at the point. See pages 45-47 of my differential geometry text.
EDIT: This is computing the normal curvature (i.e., agrees with the actual space curvature of the plane curve only when the slicing plane is normal to the surface at the point).