On the book Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo the following definition can be found:

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My question is very concrete: is there a typographical mistake in the formula boxed in red?


No, this is absolutely correct. You're looking at the portion of the curvature vector $k\mathbf n$ that is normal to the surface (i.e., in the direction of $\mathbf N$). That is, you take $(k\mathbf n)\cdot \mathbf N = k(\mathbf n\cdot \mathbf N) = k\cos\theta$.


This complicated setup starts with a surface $S$ in $\mathbb R^3$ and a point $P$ with a normal unit vector to the surface $\vec N_S.$

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Each point on the surface has an associated orthogonal vector (shortened in the diagram above to let the vector at $P$ stand out). The surface $S$ has domain boundaries between $-1<x<1$ and $-1<y<1$ and is governed by the equation:


The normal vector to the surface at any given point $\vec N_S(P)$ was calculated as:

$$\begin{align} \vec N(t)&=\left (-\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} ,-\frac{\partial f}{\partial y}\frac{\partial y}{\partial t},1\right)\\[3ex] &=\left(2x+\sin(x),\sin(y),1\right) \end{align}$$

On $S$ a space curve $C\in \mathbb R^3$ was parameterized by $t$ with $-1<t<1$ as:


On this space curve a tangent vector can be defined at each point as the curve derivative:

$$\vec T(t)=(1,2t,-2t-\sin(t)-2t\sin(t^2))$$

In the Frenet Serret or TNB frame $\vec T$ would be of unit $1,$ and defined as $\vec T=\frac{\vec r'(t)}{\vert \vec r'(t)\vert}=\vec r'(s),$ the latter part indicating that there is no need to normalize if the curve is parameterized by arc length $(s).$

A second orthogonal vector called the normal vector to the curve $C$ at $P$ can be calculated as the derivative to the tangent vector, provided that is parameterized by arc length as

$$\vec n(s)=\frac{\vec T'(s)}{\vert T'(s)\vert}$$

with $k(s)=\vert T'(s)\vert=\frac{\vert T'(t)\vert}{\vert r'(t)\vert}$ corresponding to the curvature of $C$ at $P.$ However, this is not straightforward to compute given the square roots to normalize the derivatives, and as reflected here. A way to circumvent this problem is to generate a vector via the wedge products

$$\vec n= \left( C'(t)\times C''(t)\right)\times C'(t)$$

and then proceeding to normalize it.

This vector can be used to generate the osculating circle, knowing that the curvature can also be calculated as $k(t)=\frac{C'(t)\wedge C''(t)}{\vert C'(t)\vert^3},$ and that the radius of the osculating circle is $r=1/k:$

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In the animation $\vec B(t)$ completes the Frenet-Serret triad. $\vec B(t)$ is the binormal unit vector, the cross product of $\vec T$ and $\vec n.$ It is worth noting that since the derivative of the tangent vector $C'(t)$ is normalized its derivative $C''(t)$ is orthogonal. Together, $\{\vec T, \vec n, \vec B\}$ form an orthonormal basis for $\mathbb R^3.$

The derivatives of $\vec T$ and $\vec B$ are in the span of $\vec n:$ The derivative of the tangent vector can be expressed as $T'(s)=k(s) \vec n.$ As a scalar product, $k(s) =\langle \vec n(s), T'(s) \rangle.$ Similary, the derivative of the binormal vector $\vec B$ can be expressed as $B'(s)=\tau (s) \vec n,$ where $\tau (s)$ is the torsion of the curve $C$ at $s,$ which can also be expressed as a scalar product as $\tau (s) =\langle \vec n(s), B'(s) \rangle.$ That it makes sense for $\tau$ to denote the torsion of the curve can be seen by noticing how it stays constant if the curve is planar:

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Here is an animation to illustrate the Frenet triad moving along the curve in relation to vector field $\vec N:$

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The normal vector to the curve is in the span of the normal vector to the surface at any points along a geodesic curve:

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Delving in the topic of the OP, $k_n=k\cos\theta$ is a scalar value with $\theta$ corresponding to the angle between $\vec N_S(t)$ and $\vec n(t):$

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Since both $\vec N$ and $\vec n$ are unitary, the $\cos(\theta)=\langle \vec N, \vec n\rangle$ is given by the scalar product of the vectors.

The unit vector $\vec n$ multiplied by the scalar value of the curvature $k$ yields a vector $k\vec n,$ whose projection on $\vec N$ is $k_n\vec N:$

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$K_n=k\cos\theta$ is called the normal curvature of $C$ at $P.$

The geodesic curvature, $k_g,$ is the curvature of the curve projected onto the surface tangent plane. The geodesic curvature measures how far the curve is from being a geodesic:

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And the punch line of the story is that if a vector $\vec v \in T_P S$ has norm $\vert v \vert=1,$ the second fundamental form applied to the vector, i.e. $\vec v^\top \mathbf{{II}_P} \vec v$ equals the normal curvature of any curve through $P$ at velocity $\vec v.$

The second fundamental form corresponds to the Hessian of a surface with a chart $f\left(u,v, h(u,v)\right),$ while the trace of the Hessian is the Laplacian. This makes intuitive sense, since the normal vector at a point $\vec N$ of a graph of a function is the gradient (vector of first derivatives), $\nabla F(x,t,f(x,y))=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1\right)(*),$ and the second fundamental form involved the derivative of the normal, $dN.$ The second fundamental form can be expressed as a symmetrical matrix, which when applied to the derivative of a curve $\alpha(0)$ coursing through $p\in S$ at $t=0$ will result in (see here):

$$\begin{align} \mathrm{II}_p(\alpha'(0),\alpha'(0))&=-\langle dN_p(\alpha'(0),\alpha'(0) \rangle\\ &=-\langle N'(0),\alpha'(0) \rangle\\ &\underset{*}{=}\bbox[5px,border:2px solid black]{\langle N(p),\alpha''(0) \rangle}\\ &=\langle N(p),k(p) \vec n(p) \rangle\\ &=k\langle \vec N_p,\vec n(p)\rangle\\ &=k_n(p) &=k\cos\theta \end{align} $$

The boxed result being Euler's theorem: the acceleration of curve $\alpha$ at point $p$ dotted with the normal to the surface at the same point is the second fundamental form.

$(*)$ in an arc length parameterized curve: $$\begin{align} &\langle N(s),\alpha'(s) \rangle =0\\ &\implies \langle N'(s),\alpha'(s) \rangle+\langle N(s),\alpha''(s) \rangle=0\\ \end{align}$$

The associated vectors to the minimum $k_1(p)$ and maximum $k_2(p)$ eigenvalues of the $\mathrm{II}_p$ restricted to vectors of norm $1$ in $T_pS$ will form an orthonormal basis of $T_pS,$ because $\mathrm{II}$ is a symmetric matrix. $\{k_1,k_2\}$ are the principal curvatures of the surface at $p.$

A unit vector $\vec v\in T_pS$ in the tangent space can be thus represented in relation to the angle with these orthonormal basis vectors with $\vec v:$

$$\vec v=\cos \varphi \vec e_1 + \sin \varphi e_2$$

and applying the quadratic of the second fundamental form to $\vec v:$

$$\begin{align} k_n=\mathrm{II}_p(\vec v)&=-\langle dN_p(\vec v), \vec v \rangle\\ &=-\langle dN_p(\cos \varphi \vec e_1 + \sin \varphi e_2), \cos \varphi \vec e_1 + \sin \varphi e_2 \rangle\\ &=\bbox[5px,border:2px solid black]{-\cos^2 \varphi k_1 - \sin^2 \varphi k_2} \end{align}$$

which is Euler's curvature formula.

The surface $z=f(x,y)$ is identical to $F(x,y,z)=0,$ where $F(x,y,z)=f(x,y)-z.$ Hence $\left(\frac{\partial}{\partial x} F, \frac{\partial}{\partial y} F, \frac{\partial}{\partial z} F \right)=\left(\frac{\partial}{\partial x} f(x,y), \frac{\partial}{\partial y} f(x,y),-1 \right).$


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