In here, it is seen that the curvature of a unit speed parametric curve $\alpha(t)$ in $\mathbb{R}^3$ can be measured by its acceleration $\ddot\alpha(t)$. In this case, the acceleration happens to be a normal vector field along the curve. Now we turn our attention to surfaces in Euclidean 3-space $\mathbb{R}^3$ and we would like to devise a way to measure the bending of a surface in $\mathbb{R}^3$, and it may be achieved by studying the change of a unit normal vector field on the surface. To study the change of a unit normal vector field on a surface, we need to be able to differentiate vector fields. But first let us review the directional derivative you learned in mutilvariable calculus. Let $f:\mathbb{R}^3\longrightarrow\mathbb{R}$ be a differentiable function and $\mathbf{v}$ a tangent vector to $\mathbb{R}^3$ at $\mathbf{p}$. Then the directional derivative of $f$ in the $\mathbf{v}$ direction at $\mathbf{p}$ is defined by
\begin{equation}
\label{eq:directderiv}
\nabla_{\mathbf{v}}f=\left.\frac{d}{dt}f(\mathbf{p}+t\mathbf{v})\right|_{t=0}.
\end{equation}
By chain rule, the directional derivative can be written as
\begin{equation}
\label{eq:directderiv2}
\nabla_{\mathbf{v}}f=\nabla f(\mathbf{p})\cdot\mathbf{v},
\end{equation}
where $\nabla f$ denotes the gradient of $f$
$$\nabla f=\frac{\partial f}{\partial x_1}E_1(\mathbf{p})+\frac{\partial f}{\partial x_2}E_2(\mathbf{p})+\frac{\partial f}{\partial x_3}E_3(\mathbf{p}),$$
where $E_1, E_2, E_3$ denote the standard orthonormal frame in $\mathbb{R}^3$. The directional derivative satisfies the following properties.

Theorem. Let $f,g$ be real-valued differentiable functions on $\mathbb{R}^3$, $\mathbf{v},\mathbf{w}$ tangent vectors to $\mathbb{R}^3$ at $\mathbf{p}$, and $a,b\in\mathbb{R}$. Then

1. $\nabla_{a\mathbf{v}+b\mathbf{w}}f=a\nabla_{\mathbf{v}}f+b\nabla_{\mathbf{w}}f$
2. $\nabla_{\mathbf{v}}(af+bg)=a\nabla_{\mathbf{v}}f+b\nabla_{\mathbf{v}}g$
3. $\nabla_{\mathbf{v}}fg=(\nabla_{\mathbf{v}}f)g(\mathbf{p})+f(\mathbf{p})\nabla_{\mathbf{v}}g$

The properties 1 and 2 are linearity and the property 3 is Leibniz rule. The directional derivative \eqref{eq:directderiv} can be generalized to the covariant derivative $\nabla_{\mathbf{v}}X$ of a vector field $X$ in the direction of a tangent vector $\mathbf{v}$ at $\mathbf{p}$:
\begin{equation}
\label{eq:covderiv}
\nabla_{\mathbf{v}}X=\left.\frac{d}{dt}X(\mathbf{p}+t\mathbf{v})\right|_{t=0}.
\end{equation}
Let $X=x_1E_1+x_2E_2+x_2E_3$ in terms of the standard orthonormal frame $E_1,E_2,E_3$. Then $\nabla_{\mathbf{v}}X$ can be written as
\begin{equation}
\label{eq:covderiv2}
\nabla_{\mathbf{v}}X=\sum_{j=1}^3\nabla_{\mathbf{v}}x_jE_j.
\end{equation}
Here, $\nabla_{\mathbf{v}}x_j$ is the directional derivative of the $j$-th component function of the vector field $X$ in the $\mathbf{v}$ direction as defined in \eqref{eq:directderiv}. The covariant derivative satisfies the following properties.

Theorem. Let $X,Y$ be vector fields on $\mathbb{R}^3$, $\mathbf{v},\mathbf{w}$ tangent vectors at $\mathbf{p}$, $f$ a real-valued function on $\mathbb{R}^3$, and $a,b$ scalars. Then

1. $\nabla_{\mathbf{v}}(aX+bY)=a\nabla_{\mathbf{v}}X+b\nabla_{\mathbf{v}}Y$
2. $\nabla_{a\mathbf{v}+b\mathbf{w}}X=a\nabla_{\mathbf{v}}X+b\nabla_{\mathbf{v}}X$
3. $\nabla_{\mathbf{v}}(fX)=(\nabla_{\mathbf{v}}f)X(\mathbf{p})+f(\mathbf{p})\nabla_{\mathbf{v}}X$
4. $\nabla_{\mathbf{v}}(X\cdot Y)=(\nabla_{\mathbf{v}}X)\cdot Y+X\cdot\nabla_{\mathbf{v}}Y$

The properties 1 and 2 are linearity and the properties 3 and 4 are Leibniz rules.

Hereafter, I assume that surfaces are orientable and have nonvanishing normal vector fields. Let $\mathcal{M}\subset\mathbb{R}^3$ be a surface and $p\in\mathcal{M}$. For each $\mathbf{v}\in T_p\mathcal{M}$, define
\begin{equation}
\label{eq:shape}
S_p(\mathbf{v})=-\nabla_{\mathbf{v}}N,
\end{equation}
where $N$ is a unit normal vector field on a neighborhood of $p\in\mathcal{M}$. Since $N\cdot N=1$, $(\nabla_{\mathbf{v}}N)\cdot N=-2S_p(\mathbf{v})\cdot N=0$. This means that $S_p(\mathbf{v})\in T_p\mathcal{M}$. Thus, \eqref{eq:shape} defines a linear map $S_p: T_p\mathcal M\longrightarrow T_p\mathcal{M}$. $S_p$ is called the shape operator of $\mathcal{M}$ at $p$ (derived from $N$).For each $p\in\mathcal{M}$, $S_p$ is a symmetric operator, i.e.,
$$\langle S_p(\mathbf{v}),\mathbf{w}\rangle=\langle S_p(\mathbf{w}),\mathbf{v}\rangle$$
for any $\mathbf{v},\mathbf{w}\in T_p\mathcal{M}$.

Let us assume that $\mathcal{M}\subset\mathbb{R}^3$ is a regular surface so that any differentiable curve $\alpha: (-\epsilon,\epsilon)\longrightarrow\mathcal{M}$ is a regular curve, i.e., $\dot\alpha(t)\ne 0$ for every $t\in(-\epsilon,\epsilon)$. If $\alpha$ is a differentiable curve in $\mathcal{M}\subset\mathbb{R}^3$, then
\begin{equation}
\label{eq:acceleration}
\langle\ddot\alpha,N\rangle=\langle S(\dot\alpha),\dot\alpha\rangle.
\end{equation}
$\langle\ddot\alpha,N\rangle$ is the normal component of the acceleration $\ddot\alpha$ to the surface $\mathcal{M}$. \eqref{eq:acceleration} says the normal component of $\ddot\alpha$ depends only on the shape operator $S$ and the velocity $\dot\alpha$. If $\alpha$ is represented by arc-length, i.e., $|\dot\alpha|=1$, then we get a measurement of the way $\mathcal{M}$ is bent in the $\dot\alpha$ direction. Hence we have the following definition:

Definition. Let $\mathbf{u}$ be a unit tangent vector to $\mathcal{M}\subset\mathbb{R}^3$ at $p$. Then the number $\kappa(\mathbf{u})=\langle S(\mathbf{u}),\mathbf{u}\rangle$ is called the normal curvature of $\mathcal{M}$ in $\mathbf{u}$ direction. The normal curvature $\kappa$ can be considered as a continuous function on the unit circle $\kappa: S^1\longrightarrow\mathbb{R}$. Since $S^1$ is compact (closed and bounded), $\kappa$ attains a maximum and a minimum values, say $\kappa_1$, $\kappa_2$, respectively. $\kappa_1$, $\kappa_2$ are called the principal curvatures of $\mathcal{M}$ at $p$. The principal curvatures $\kappa_1$, $\kappa_2$ are the eigenvalues of the shape operator $S$ and $S$ can be written as the $2\times 2$ matrix
\begin{equation}
\label{eq:shape2}
S=\begin{pmatrix}
\kappa_1 & 0\\
0 & \kappa_2
\end{pmatrix}.
\end{equation}
The arithmetic mean $H$ and the squared Gau{\ss}ian mean $K$ of $\kappa_1$, $\kappa_2$
\begin{align}
\label{eq:mean}
H&=\frac{\kappa_1+\kappa_2}{2}=\frac{1}{2}\mathrm{tr}S,\\
\label{eq:gauss}
K&=\kappa_1\kappa_2=\det S
\end{align}
are called, respectively, the mean and the Gaußian curvatures of $\mathcal{M}$. The definitions \eqref{eq:mean} and \eqref{eq:gauss} themselves however are not much helpful for calculating the mean and the Gaußian curvatures of a surface. We can compute the mean and the Gaußian curvatures of a parametric regular surface $\varphi: D(u,v)\longrightarrow\mathbb{R}^3$ using Gauß’ celebrated formulas
\begin{align}
\label{eq:mean2}
H&=\frac{G\ell+En-2Fm}{2(EG-F^2)},\\
\label{eq:gauss2}
K&=\frac{\ell n-m^2}{EG-F^2},
\end{align}
where
\begin{align*}
E&=\langle\varphi_u,\varphi_u\rangle,\ F=\langle\varphi_u,\varphi_v\rangle,\ G=\langle\varphi_v,\varphi_v\rangle,\\
\ell&=\langle N,\varphi_{uu}\rangle,\ m=\langle N,\varphi_{uv}\rangle,\ n=\langle N,\varphi_{vv}\rangle.
\end{align*}
It is straightforward to verify that
\begin{equation}
\label{eq:normal}
|\varphi_u\times\varphi_v|^2=EG-F^2.
\end{equation}

Example. Compute the Gaußian and the mean curvatures of helicoid
$$\varphi(u,v)=(u\cos v,u\sin v, bv),\ b\ne 0.$$

Helicoid

Solution. \begin{align*}
\varphi_u&=(\cos v,\sin v,0),\ \varphi_v=(-u\sin v,u\cos v,b),\\
\varphi_{uu}&=0,\ \varphi_{uv}=(-\sin v,\cos v,0),\ \varphi_{vv}=(-u\cos v,-u\sin v,0).
\end{align*}
$E$, $F$ and $G$ are calculated to be
$$E=1,\ F=0,\ G=b^2+u^2.$$
$\varphi_u\times\varphi_v=(b\sin v,-b\cos v,u)$, so the unit normal vector field $N$ is given by
$$N=\frac{\varphi_u\times\varphi_v}{\sqrt{EG-F^2}}=\frac{(b\sin v,-b\cos v,u)}{\sqrt{b^2+u^2}}.$$
Next, $\ell, m,n$ are calculated to be
$$\ell=0,\ m=-\frac{b}{\sqrt{b^2+u^2}},\ n=0.$$
Finally we find the Gaußian curvature $K$ and the mean curvature $H$:
\begin{align*}
K&=\frac{\ell n-m^2}{EG-F^2}=-\frac{b^2}{(b^2+u^2)^2},\\
H&=\frac{G\ell+En-2Fm}{2(EG-F^2)}=0.
\end{align*}
Surfaces with $H=0$ are called minimal surfaces.

For further reading on the topic I discussed here, I recommend:

Barrett O’Neil, Elementary Differential Geometry, Academic Press, 1967

The quantity curvature is intended to be a measurement of the bending or turning of a curve. Let $\alpha: I\longrightarrow\mathbb{R}^3$ be a regular curve (i.e. a smooth curve whose derivative never vanishes). If $\alpha$ were to have the unit speed, i.e.
\begin{equation}
\label{eq:unitspped}
||\dot\alpha(t)||^2=\alpha(t)\cdot\alpha(t)=1.
\end{equation}
Differentiating \eqref{eq:unitspped}, we see that $\dot\alpha(t)\cdot\ddot\alpha(t)=0$, i.e. the acceleration is normal to the velocity which is tangent to $\alpha$. Hence, measuring the acceleration is measuring the curvature. So, if we denote the curvature by $\kappa$, then
\begin{equation}
\label{eq:curvature}
\kappa=||\ddot\alpha(t)||.
\end{equation}
Remember that the definition of curvature \eqref{eq:curvature} requires the curve $\alpha$ to be a unit speed curve, but it is not necessarily always the case. What we know is that we can always reparametrize a curve and reparametrization does not change the curve itself but only changes its speed. There is one particular parametrization that we are interested in as it results a unit speed curve. It is called paramtrization by arc-length. This time let us assume that $\alpha$ is not a unit speed curve and define
\begin{equation}
\label{eq:arclength}
s(t)=\int_a^t||\dot\alpha(u)||du,
\end{equation}
where $a\in I$. Since $\frac{ds}{dt}>0$, $s(t)$ is an increasing function and so it is one-to-one. This means that we can solve \eqref{eq:arclength} for $t$ and this allows us to reparametrize $\alpha(t)$ by the arc-length parameter $s$.

Example. Let $\alpha: (-\infty,\infty)\longrightarrow\mathbb{R}^3$ be given by
$$\alpha(t)=(a\cos t,a\sin t,bt)$$
where $a>0$, $b\ne 0$. $\alpha$ is a right circular helix. Its speed is
$$||\dot\alpha(t)||=\sqrt{a^2+b^2}\ne 1.$$
$s(t)=\sqrt{a^2+b^2}t$, so $t=\frac{s}{\sqrt{a^2+b^2}}$. The reparametrization of $\alpha(t)$ by $s$ is given by
$$\alpha(s)=\left(a\cos\frac{s}{\sqrt{a^2+b^2}},b\sin\frac{s}{\sqrt{a^2+b^2}},\frac{bs}{\sqrt{a^2+b^2}}\right).$$
Hence the curvature $\kappa$ is
$$\kappa=\frac{a}{a^2+b^2}.$$

Definition. The dual 1-forms $\theta_1,\theta_2,\theta_3$ of a frame $E_1,E_2,E_3$ on $\mathbb{E}^3$ are defined by
$$\theta_i(v)=v\cdot E_i(p),\ v\in T_p\mathbb{E}^3.$$
Clearly $\theta_i$ is linear.

Example. The dual 1-forms of the natural frame $U_1,U_2,U_3$ are $dx_1$, $dx_2$, $dx_3$ since
$$dx_i(v)=v_i=v\cdot U_i(p)$$
for each $v\in T_p\mathbb{E}^3$.

For any vector field $V$ on $\mathbb{E}^3$,
$$V=\sum_i\theta_i(V)E_i.$$
To see this, let us calculate for each $V(p)\in T_p\mathbb{E}^3$
\begin{align*}
\sum_i\theta_i(V(p))E_i(p)&=\sum_i(V(p)\cdot E_i(p))E_i(p)\\
&=\sum_iV_i(p)E_i(p)\\
&=V(p).
\end{align*}

Lemma. Let $\theta_1,\theta_2,\theta_3$ be the dual 1-forms of a frame $E_1, E_2, E_3$. Then any 1-form $\phi$ on $\mathbb{E}^3$ has a unique expression
$$\phi=\sum_i\phi(E_i)\theta_i.$$

Proof. Let $V$ be any vector field on $\mathbb{E}^3$. Then
\begin{align*}
\sum_i\phi(E_i)\theta_i(V)&=\sum_i\phi(E_i)\theta_i(V)\\
&=\phi(\sum_i\theta_i(V)E_i)\ \mbox{by linearity of $phi$}\\
&=\phi(V).
\end{align*}
Let $A=(a_{ij})$ be the attitude matrix of a frame field $E_1$, $E_2$, $E_3$, i.e.
\begin{equation}\label{eq:frame}E_i=\sum_ja_{ij}U_j,\ i=1,2,3.\end{equation}
Clearly $\theta_i=\sum_j\theta_i(U_j)dx_j$. On the other hand,
$$\theta_i(U_j)=E_i\cdot U_j=\left(\sum_ka_{ik}U_k\right)\cdot U_j=a_{ij}.$$ Hence the dual formulation of \eqref{eq:frame} is
\begin{equation}\label{eq:dualframe}\theta_i=\sum_ja_{ij}dx_j.\end{equation}

Theorem. [Cartan Structural Equations] Let $E_1$, $E_2$, $E_3$ be a frame field on $\mathbb{E}^3$ with dual 1-forms $\theta_1$, $\theta_2$, $\theta_3$ and connection forms $\omega_{ij}$, $i,j=1,2,3$. Then

1. The First Structural Equations: $$d\theta_i=\sum_j\omega_{ij}\wedge\theta_j.$$
2. The Second Structural Equations: $$d\omega_{ij}=\sum_k\omega_{ik}\wedge\omega_{kj}.$$

Proof. The exterior derivative of \eqref{eq:dualframe} is
$$d\theta_i=\sum_jda_{ij}\wedge dx_j.$$ Since $\omega=dA\cdot{}^tA$ and ${}^tA=A^{-1}$ (recall that $A$ is an orthogonal matrix), $dA=\omega\cdot A$, i.e.
$$da_{ij}=\sum_k\omega_{ik}a_{kj}.$$
So,
\begin{align*}
d\theta_i&=\sum_j\left\{\left(\sum_k\omega_{ik}a_{kj}\right)\wedge dx_j\right\}\\
&=\sum_k\left\{\omega_{ik}\wedge\sum_j a_{kj}dx_j\right\}\\
&=\sum_k\omega_{ik}\wedge\theta_k.
\end{align*}

From $\omega=dA\cdot{}^tA$,
\begin{equation}\label{eq:connectform}\omega_{ij}=\sum_kda_{ik}a_{jk}.\end{equation}
The exterior derivative of \eqref{eq:connectform} is
\begin{align*}
d\omega_{ij}&=\sum_k da_{jk}\wedge d_{ik}\\
&=-\sum_k da_{ik}\wedge da_{jk},
\end{align*}
i.e.
\begin{align*}
d\omega&=-dA\wedge{}^t(dA)\\
&=-(\omega\cdot A)\cdot({}^tA\cdot{}^t\omega)\\
&=-\omega\cdot (A\cdot{}^tA)\cdot{}^t\omega\\
&=-\omega\cdot{}^t\omega\ \ \ (A\cdot{}^tA=I)\\
&=\omega\cdot\omega.\ \ \ (\mbox{$\omega$ is skew-symmetric.})
\end{align*}
This is equivalent to the second structural equations.

Example. [Structural Equations for the Spherical Frame Field] Let us first calculate the dual forms and connection forms.

From the spherical coordinates
\begin{align*}
x_1&=\rho\cos\varphi\cos\theta,\\
x_2&=\rho\cos\varphi\sin\theta,\\
x_3&=\rho\sin\varphi,
\end{align*}
we obtain differentials
\begin{align*}
dx_1&=\cos\varphi\cos\theta d\rho-\rho\sin\varphi\cos\theta d\varphi-\rho\cos\varphi\sin\theta d\theta,\\
dx_2&=\cos\varphi\sin\theta d\rho-\rho\sin\varphi\sin\theta d\varphi+\rho\cos\varphi\cos\theta d\theta,\\
dx_3&=\sin\varphi d\rho+\rho\cos\varphi d\varphi.
\end{align*}
From the spherical frame field $F_1$, $F_2$, $F_3$ discussed here, we find its attitude matrix
$$A=\begin{pmatrix}
\cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\
-\sin\theta & \cos\theta & 0\\
-\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi
\end{pmatrix}.$$
Thus by (2) we find the dual 1-forms
\begin{align*}
\begin{pmatrix}
\theta_1\\
\theta_2\\
\theta_3
\end{pmatrix}&=\begin{pmatrix}
\cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\
-\sin\theta & \cos\theta & 0\\
-\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi
\end{pmatrix}\begin{pmatrix}
dx_1\\
dx_2\\
dx_3
\end{pmatrix}\\
&=\begin{pmatrix}
d\rho\\
\rho\cos\theta d\theta\\
\rho d\varphi
\end{pmatrix}.
\end{align*}
\begin{align*}
&dA=\\
&\begin{bmatrix}
-\sin\varphi\cos\theta d\varphi-\cos\varphi\sin\theta d\theta & -\sin\varphi\sin\theta d\varphi+\cos\varphi\cos\theta d\theta & \cos\varphi d\varphi\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
-\cos\varphi\cos\theta d\varphi+\sin\varphi\sin\theta d\theta & -\cos\varphi\sin\theta d\varphi-\sin\varphi\sin\theta d\theta & -\sin\varphi d\varphi
\end{bmatrix}\end{align*}
and so,
\begin{align*}
\omega&=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 & \omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}\\
&=dA\cdot{}^tA\\
&=\begin{pmatrix}
0 & \cos\varphi d\theta & d\varphi\\
-\cos\varphi d\theta & 0 & \sin\varphi d\theta\\
-d\varphi & -\sin\varphi d\theta & 0
\end{pmatrix}.
\end{align*}
From these dual 1-forms and connections forms one can immediately verify the first and the second structural equations.

Let $E_1, E_2, E_3$ be an arbitrary frame field on $\mathbb{E}^3$. At each $v\in T_p\mathbb{E}^3$, $\nabla_v E_i\in T_p\mathbb{E}^3$, $i=1,2,3$. So, there exists uniquely 1-forms $\omega_{ij}:T_p\mathbb{E}^3\longrightarrow\mathbb{R}$, $i,j=1,2,3$ such that
\begin{align*}
\nabla_vE_1&=\omega_{11}(v)E_1(p)+\omega_{12}(v)E_2(p)+\omega_{13}(v)E_3(p),\\
\nabla_vE_2&=\omega_{21}(v)E_1(p)+\omega_{22}(v)E_2(p)+\omega_{23}(v)E_3(p),\\
\nabla_vE_3&=\omega_{31}(v)E_1(p)+\omega_{32}(v)E_2(p)+\omega_{33}(v)E_3(p)
\end{align*}
for each $v\in T_p\mathbb{E}^3$. These equations are called the connection equations of the frame field $E_1$, $E_2$, $E_3$. One can clearly see that $\omega_{ij}$ is determined by
$$\omega_{ij}(v)=\nabla_v E_i\cdot E_j(p).$$ The 1-forms $\omega_{ij}$ are called the connection forms of the frame field $E_1,E_2,E_3$. Often the matrix $\omega=(\omega_{ij})$ is called the connection 1-form of the frame field $E_1,E_2,E_3$. The linearity of $\omega_{ij}$ is due to the linearity of the covariant derivative $\nabla E_i$.

Proposition. The matrix $\omega$ is a skew symmetric matrix, i.e. $\omega+{}^t\omega=0$.

Proof. Since $E_i\cdot E_j=0$, the directional derivative $v[E_i\cdot E_j]=0$. On the other hand, by Leibniz rule,
\begin{align*}
v[E_i\cdot E_j]&=\nabla_vE_i\cdot E_j(p)+E_i(p)\cdot \nabla_vE_j\\
&=\omega_{ij}(v)+\omega_{ji}(v).
\end{align*}
Hence,
\begin{equation}\label{eq:skewsymm}\omega_{ij}+\omega_{ji}=0.\end{equation}

If $i=j$ in \eqref{eq:skewsymm}, we get $\omega_{ii}=0$. So, the connection 1-form $\omega$ is written as
$$\omega=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 &\omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}.$$

Remark. The set of all $3\times 3$ skew symmetric matrices is denoted by $\mathfrak{o}(3)$. It is the Lie algebra of the orthogonal group $\mathrm{O}(3)$. The orthogonal group $\mathrm{O}(3)$ is the set of all $3\times 3$ orthogonal matrices and it is a Lie group. Recall that a square matrix $A$ is orthogonal if and only if $A\cdot{}^tA=I$, i.e. $A^{-1}={}^tA$.

The connection equations of the frame field $E_1$, $E_2$, $E_3$
\begin{equation}\label{eq:connecteqns}\nabla_VE_i=\sum_i\omega_{ij}(V)E_j,\ i=1,2,3\end{equation}
where $V$ is a vector field on $\mathbb{E}^3$ become
$$\begin{array}{ccccccc}
\nabla_VE_1&=&&&\omega_{12}(V)E_2&+&\omega_{13}(V)E_3,\\
\nabla_VE_2&=&-\omega_{12}(V)E_1& & &+&\omega_{23}(V)E_3,\\
\nabla_VE_3&=&-\omega_{13}(V)E_1&-&\omega_{23}(V)E_2.
\end{array}
$$
The connections equations are in fact a generalization of the Frenet-Serret formulas.

Let $Y$ be a vector field defined on a region containing a curve $\alpha(t)$. Then $Y_\alpha(t):=Y(\alpha(t))$ defined a vector field on the curve $\alpha(t)$. Then one can easily see that
$$\nabla_{\dot\alpha(t)}Y=\frac{d}{dt}Y_\alpha(t).$$
Let $\alpha(t)$ be a curve with unit speed. Let $E_1=T$, $E_2=N$, $E_3=B$. Then
\begin{align*}
\omega_{12}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_2=\dot T\cdot N=(\kappa N)\cdot N=\kappa,\\
\omega_{13}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_3=\dot T\cdot B=0,\\
\omega_{23}&=\nabla_{\dot\alpha_(t)}E_2\cdot E_3=\dot N\cdot B=(-\kappa T+\tau B)=\tau.
\end{align*}
The connection equations \eqref{eq:connecteqns} are then nothing but the Frenet-Serret formulas
$$\begin{array}{ccccccc}
\dot T&=&&&\kappa N&&\\
\dot N&=&-\kappa T& & &+&\tau B\\
\dot B&=&&-&\tau N.
\end{array}
$$

The frame $E_1,E_2,E_3$ can be written in terms of the natural frame $U_1,U_2,U_3$ as
\begin{align*}
E_1&=a_{11}U_1+a_{12}U_2+a_{13}U_3,\\
E_2&=a_{21}U_1+a_{22}U_2+a_{23}U_3,\\
E_3&=a_{31}U_1+a_{32}U_2+a_{33}U_3.
\end{align*}
Each real-valued function $a_{ij}:\mathbb{E}^3\longrightarrow\mathbb{R}$ is uniquely determined by $a_{ij}=E_i\cdot U_j$. The matrix $A=(a_{ij})$ is called the attitude matrix (also called rotation matrix or orientation matrix) of the frame field $E_1,E_2,E_3$. One can clearly see that the attitude matrix $A$ is an orthogonal matrix. In the above remark, I mentioned that the set of all $3\times $ skew symmetric matrices is the Lie algebra $\mathfrak{o}(3)$. The Lie algebra $\mathfrak{g}$ of a Lie group $G$ is defined to be the tangent space $T_e G$ to $G$ at the identity element $e$. (A Lie group is a differentiable manifold, so it make sense to talk about tangent spaces to $G$.)

Let us define a curve $\gamma: \mathbb{R}\longrightarrow\mathrm{O}(3)$ by
$$\gamma(t)=A(t)\cdot{}^tA(0).$$
Then $\gamma(0)=I$.
Hence $\dot{\gamma}(0)=\frac{dA(t)}{dt}|_{t=0}\cdot{}^tA(0)$ is a tangent vector to $\mathrm{O}(3)$ at the identity matrix $I$. That is, $\dot{\gamma}(0)\in\mathfrak{o}(3)$. Hence one can easily expect that the following theorem holds.

Theorem. If $A=(a_{ij})$ is the attitude matrix and $\omega=(\omega_{ij})$ the connection 1-form of a frame field $E_1, E_2, E_3$, then
$$\omega=dA\cdot{}^tA$$
or equivalently
$$\omega_{ij}=\sum_k da_{ik} \cdot a_{jk}\ \mbox{for}\ i,j=1,2,3.$$

Proof. For each $v\in T_p\mathbb{E}^3$,
$$\omega_{ij}(v)=\nabla_vE_i\cdot E_j(p).$$
In terms of the natural field $U_i$, $i=1,2,3$,
$$E_i=\sum_ka_{ik}U_k,\ i=1,2,3.$$
So,
\begin{align*}
\nabla_vE_i&=\sum_k v[a_{ik}]U_k(p)\\
&=\sum_k da_{ik} U_k(p).
\end{align*}
Hence,
$$\omega_{ij}=\sum_k da_{ik}a_{jk},$$
i.e.
$$\omega=dA\cdot{}^tA.$$

Remark. In general, if $G$ is a Lie group then its Lie algebra $\mathfrak{g}$ is given by the set of differential $1$-forms
$$\mathfrak{g}=\{g^{-1}dg:\ g\in G\}=\{(dg^{-1})g:\ g\in G\}.$$

Example. Let us compute the connection forms of the cylindrical frame field. The attitude matrix is
$$A=\begin{pmatrix}
\cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0\\
0 & 0 & 1
\end{pmatrix}.$$ Thus
$$dA=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}.$$
Hence,
\begin{align*}
\omega&=dA\cdot{}^tA\\
&=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0 & 0 & 1\end{pmatrix}\\
&=\begin{pmatrix}
0 & d\theta & 0\\
-d\theta & 0 & 0\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
The connection equations of the cylindrical frame field are then
\begin{align*}
\nabla_VE_1&=d\theta(V)E_2=V[\theta]E_2,\\
\nabla_VE_2&=-d\theta(V)E_1=-V[\theta]E_1,\\
\nabla_VE_3&=0
\end{align*}
for all vector fields $V$. As expected the vector field $E_3$ is parallel.