Example with Figures on @JEB 's answer.

Let the regular (smooth) curve with parametric equation
\begin{equation}
\mathbf{x}\left(t\right)=\bigl[x_{1}\left(t\right),x_{2}\left(t\right),x_{3}\left(t\right)\bigr]=\left(5\cos t,5\sin t,2t\right)
\tag{01}
\end{equation}
The parameter $\:t\:$ would represent time in case that this curve is the trajectory of a particle.
Now, the vector
\begin{equation}
\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}=\Biggl(\dfrac{\mathrm dx_{1}}{\mathrm dt},\dfrac{\mathrm dx_{2}}{\mathrm dt},\dfrac{\mathrm dx_{3}}{\mathrm dt}\Biggr)=\left(-5\sin t,5\cos t,2\right)
\tag{02}
\end{equation}
is tangent to the curve at the point $\:\mathbf{x}\left(t\right)\:$ and well-defined without any indeterminacy. In case of particle motion this is the velocity vector of the particle.
In order to normalize this vector we have
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert=\sqrt{29}
\tag{03}
\end{equation}
This norm, the speed of the particle, is a function of $\:t\:$ in general. Here accidentally is constant. From (02) and (03) we produce the unit vector
\begin{equation}
\mathbf{t}=\dfrac{\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right)
\tag{04}
\end{equation}
The vector $\:\mathbf{t}\left(t\right)\:$ is the unit tangent vector to the curve at point $\:\mathbf{x}\left(t\right)$.
\begin{equation}
\boxed{\:\mathbf{t}=\sqrt{\frac{1}{29}}\left(-5\sin t,5\cos t,2\right)\:}
\tag{05}
\end{equation}
Differentiating again we have
\begin{equation}
\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}=\sqrt{\frac{1}{29}}\left(-5\cos t,-5\sin t,0\right)
\tag{06}
\end{equation}
a vector normal to $\:\mathbf{t}\:$ with norm
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert=5\sqrt{\frac{1}{29}}
\tag{07}
\end{equation}
Again this norm is a function of $\:t\:$ in general. From (06) and (07) we produce the unit vector
\begin{equation}
\mathbf{n}=\dfrac{\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}}{\left\Vert\dfrac{\mathrm d\mathbf{t}}{\mathrm dt}\right\Vert}=\left(-\cos t,-\sin t,0\right)
\tag{08}
\end{equation}
The vector $\:\mathbf{n}\left(t\right)\:$ is the principal normal unit vector to the curve at point $\:\mathbf{x}\left(t\right)$.
\begin{equation}
\boxed{\:\mathbf{n}=\left(-\cos t,-\sin t,0\right)\vphantom{\sqrt{\frac{1}{29}}}\:}
\tag{09}
\end{equation}
Finally we construct the unit vector
\begin{equation}
\mathbf{b}=\mathbf{t}\boldsymbol{\times}\mathbf{n}=\sqrt{\frac{1}{29}}
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
-5\sin t & 5\cos t & 2\vphantom{\dfrac{\dfrac{}{}}{}}\\
-\cos t & -\sin t & 0 \vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right)
\tag{10}
\end{equation}
so
\begin{equation}
\boxed{\:\mathbf{b}=\sqrt{\frac{1}{29}}\left(2\sin t,-2\cos t,5\right)\vphantom{\sqrt{\frac{1}{29}}}\:}
\tag{11}
\end{equation}
The vector $\:\mathbf{b}\left(t\right)\:$ is the unit binormal vector to the curve at point $\:\mathbf{x}\left(t\right)$.
The triad of vectors $\:\left(\mathbf{t},\mathbf{n},\mathbf{b}\right)\:$ forms a right-handed orthonormal triplet, as shown in Figures, called the moving trihedron. For the three planes, sides of the trihedron, we have the following terminology
\begin{align}
\text{plane }\left(\mathbf{t},\mathbf{n}\right) & =\textit{Osculating plane}
\tag{12a}\\
\text{plane }\! \left(\mathbf{n},\mathbf{b}\right) & =\textit{Normal plane}
\tag{12b}\\
\text{plane }\left(\mathbf{b},\mathbf{t}\right) & =\textit{Rectifying plane}
\tag{12c}
\end{align}
----- 3D image 1
----- 3D image 2
----- 2D video
----- 3D video
----- The moving trihedron 3D video