\noindent
\begin{figure}[htbp]
\centering
\begin{subfigure}[t]{0.45\textwidth}
\centering
\fbox{\resizebox{2.4in}{!}{\includegraphics{../images/instructors_guide_illustrations/tangent_vectors_curve}}}
\caption{Derivatives associate a curve to its tangent vector. Dividing by the magnitude turns both infinitesimal tangent vectors ($d\vec{r}$) or tangent vectors formed from parametric equations into unit vectors.}
\label{fig:tangent_vectors_curve}
\end{subfigure}
\hfill
\begin{subfigure}[t]{0.45\textwidth}
\centering
\fbox{\resizebox{2.4in}{!}{\includegraphics{../images/instructors_guide_illustrations/tangent_vectors_surface}}}
\caption{Unit Tangent vectors to a surface can be defined using $\frac{d\vec{r}}{\left| d\vec{r} \right| }$ or by finding tangent vectors to a curve within the surface.}
\label{fig:tangent_vectors_surface}
\end{subfigure}
\caption[Tangent vectors to curves and surfaces]{Tangent vectors and unit tangent vectors to curves and surfaces}
\label{fig:tangent_vectors}
\end{figure}
Geometrically, the direction (not the magnitude) of a vector determines whether it is tangent to a surface or curve. The vector's magnitude can often encode other information, such as the speed at which a dot moves along the curve according to some parametrization. This is not always helpful in understanding how the curve or surface changes - especially in situations where the parametrization causes the dot to turn around or momentarily stop, resulting in the tangent vector vanishing!
There are (at least) two possibilities for eliminating this dependence upon the parametrization. One option is to always work with unit tangent vectors by dividing any tangent vector by it's magnitude. This turns every tangent vector into pure direction, which is (usually) sufficient for understanding the (local) shape of a curve or a surface. The second alternative is to eliminate parametrization and construct tangent vectors whose components are in the correct ratios as determined by partial derivatives or differentials. One way to accomplish this is to use differentials and the vector differential $d\vec{r}$.
An (infinitesimal) small change between two points on a curve can be described by
\[d\vec{r} = dx \; \hat{\imath} + dy \; \hat{\jmath} \]
where the differential of the curve describes how the quantities $dx$ and $dy$ are related at a point. If the curve is given by $y = x^2$, then
\[dy = 2x\; dx = 8\; dx \textrm{at the point } (4,16).\]
This leads to the infinitesimal tangent vector
\[d\vec{r} = \left(1 \; \hat{\imath} + 8 \; \hat{\jmath} \right) dx. \]
This vector can become a unit tangent vector simply by dividing by its magnitude $ds = \left| d\vec{r} \right|$:
\[ \frac{d\vec{r}}{ds} = \frac{d\vec{r}}{\left| d\vec{r} \right|} = \frac{(\hat{\imath} + 8\; \hat{\jmath})\; dx}{\sqrt{65}\; dx} = \frac{\sqrt{65}}{65} \hat{\imath} + \frac{8\sqrt{65}}{65} \hat{\jmath}\]
\noindent
Notice that first dividing out by $dx$ or $dy$ gives tangent vectors $\frac{d\vec{r}}{dx} = \hat{\imath} + \frac{\partial y}{\partial x} \hat{\jmath}$ and $\frac{d\vec{r}}{dy} = \frac{\partial x}{\partial y}\hat{\imath} + \hat{\jmath}$ which are not unit vectors.
A (family of) tangent vectors to a surface can be defined in the same way. In this case,
\[d\vec{r} = dx \; \hat{\imath} + dy \; \hat{\jmath} + dz \; \hat{k}.\]
The equation for the surface (e.g. $z = x^2 + y^2$) gives a relationship between the differentials $dx, dy,$ and $dz$. A particular tangent vector from this family is then chosen by giving a second relationship between these quantities, either explicitly with differentials (e.g. $dy = 2x dx$ or $dy = 0$) or by finding the differential of a curve (e.g. $y = x^2$ or $y = 5$).
The expressions for $d\vec{r}$ in polar, cylindrical, and spherical coordinates are:
\[
\begin{array}{ll}
\textrm{polar:} & d\vec{r} = dr \; \hat{r} + r d\varphi \; \hat{\varphi} \\[8pt]
\textrm{cylindrical coordinates:} & d\vec{r} = dr \; \hat{r} + r d\varphi \; \hat{\varphi} + dz\; \hat{z} \\[8pt]
\textrm{spherical coordinates:} & d\vec{r} = dr \; \hat{r} + r d\theta \; \hat{\theta} + r \sin(\theta) \; \hat{\varphi} \\[8pt]
\end{array}
\]