Answer
The curvature is related to the tangent and normal vectors by the equation
$$
\frac{d \mathbf{T}}{d s}=\kappa \mathbf{N} .
$$
See proof below.
Work Step by Step
Since the curvature $\kappa $ is given by:
$$
\kappa =\left|\frac{d \mathbf{T}}{d s}\right|=\left|\frac{d \mathbf{T} / d t}{d s / d t}\right|=\frac{|d \mathbf{T} / d t|}{d s / d t}
$$
and the unit normal is given by:
$$
\mathbf{N}=\frac{d \mathbf{T} / d t}{|d \mathbf{T} / d t|},
$$
then
$$
\begin{aligned}
\kappa \mathbf{N} &=\frac{|d \mathbf{T} / d t|}{d s / d t} . \frac{d \mathbf{T} / d t}{|d \mathbf{T} / d t|} \\
&=\frac{\left|\frac{d \mathbf{T}}{d t}\right| \frac{d \mathbf{T}}{d t}}{\left|\frac{d \mathbf{T}}{d t}\right| \frac{d s}{d t}} \\
&=\frac{d \mathbf{T} / d t}{d s / d t} \ \ \ \ \text { by using the Chain Rule. }\\
&=\frac{d \mathbf{T}}{d s} .
\end{aligned}
$$
Therefore, the curvature is related to the tangent and normal vectors by the equation
$$
\frac{d \mathbf{T}}{d s}=\kappa \mathbf{N}
$$
