First, we need an intuitive definition for the curvature. For flatter surface, the unit tangent doesn't change much when increasing the arc length. However, for curvier surfaces, the unit tangent vector changes a lot:
Let \(C\) be a smooth curve in the plane or in space given. The curvature \(k\) of the curve is defined as:
\[ k(t) = \left\Vert \frac{d \ \textbf{T}(t)}{d \ s(t)} \right\Vert \]
where \(\textbf{T}(t)\) is the unit tangent vector. This is our first formula for the curvature. Using the chain rule, we know:
\[ \frac{d \ \textbf{T}(t)}{dt} = \frac{d \ \textbf{T}(t)}{d \ s(t)} \frac{d \ s(t)}{dt} \]
Since \(\frac{d \ s(t)}{dt} = \Vert \textbf{r}'(t) \Vert\):
\[ \begin{gather} \textbf{T}'(t)= \frac{d \ \textbf{T}(t)}{d \ s(t)} \Vert \textbf{r}'(t) \Vert \\ \frac{\textbf{T}'(t)}{\Vert \textbf{r}'(t) \Vert}=\frac{d \ \textbf{T}(t)}{d \ s(t)}\end{gather}\]
Substituting this into our formula for the curvature:
\[ k(t) = \left\Vert \frac{\textbf{T}'(t)}{\Vert \textbf{r}'(t) \Vert}\right\Vert = \frac{\Vert \textbf{T}'(t) \Vert}{\Vert \textbf{r}'(t) \Vert} \]
This is our second formula for the curvature. Since \(\textbf{T}(t) = \frac{\textbf{r}'(t)}{\Vert \textbf{r}(t) \Vert}\):
\[ \textbf{r}'(t) = \Vert \textbf{r}'(t) \Vert \textbf{T}(t) = \frac{d \ s(t)}{dt}\textbf{T}(t) \]
Using the product rule for derivatives:
\[ \textbf{r}''(t) = \frac{d^2 \ s(t)}{(dt)^2}\textbf{T}(t) + \frac{d \ s(t)}{dt}\textbf{T}'(t) \]
If we take the cross product of \(\textbf{r}'(t)\) and \(\textbf{r}''(t)\):
\[ \textbf{r}'(t) \times \textbf{r}''(t) = \left[ \frac{d \ s(t)}{dt}\textbf{T}(t) \right] \times \left[ \frac{d^2 \ s(t)}{(dt)^2}\textbf{T}(t) + \frac{d \ s(t)}{dt}\textbf{T}'(t) \right] \]
Since cross product is distributive:
\[ \textbf{r}'(t) \times \textbf{r}''(t) = \left[ \frac{d \ s(t)}{dt}\textbf{T}(t) \times \frac{d^2 \ s(t)}{(dt)^2}\textbf{T}(t)\right] + \left[ \frac{d \ s(t)}{dt}\textbf{T}(t) \times \frac{d \ s(t)}{dt}\textbf{T}'(t)\right]\]
Since \(\textbf{T} \times \textbf{T} = 0\):
\[ \textbf{r}'(t) \times \textbf{r}''(t) = \frac{d \ s(t)}{dt}\textbf{T}(t) \times \frac{d \ s(t)}{dt}\textbf{T}'(t) \]
According to the "multiplication by a constant" rule of cross product:
\[ \textbf{r}'(t) \times \textbf{r}''(t) = \frac{d^2 \ s(t)}{(dt)^2} \left[ \textbf{T}(t) \times \textbf{T}'(t) \right] \]
Since \(\textbf{T}(t)\) is a unit tangent vector, then it's magnitude is always 1. If a vector has constant magnitude, then the derivative vector will always be orthogonal. This means \(\textbf{T}(t)\) and \(\textbf{T}'(t)\) are orthogonal, so \(\Vert \textbf{T}(t)\times \textbf{T}'(t) \Vert = \Vert \textbf{T}(t) \Vert \ \Vert \textbf{T}'(t) \Vert\). This means:
\[\Vert \textbf{r}'(t) \times \textbf{r}''(t) \Vert = \frac{d^2 \ s(t)}{(dt)^2} \Vert \textbf{T}(t) \times \textbf{T}'(t) \Vert = \frac{d^2 \ s(t)}{(dt)^2} \Vert \textbf{T}(t) \Vert \ \Vert \textbf{T}'(t) \Vert \]
Since \(\Vert \textbf{T}(t) \Vert = 1\):
\[\Vert \textbf{r}'(t) \times \textbf{r}''(t) \Vert = \frac{d^2 \ s(t)}{(dt)^2} \Vert \textbf{T}'(t) \Vert \]
This means:
\[\begin{gather} \Vert \textbf{r}'(t) \times \textbf{r}''(t) \Vert = \Vert \textbf{r}'(t) \Vert^2 \Vert \textbf{T}'(t) \Vert \\ \Vert \textbf{r}'(t) \times \textbf{r}''(t) \Vert = \Vert \textbf{r}'(t) \Vert^3 k(t) \end{gather}\]
This leads to the third formula for the curvature:
\[k(t) = \frac{\Vert \textbf{r}'(t) \times \textbf{r}''(t) \Vert}{\Vert \textbf{r}'(t) \Vert^3} \]
Suppose the curve is on a two dimensional plane, and the curve follows is defined by the function \(f(x)\). Then we can define \(\textbf{r}(x) = x \textbf{i} +f(x)\textbf{j} + 0\textbf{k}\). This means:
\[\begin{gather} \textbf{r}'(x) = \textbf{i} + f'(x) \textbf{j} \\ \textbf{r}''(x) = f''(x)\textbf{j}\end{gather}\]
If we take the cross product:
\[ \textbf{r}'(x) \times \textbf{r}''(x) =f''(x) \textbf{k} \]
Therefore:
\[k(x) = \frac{\Vert \textbf{r}'(x) \times \textbf{r}''(x) \Vert}{\Vert \textbf{r}'(x) \Vert^3} = \frac{\Vert f''(x) \textbf{k} \Vert}{\Vert \textbf{r}'(x) \Vert^3}= \frac{| f''(x) |}{\Vert \textbf{r}'(x) \Vert^3} \]
If \(\textbf{r}'(x) = \textbf{i} + f'(x)\textbf{j} \), then \(\Vert \textbf{r}'(x) \Vert = \sqrt{1+ [f'(x)]^2}\):
\[k(x) = \frac{| f''(x) |}{( 1+ [f'(x)]^2 )^{\frac{3}{2}}} \]
This is the fourth formula for the curvature.
%2Fvector-valued_functions%2F2.png&w=3840&q=75)