# Curvature

There is a natural way to compute the curvature for a curve whose position is given as a function of time, $\vec{r} = \vec{r}(t)$. The following is the vector proof which shows how the velocity and acceleration vectors can be used to compute the curvature at any time.

Is there a natural way to describe the curvature of a function without using the parametric form? For example, suppose a curve is the consequence of the intersection of three volumes defined using an extended Cartesian system with four coordinates: $x, y, z, w$. So, $w = w_1(x,y,z) = w_2(x,y,z) = w_3(x,y,z)$, and we wish to know the curvature at $(x,y,z)$. This is a well-defined question with no reference to a parameter, $t$. It is natural to wonder whether there is a way to find curvature without imposing a new parameter.

My initial thoughts are that we can still define the differential arc length in this approach $ds = \sqrt{dx^2+dy^2+dz^2}$; however, at the point $(x,y,z)$, one’s movement is restricted by the requirement that $w_1(x,y,z)= w_2(x,y,z)=w_3(x,y,z)$. So the differential arc length and the unit tangent vector will reflect this in the solution.

Help me out. Add a solution route in a comment below if you have a solution. I have a compact form for the tangent vector, but I haven’t found a method to get the curvature yet.