# Curvature (part 2)

In my last post Curvature, I showed how the curvature, $\kappa = \|\frac{d\vec{T}}{ds}\|$ along a path is equal to: $\kappa = \frac{\|\vec{a}\times\vec{v}\|}{\|\vec{v}\|^3}$. I then asked a “challenge” question about how one could compute the curvature without referencing a velocity or acceleration vector. I have slightly rephrased the question below:

Is there a natural way to describe the curvature of a function without using the parametric form? For example, suppose there exists a curve defined by the intersection of three volumes, which are 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 a point $(x,y,z)$. 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.

## Spoiler (my work on the problem)

See if your work agrees with mine:

Part 1: Show that: $\vec{T} = \frac{\nabla w_3 \times \nabla w_1 +\nabla w_1 \times \nabla w_2 +\nabla w_2 \times \nabla w_3}{\|\nabla w_3 \times \nabla w_1 +\nabla w_1 \times \nabla w_2 +\nabla w_2 \times \nabla w_3\|}$

Part 2: Show that: $\kappa = \|\vec{T} \cdot \nabla \vec{T}\|$, where $\vec{T}$ is defined above.

Part 3: Use Parts 1 and 2 to show that the path which satisfies the system of equations: $w_1= x^2 + y^2 + z^2=w_2=x^2 + y^2=w_3=4+z$ is a circle of radius two; and the curvature at any point on this path is $\kappa = \frac{1}{2}$, as expected since $\kappa = \frac{1}{R}$ for a circle of radius, $R$.