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.


I am a math and science teacher at a boarding school in Delaware.

Posted in Multivariable Calculus

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