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.

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

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Posted in Multivariable Calculus
One comment on “Curvature
  1. […] my last post Curvature, I showed how the curvature, along a path is equal to: . I then asked a […]

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