Thread: plane curvature

Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!

Originally Posted by gralla55

Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!

We can express the curvature of a plane curve as , so evaluate this taking into consideration that if we use the arclength parametrization.

Tonio

Could you elaborate? Thanks

Originally Posted by gralla55

Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!

Where,



X

Which is just the cross product from what we wrote above, which is