Vector Calculus: Curvature

1. May 3, 2013

Nikitin

Hey. so you have two formulas for curvature:

The ordinary: |dT/ds| = |a|/|v|2

And the advanced: |v x a|/|v|3 = |a|*sin(α)/|v|2 = |aN|/|v|2

But the problem is, those two formulas aren't the same? The top one has acceleration divided by speed squared, while the bottom one has normal component of acceleration divided by speed squared? where is my mistake?

2. May 3, 2013

jambaugh

The tangential vector T has been normalized which means its length doesn't change and so no tangential component of acceleration appears in its derivative. Remember that the rate of change of a unit vector or any vector of constant length is a vector orthogonal to it. Any derivative of a fixed length vector valued function will be orthogonal.

The two formulas represent two distinct ways at getting just the normal component of acceleration in the expressions.

3. May 3, 2013

Nikitin

Yeah, but that's already baked into the formula I gave...

T = v/|v|. dT/ds = dT*dt/ds*dt = |v|-1*dT/dt = |v|-1*(dv/dt)/|v| = a/|v|2. If I take the absolute value it becomes the formula for curvature. But there, a = d2r/d2t, ie a equals the total acceleration.

Last edited: May 3, 2013
4. May 4, 2013

Nikitin

did I formulate myself unclear in post #3? can some1 pls explain this to me? my exams are coming too fast :((

EDIT: nvm, problem's solved.

Last edited: May 4, 2013