Orthogonal derivative of binormal vector.

[Locoism]

Homework Statement

Suppose that r(s) defines a curve parametrically with respect to arc length and the r′(s) is nonzero on the curve. Show that dB/ds is orthogonal to both
B(s) and T(s). Conclude that there is a scalar function τ(s) such that
dB/ds = −τ (s)N . (This function τ is known as the torsion of the curve.)

The Attempt at a Solution

Not really sure how to approach this one. I'd say we need to show some way that dB/ds x B = 0 and the same with T, but I'm not sure how to do it completely algebraically... As for τ, is it enough just to show that dB/ds is parallel to N? This would simply follow from the proof that it is orthogonal to B and T, since T x B = N (or T x N = B)

 

[mighty2000]

and therefore dB/ds x N = -dB/ds x T = τ(s).


Hello, thank you for your post. Let's start by defining some terms. The curve r(s) is a parametric curve defined with respect to arc length, which means that the parameter s represents the distance along the curve. The vector r'(s) is the tangent vector to the curve at any given point, and it is nonzero on the curve because the curve is smooth.

To show that dB/ds is orthogonal to both B(s) and T(s), we can use the fact that the derivative of a vector function is always orthogonal to the vector itself. In this case, we have:

(dB/ds) · B = dB/ds · (r' x r'') = (dB/ds x r') · r'' = 0

Similarly, we can show that (dB/ds) · T = 0. This means that dB/ds is orthogonal to both B and T, and therefore parallel to N.

To show that there is a scalar function τ(s) such that dB/ds = -τ(s)N, we can use the fact that the derivative of B(s) is equal to -τ(s)N. This means that τ(s) = -|dB/ds|, where |dB/ds| is the magnitude of the derivative of B. This function τ(s) is known as the torsion of the curve, and it measures the rate of change of the direction of the curve.

I hope this helps! Let me know if you have any further questions.