Question

Use Exercise 52 to show that $\frac{d\mathbf{N}}{ds}=-k\mathbf{T}+\zeta \mathbf{B},$ where κ is the curvature. (Hint: Differentiate N = B × T with respect to arc length.)

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

The Relationship between$\frac{dT}{ds},\kappa ,\mathrm{and}N$can be defined as letting C be the graph of a vector function r(s) defined on an intervalI and parametrized by

arc length s. If the unit tangent vector T has a nonzero derivative, then $\frac{dT}{ds}=\kappa N.$

Step 2. Showing that dNds=-kT+ζB.

To show that $\frac{d\mathbf{N}}{ds}=-k\mathbf{T}+\zeta \mathbf{B}\mathbf{,}$ where κ is the curvature, we will use the relationship between $\frac{dT}{ds},\kappa ,\mathrm{and}N.$

By using the relationship,

role="math" localid="1649676899612" $$\begin{gathered} \frac{dT}{ds} \\ =\frac{d}{ds}\left(N\right) \\ =\frac{d}{ds}\left(B\times T\right)\left[N=B\times T\right] \\ =\frac{dB}{ds}\times T+B\times \frac{dT}{ds} \\ =-\zeta N\times T+B\times kN\left[\frac{dB}{ds}=-\zeta N\mathrm{and}\frac{dT}{ds}=kN\right] \\ =\zeta B-kT\left[B\times N=-Tand-N\times T=B\right] \end{gathered}$$

Hence proved.