Question

Exercises 51 and 52 derive the equations necessary to define the torsion τ of a space curve. (Torsion measures the rate at which a space curve twists away from the osculating plane.) In each of these exercises, let C be a space curve and let r(s) be an arc length parametrization for C such that $\frac{d\mathit{T}}{ds}\mathrm{and}\frac{d\mathit{N}}{ds}$exist at every point on C.

Explain why $\frac{d\mathit{B}}{ds}$exists at every point on C. (Hint: Differentiate B = T × N with respect to arc length.)

Short answer

$\frac{dB}{ds}$exists at every point on C because by using the binormal vector we get $\frac{dT}{ds}\mathrm{and}\frac{dN}{ds}$ which exist on curve C.

Step-by-step solution

Step 1. Given Information.

It is given that C is a space curve and letting r(s) be an arc length parametrization for C such that $\frac{d\mathit{T}}{ds}\mathrm{and}\frac{d\mathit{N}}{ds}$exist at every point on C.

Step 2. Explanation.

To explain $\frac{dB}{ds}$exists at every point on C, we will use the binormal vector.

Binormal vector can be defined as let $r\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)$be a differentiable vector function on some interval I ⊆ R such that the derivative of the unit tangent vector $T\text{'}\left(t_0\right)\ne 0where{t}_0\in I.$The binormal vector B at$r\left(t_0\right)$is defined to be $B\left(t_0\right)=T\left(t_0\right)\times N\left(t_0\right).$

Let's solve $\frac{dB}{ds},$

$$\begin{gathered} =\frac{d}{ds}\left(B\right) \\ =\frac{d}{ds}\left(T\times N\right) \\ =\frac{dT}{ds}\times N+\frac{dN}{ds}\times T \end{gathered}$$

Now, as we know $\frac{dT}{ds}\mathrm{and}\frac{dN}{ds}$are the derivatives of the unit tangent vector and principal unit normal vector with respect to the arc lengths.

All these points exist on curve C. Thus,$\frac{dB}{ds}$exist at every point on C.