Question

What makes Definition 11.21 for curvature easy to understand? What makes it difficult to use?

Short answer

The reason has been explained.

Step-by-step solution

Step 1. Given information.

We have to explain What makes Definition 11.21 for curvature easy to understand and What makes it difficult to use.

Step 2. Definition

The definition of curvature is, let c be the graph of a vector function r(s) defined on an interval I, parametrized by arc length s, and with unit tangent vector T.

The curvature k of c at a point on the curve is the scalar given by $k=∥\frac{dT}{ds}∥$i.e. the rate of change of the tangent vector with respect to arc length is the curvature.

Step 3. Explanation

Let a helix is defined by $\mathrm{r}\left(\mathrm{t}\right)=⟨\sin \alpha t,\cos \alpha t,\beta t⟩\text{on}[\mathrm{a},\mathrm{b}]\text{where}\alpha ,\beta ,a\text{and}b$

The derivative is $r^{\mathrm{\prime }}\left(t\right)=⟨\alpha \cos \alpha t,-\alpha \sin \alpha t,\beta ⟩$

The arc length is :

$\begin{array}{r}{\int }_a^b ∥r^{\mathrm{\prime }}\left(t\right)∥dt={\int }_a^b \sqrt{{\alpha }^2{\cos }^2\alpha t+{\alpha }^2{\sin }^2\alpha t+{\beta }^2}dt\\ ={\int }_a^b \sqrt{{\alpha }^2+{\beta }^2}dt\\ {\left.=\sqrt{\left({\alpha }^2+{\beta }^2\right)t}\right]}_a^b\\ =\sqrt{\left({\alpha }^2+{\beta }^2\right)}(b-a)\end{array}$

Thus, the arc length of r(t) is$(b-a)\sqrt{{\alpha }^2+{\beta }^2}$