Curvature — Question 10

Question 10

Let $\mathbf r(s)$ be a twice-differentiable curve parametrized by arc length, so $\|\mathbf r'(s)\|=1$.

Tasks

1. Prove that $\mathbf r''(s)$ is perpendicular to $\mathbf r'(s)$.

2. Prove that $\kappa(s)=\|\mathbf r''(s)\|$.

3. Explain what $\mathbf r''(s)=\mathbf 0$ means locally.

Question 10 – Solution

Strategy. Differentiate the unit-speed invariant, then compare the definition $\kappa=\|d\mathbf T/ds\|$ with $\mathbf T=\mathbf r'$.

Step 1: Orthogonality Unit speed means $$\mathbf r'(s)\cdot\mathbf r'(s)=1.$$ Differentiate both sides: $$2\mathbf r'(s)\cdot\mathbf r''(s)=0.$$ Therefore $$\boxed{\mathbf r'(s)\cdot\mathbf r''(s)=0}.$$ Thus the change of the unit tangent has no tangential component.

Step 2: Curvature formula Because the parameter is arc length, $$\mathbf T(s)=\frac{\mathbf r'(s)}{\|\mathbf r'(s)\|}=\mathbf r'(s).$$ Consequently $$\kappa(s)=\left\|\frac{d\mathbf T}{ds}\right\| =\|\mathbf r''(s)\|,$$ so $$\boxed{\kappa(s)=\|\mathbf r''(s)\|}.$$

Step 3: Zero second derivative If $\mathbf r''=\mathbf 0$ throughout a neighborhood, then $\mathbf r'$ is a constant unit vector there. Integrating gives $\mathbf r(s)=\mathbf p+s\mathbf u$ locally, a straight line traversed at unit speed. At a single point, $\mathbf r''(s_0)=0$ means zero curvature at that point, but does not by itself force an entire neighboring segment to be straight.