Curvature — Question 6

Question 6

At a point of a regular curve, $$\mathbf r'=\left\langle 2,-1,2\right\rangle,\qquad \mathbf r''=\left\langle 1,2,0\right\rangle.$$ Tasks

1. Compute the curvature.

2. Decompose $\mathbf r''$ into components parallel and perpendicular to $\mathbf r'$.

3. Verify curvature using $\kappa=\|\mathbf r''_\perp\|/\|\mathbf r'\|^2$.

Question 6 – Solution

Strategy. Compare the cross-product formula with the geometric perpendicular-acceleration formula.

Step 1: Cross-product computation $$\mathbf r'\times\mathbf r'' =\left\langle 2,-1,2\right\rangle\times\left\langle 1,2,0\right\rangle=\left\langle -4,2,5\right\rangle.$$ Its length is $\sqrt{16+4+25}=3\sqrt 5$, while $\|\mathbf r'\|=3$. Therefore $$\boxed{\kappa=\frac{3\sqrt 5}{3^3}=\frac{\sqrt 5}{9}}.$$

Step 2: Parallel component Since $$\mathbf r''\cdot\mathbf r'=2-2+0=0,$$ the projection of $\mathbf r''$ onto $\mathbf r'$ is $$\mathbf r''_{\parallel} =\frac{\mathbf r''\cdot\mathbf r'}{\|\mathbf r'\|^2}\mathbf r' =\mathbf 0.$$ Thus $\mathbf r''_\perp=\mathbf r''=\left\langle 1,2,0\right\rangle$ and $\|\mathbf r''_\perp\|=\sqrt 5$.

Step 3: Verification $$\frac{\|\mathbf r''_\perp\|}{\|\mathbf r'\|^2} =\frac{\sqrt 5}{9},$$ which agrees exactly with the cross-product result.