Curvature β€” Question 6

PDF β†—

Question 6

At a point of a regular curve, 𝒓′=⟨2,βˆ’1,2⟩,𝒓″=⟨1,2,0⟩.\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 𝒓″\mathbf r'' into components parallel and perpendicular to 𝒓′\mathbf r'.

  3. Verify curvature using ΞΊ=βˆ₯π’“βŸ‚β€³βˆ₯/βˆ₯𝒓′βˆ₯2\kappa=\|\mathbf r''_\perp\|/\|\mathbf r'\|^2.

Original worksheet page 1: question and worked solution for 1-10-006
Show solutionHide solution

Question 6 – Solution

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

Step 1: Cross-product computation 𝒓′×𝒓″=⟨2,βˆ’1,2βŸ©Γ—βŸ¨1,2,0⟩=βŸ¨βˆ’4,2,5⟩.\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 16+4+25=35\sqrt{16+4+25}=3\sqrt 5, while βˆ₯𝒓′βˆ₯=3\|\mathbf r'\|=3. Therefore ΞΊ=3533=59.\boxed{\kappa=\frac{3\sqrt 5}{3^3}=\frac{\sqrt 5}{9}}.

Step 2: Parallel component Since 𝒓″⋅𝒓′=2βˆ’2+0=0,\mathbf r''\cdot\mathbf r'=2-2+0=0, the projection of 𝒓″\mathbf r'' onto 𝒓′\mathbf r' is 𝒓βˆ₯β€³=𝒓″⋅𝒓′βˆ₯𝒓′βˆ₯2𝒓′=𝟎.\mathbf r''_{\parallel} =\frac{\mathbf r''\cdot\mathbf r'}{\|\mathbf r'\|^2}\mathbf r' =\mathbf 0. Thus π’“βŸ‚β€³=𝒓″=⟨1,2,0⟩\mathbf r''_\perp=\mathbf r''=\left\langle 1,2,0\right\rangle and βˆ₯π’“βŸ‚β€³βˆ₯=5\|\mathbf r''_\perp\|=\sqrt 5.

Step 3: Verification βˆ₯π’“βŸ‚β€³βˆ₯βˆ₯𝒓′βˆ₯2=59,\frac{\|\mathbf r''_\perp\|}{\|\mathbf r'\|^2} =\frac{\sqrt 5}{9}, which agrees exactly with the cross-product result.

Original worksheet page 2: question and worked solution for 1-10-006

Original worksheet layout. Use Enlarge or open the PDF for a closer view.