Curvature β€” Question 10

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Question 10

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

Tasks

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

  2. Prove that ΞΊ(s)=βˆ₯𝒓″(s)βˆ₯\kappa(s)=\|\mathbf r''(s)\|.

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

Original worksheet page 1: question and worked solution for 1-10-010
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Question 10 – Solution

Strategy. Differentiate the unit-speed invariant, then compare the definition ΞΊ=βˆ₯d𝑻/dsβˆ₯\kappa=\|d\mathbf T/ds\| with 𝑻=𝒓′\mathbf T=\mathbf r'.

See the diagram in the original worksheet below.

Step 1: Orthogonality Unit speed means 𝒓′(s)⋅𝒓′(s)=1.\mathbf r'(s)\cdot\mathbf r'(s)=1. Differentiate both sides: 2𝒓′(s)⋅𝒓″(s)=0.2\mathbf r'(s)\cdot\mathbf r''(s)=0. Therefore 𝒓′(s)⋅𝒓″(s)=0.\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, 𝑻(s)=𝒓′(s)βˆ₯𝒓′(s)βˆ₯=𝒓′(s).\mathbf T(s)=\frac{\mathbf r'(s)}{\|\mathbf r'(s)\|}=\mathbf r'(s). Consequently ΞΊ(s)=βˆ₯d𝑻dsβˆ₯=βˆ₯𝒓″(s)βˆ₯,\kappa(s)=\left\|\frac{d\mathbf T}{ds}\right\| =\|\mathbf r''(s)\|, so ΞΊ(s)=βˆ₯𝒓″(s)βˆ₯.\boxed{\kappa(s)=\|\mathbf r''(s)\|}.

Step 3: Zero second derivative If 𝒓″=𝟎\mathbf r''=\mathbf 0 throughout a neighborhood, then 𝒓′\mathbf r' is a constant unit vector there. Integrating gives 𝒓(s)=𝒑+s𝒖\mathbf r(s)=\mathbf p+s\mathbf u locally, a straight line traversed at unit speed. At a single point, 𝒓″(s0)=0\mathbf r''(s_0)=0 means zero curvature at that point, but does not by itself force an entire neighboring segment to be straight.

Original worksheet page 2: question and worked solution for 1-10-010

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