Tangent, Normal and Binormal Vectors β€” Question 3

Section 14-8 Β· Calculus pages 2785–2786

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Question 3

Consider the plane curve 𝒓(t)=⟨t,t2,0⟩.\mathbf r(t)=\left\langle t,\ t^2,\ 0\right\rangle. At t=1t=1, determine 𝑻\mathbf T, 𝑡\mathbf N, and 𝑩\mathbf B.

Tasks

  1. Differentiate and normalize to find 𝑻(1)\mathbf T(1).

  2. Differentiate the general 𝑻(t)\mathbf T(t) before evaluating to find 𝑡(1)\mathbf N(1).

  3. Compute 𝑩(1)\mathbf B(1) and interpret its direction.

Question 3 – Solution

Strategy. Computing 𝑻′(1)\mathbf T'(1) requires differentiating the normalized tangent, not merely normalizing 𝒓″(1)\mathbf r''(1).

Step 1: Unit tangent 𝒓′(t)=⟨1,2t,0⟩,𝑻(t)=⟨1,2t,0⟩1+4t2.\mathbf r'(t)=\left\langle 1,2t,0\right\rangle,\qquad \mathbf T(t)=\frac{\left\langle 1,2t,0\right\rangle}{\sqrt{1+4t^2}}. Hence 𝑻(1)=15⟨1,2,0⟩.\boxed{\mathbf T(1)=\frac 1{\sqrt 5}\left\langle 1,2,0\right\rangle}.

Step 2: Differentiate 𝑻\mathbf T Componentwise differentiation gives 𝑻′(t)=1(1+4t2)3/2βŸ¨βˆ’4t,2,0⟩.\mathbf T'(t)=\frac 1{(1+4t^2)^{3/2}}\left\langle -4t,2,0\right\rangle. At t=1t=1 this vector is a positive multiple of βŸ¨βˆ’2,1,0⟩\left\langle -2,1,0\right\rangle, whose length is 5\sqrt 5. Therefore 𝑡(1)=15βŸ¨βˆ’2,1,0⟩.\boxed{\mathbf N(1)=\frac 1{\sqrt 5}\left\langle -2,1,0\right\rangle}.

Step 3: Binormal 𝑩(1)=𝑻(1)×𝑡(1)=15⟨1,2,0βŸ©Γ—βŸ¨βˆ’2,1,0⟩=⟨0,0,1⟩.\mathbf B(1)=\mathbf T(1)\times\mathbf N(1) =\frac 15\left\langle 1,2,0\right\rangle\times\left\langle -2,1,0\right\rangle =\boxed{\left\langle 0,0,1\right\rangle}. The curve lies in the xyxy-plane, so its binormal points perpendicular to that plane. The three vectors are unit length and mutually perpendicular.