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Principal Unit Normal Vector

Formulas & Definitions
Core Definition

Let r⃗(t)\vec{r}(t) be a smooth space curve with non-zero curvature (κ>0\kappa > 0) and principal unit tangent vector T^(t)\hat{T}(t) such that T^′(t)≠0⃗\hat{T}'(t) \ne \vec{0}. The principal unit normal vector N^(t)\hat{N}(t) is defined as the derivative of the unit tangent vector normalized by its own magnitude:

N^(t)=T^′(t)∥T^′(t)∥\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|}
By construction, ∥N^(t)∥=1\|\hat{N}(t)\| = 1, and N^(t)\hat{N}(t) is strictly perpendicular to T^(t)\hat{T}(t), pointing inward toward the center of curvature in the direction that the curve is turning.

Concept Explanation

Proof of Orthogonality (T^⋅N^=0\hat{T} \cdot \hat{N} = 0)

Because T^(t)\hat{T}(t) is a unit vector for all tt, its dot product with itself is constant:

∥T^(t)∥2=T^(t)⋅T^(t)=1\|\hat{T}(t)\|^2 = \hat{T}(t) \cdot \hat{T}(t) = 1

Differentiating both sides with respect to tt using the vector product rule:

ddt[T^(t)⋅T^(t)]=T^′(t)⋅T^(t)+T^(t)⋅T^′(t)=0\frac{d}{dt}[\hat{T}(t) \cdot \hat{T}(t)] = \hat{T}'(t) \cdot \hat{T}(t) + \hat{T}(t) \cdot \hat{T}'(t) = 0
2(T^(t)⋅T^′(t))=0  ⟹  T^(t)⋅T^′(t)=02\left(\hat{T}(t) \cdot \hat{T}'(t)\right) = 0 \implies \hat{T}(t) \cdot \hat{T}'(t) = 0

Since N^(t)\hat{N}(t) is a positive scalar multiple of T^′(t)\hat{T}'(t):

T^(t)⋅N^(t)=0\hat{T}(t) \cdot \hat{N}(t) = 0
This proves that the rate of change of any constant-length vector is always orthogonal to the vector itself.

                      z ^
                        |              ^ N(t) (Principal Normal, points inward)
                        |              |
                        |              |
                        |        P(t)  *--------> T(t) (Unit Tangent)
                        |             /
                        |            /  Trajectory r(t)
                        +-----------+----------> y
                       /
                      /
                     v x

The TNB (Frenet-Serret) Moving Frame

Together with the Binormal Vector B^(t)=T^(t)×N^(t)\hat{B}(t) = \hat{T}(t) \times \hat{N}(t), the three mutually perpendicular unit vectors {T^(t),N^(t),B^(t)}\{\hat{T}(t), \hat{N}(t), \hat{B}(t)\} form a right-handed orthonormal coordinate system known as the Frenet-Serret Frame (or TNB frame) that moves along the space curve:

  1. T^\hat{T} (Tangent): Points in the instantaneous direction of motion.
  2. N^\hat{N} (Normal): Points toward the center of curvature in the osculating plane.
  3. B^\hat{B} (Binormal): Points perpendicular to the osculating plane (B^=T^×N^\hat{B} = \hat{T} \times \hat{N}).

Engineering Applications

Example Problems

Example 1: Calculating N^(t)\hat{N}(t) for a 2D Circular Trajectory

Find the principal unit normal vector N^(t)\hat{N}(t) for r⃗(t)=⟨Rcos⁡(t),Rsin⁡(t)⟩\vec{r}(t) = \langle R\cos(t), R\sin(t) \rangle.

Solution
  1. Compute Velocity and Unit Tangent Vector T^(t)\hat{T}(t):
    r⃗′(t)=⟨−Rsin⁡(t),Rcos⁡(t)⟩  ⟹  ∥r⃗′(t)∥=R\vec{r}'(t) = \langle -R\sin(t), R\cos(t) \rangle \implies \|\vec{r}'(t)\| = R
    T^(t)=r⃗′(t)∥r⃗′(t)∥=⟨−sin⁡(t),cos⁡(t)⟩\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|} = \langle -\sin(t), \cos(t) \rangle
  2. Differentiate T^(t)\hat{T}(t):
    T^′(t)=⟨−cos⁡(t),−sin⁡(t)⟩\hat{T}'(t) = \langle -\cos(t), -\sin(t) \rangle
  3. Compute Magnitude ∥T^′(t)∥:\|\hat{T}'(t)\|:
    ∥T^′(t)∥=(−cos⁡(t))2+(−sin⁡(t))2=1\|\hat{T}'(t)\| = \sqrt{(-\cos(t))^2 + (-\sin(t))^2} = 1
  4. Normalize to Find N^(t)\hat{N}(t):
    N^(t)=T^′(t)∥T^′(t)∥=⟨−cos⁡(t),−sin⁡(t)⟩\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|} = \langle -\cos(t), -\sin(t) \rangle
    (Notice N^(t)=−r⃗(t)R\hat{N}(t) = -\frac{\vec{r}(t)}{R}, which points directly inward toward the center of the circle).

Connections & References