# 8. Properties of Curves

## 1. Frenet Frames & Derivatives of TNB

$$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ make up what is called the Frenet Frame. This is a right handed set of $$3$$ mutually orthogonal unit vectors that moves along the curve, as opposed to the $$\hat{\imath},\hat{\jmath},\hat{k}$$ unit vectors which make up the fixed standard reference frame for $$\mathbb{R}^3$$.

### Algebraic Properties

The Frenet Frame can be used much like the standard frame. As a right handed set of $$3$$ orthogonal unit vectors, $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ satisfy: \begin{aligned} \hat{T}\cdot\hat{T}&=1 \qquad &\hat{N}\cdot\hat{N}&=1 \qquad &\hat{B}\cdot\hat{B}&=1 \\ \hat{T}\cdot\hat{N}&=0 \qquad &\hat{N}\cdot\hat{B}&=0 \qquad &\hat{B}\cdot\hat{T}&=0 \qquad (*) \\ \hat{T}\times\hat{N}&=\hat{B} \qquad &\hat{N}\times\hat{B}&=\hat{T} \qquad &\hat{B}\times\hat{T}&=\hat{N} \end{aligned}

Suppose $$\vec u=u_T\hat T + u_N\hat N + u_B\hat B$$ and $$\vec v=v_T\hat T + v_N\hat N + v_B\hat B$$.
Use the distributive rule, the commutative rule and conditions (*) to compute $$\vec u\cdot\vec v$$. What do you notice?

$$\displaystyle \vec u\cdot\vec v =u_Tv_T + u_Nv_N + u_Bv_B$$
As with using components relative to the standard basis, the dot product is the sum of the pairwise products of the components relative to the Frenet basis.

Using the distributive rule: \begin{aligned} \vec u\cdot\vec v &=(u_T\hat T + u_N\hat N + u_B\hat B)\cdot(u_T\hat T + u_N\hat N + u_B\hat B) \\ &= u_Tv_T\hat T\cdot\hat T + u_Tv_N\hat T\cdot\hat N + u_Tv_B\hat T\cdot\hat B \\ &\quad + u_Nv_T\hat N\cdot\hat T + u_Nv_N\hat N\cdot\hat N + u_Nv_B\hat N\cdot\hat B \\ &\quad + u_Bv_T\hat B\cdot\hat T + u_Bv_N\hat B\cdot\hat N + u_Bv_B\hat B\cdot\hat B \\ \end{aligned} Then using (*) and the commutative rule: \begin{aligned} \vec u\cdot\vec v &= u_Tv_T(1) + u_Tv_N(0) + u_Tv_B(0) \\ &\quad + u_Nv_T(0) + u_Nv_N(1) + u_Nv_B(0) \\ &\quad + u_Bv_T(0) + u_Bv_N(0) + u_Bv_B(1) \\ &=u_Tv_T + u_Nv_N + u_Bv_B \\ \end{aligned} As with using components relative to the standard basis, the dot product is the sum of the pairwise products of the components relative to the Frenet basis.

Suppose $$\vec u=u_T\hat T + u_N\hat N + u_B\hat B$$ and $$\vec v=v_T\hat T + v_N\hat N + v_B\hat B$$.

1. Use the distributive rule, the anti-commutative rule and conditions (*) to compute $$\vec u\times\vec v$$.

$$\vec u\times\vec v =(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B$$

Using the distributive rule: \begin{aligned} \vec u\times\vec v &=(u_T\hat T + u_N\hat N + u_B\hat B)\times(u_T\hat T + u_N\hat N + u_B\hat B) \\ &= u_Tv_T\hat T\times\hat T + u_Tv_N\hat T\times\hat N + u_Tv_B\hat T\times\hat B \\ &\quad + u_Nv_T\hat N\times\hat T + u_Nv_N\hat N\times\hat N + u_Nv_B\hat N\times\hat B \\ &\quad + u_Bv_T\hat B\times\hat T + u_Bv_N\hat B\times\hat N + u_Bv_B\hat B\times\hat B \\ \end{aligned} Then using (*) and the anti-commutative rule: \begin{aligned} \vec u\times\vec v &= u_Tv_T(\vec 0) + u_Tv_N(\vec B) + u_Tv_B(-\vec N) \\ &\quad + u_Nv_T(-\vec B) + u_Nv_N(\vec 0) + u_Nv_B(\vec T) \\ &\quad + u_Bv_T(\vec N) + u_Bv_N(-\vec T) + u_Bv_B(\vec 0) \\ &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \end{aligned}

2. Compute \begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} \end{aligned} by expanding on the first row. What do you notice?

\begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \\ \end{aligned}
We notice, this is the same as $$\vec u\times\vec v$$.

\begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} &=\hat T \begin{vmatrix} u_N & u_B \\ v_N & v_B \end{vmatrix} -\hat N \begin{vmatrix} u_T & u_B \\ v_T & v_B \end{vmatrix} +\hat B \begin{vmatrix} u_T & u_N \\ v_T & v_N \end{vmatrix} \\ &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \\ \end{aligned} We notice, this is the same as $$\vec u\times\vec v$$.

These two exercises show that the dot and cross products can be computed using components relative to $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ in exactly the same way as they are computed using components relative to $$\hat\imath$$, $$\hat\jmath$$ and $$\hat k$$.

### Differential Properties

We now turn to the derivative properties of Frenet frames. First notice that a vector field, $$\vec F(t)$$, along a curve, $$\vec r(t)$$, can be expanded in either the $$\hat\imath\hat\jmath\hat k$$-basis or the $$\hat{T}\hat{N}\hat{B}$$-basis: \begin{aligned} \vec F&=F_1\hat\imath + F_2\hat\jmath + F_3\hat k \\ &=F_T\hat T + F_N\hat N + F_B\hat B \end{aligned} We want to differentiate $$\vec F$$. In the $$\hat\imath\hat\jmath\hat k$$ form, since $$\hat\imath$$, $$\hat\jmath$$ and $$\hat k$$ are constant, we only need to differentiate the coefficients: $\dfrac{d\vec F}{dt} =\dfrac{dF_1}{dt}\hat\imath + \dfrac{dF_2}{dt}\hat\jmath + \dfrac{dF_3}{dt}\hat k$ However, in the $$\hat{T}\hat{N}\hat{B}$$ form, $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ are not constant, and so we also need to differentiate them. By the Product Rule: \begin{aligned} \dfrac{d\vec F}{dt} &=\dfrac{dF_T}{dt}\hat T + F_T\dfrac{d\hat T}{dt} \\ &\quad+ \dfrac{dF_N}{dt}\hat N + F_N\dfrac{d\hat N}{dt} \\ &\quad\qquad+ \dfrac{dF_B}{dt}\hat B + F_B\dfrac{d\hat B}{dt} \end{aligned} The Frenet formulas tell us the derivatives of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ (with respect to arclength, $$s$$) written as linear combinations of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$. The coefficients turn out to be the curvature, $$\kappa$$, and the torsion, $$\tau$$. The derivatives with respect to time, $$t$$, are given as a corollary.

The derivatives of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ satisfy: \begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\,\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\,\tau \hat{N} \end{aligned}

First, note that each of the derivatives can be written as a linear combination of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$: \begin{aligned} \dfrac{d\hat{T}}{ds}&=a\hat{T}+b\hat{N}+c\hat{B} \\ \dfrac{d\hat{N}}{ds}&=d\hat{T}+e\hat{N}+f\hat{B} \\ \dfrac{d\hat{B}}{ds}&=g\hat{T}+h\hat{N}+i\hat{B} \end{aligned} for some coefficients $$a$$ through $$i$$ which we would like to determine. Since $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ are mutually perpendicular unit vectors, the coefficients can be found from: $\begin{array}{ccc} a=\hat{T}\cdot\dfrac{d\hat{T}}{ds}& b=\hat{N}\cdot\dfrac{d\hat{T}}{ds}& c=\hat{B}\cdot\dfrac{d\hat{T}}{ds} \\ d=\hat{T}\cdot\dfrac{d\hat{N}}{ds}& e=\hat{N}\cdot\dfrac{d\hat{N}}{ds}& f=\hat{B}\cdot\dfrac{d\hat{N}}{ds} \\ g=\hat{T}\cdot\dfrac{d\hat{B}}{ds}& h=\hat{N}\cdot\dfrac{d\hat{B}}{ds}& i=\hat{B}\cdot\dfrac{d\hat{B}}{ds} \end{array}$ To find the diagonal coefficients, suppose $$\hat{u}$$ is a unit vector. Then $$\hat{u}\cdot\hat{u}=1$$ and consequently: $0=\dfrac{d}{ds}(\hat{u}\cdot\hat{u}) =\dfrac{d\hat{u}}{ds}\cdot\hat{u}+\hat{u}\cdot\dfrac{d\hat{u}}{ds} =2\hat{u}\cdot\dfrac{d\hat{u}}{ds}$ Applying this with $$\hat{u}=\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ gives $$a=e=i=0$$. Next suppose $$\hat{u}$$ and $$\hat{v}$$ are perpendicular. Then $$\hat{u}\cdot\hat{v}=0$$ and consequently: $0=\dfrac{d}{ds}(\hat{u}\cdot\hat{v}) =\dfrac{d\hat{u}}{ds}\cdot\hat{v}+\hat{u}\cdot\dfrac{d\hat{v}}{ds} \quad \text{or} \quad \hat{u}\cdot\dfrac{d\hat{v}}{ds} =-\,\dfrac{d\hat{u}}{ds}\cdot\hat{v}$ Applying this with $$\hat{u}$$ and $$\hat{v}$$ being two of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ gives $$b=-d$$ and $$c=-g$$ and $$f=-h$$. This reduces the system of equations to: \begin{aligned} \dfrac{d\hat{T}}{ds}&=\,\qquad\; \quad b \hat{N} + c \hat{B} \qquad\text{(1)} \\ \dfrac{d\hat{N}}{ds}&=-\,b \hat{T} \quad\; + \quad\; f \hat{B} \qquad\text{(2)} \\ \dfrac{d\hat{B}}{ds}&=-\,c \hat{T} - f \hat{N} \qquad\qquad\;\,\text{(3)} \end{aligned} Next, we recall the definition of the curvature and one of the formulas for $$\hat{N}$$: $\kappa=\left|\dfrac{d\hat{T}}{ds}\right| \qquad \hat{N}=\dfrac{\dfrac{d\hat{T}}{ds}}{\left|\dfrac{d\hat{T}}{ds}\right|}$ Consequently: $\dfrac{d\hat{T}}{ds}=\kappa\hat{N}$ Comparing this to equation (1), we find $$b=\kappa$$ and $$c=0$$. Finally, from the definition of the torsion, we have: $\tau=-\,\dfrac{d\hat{B}}{ds}\cdot\hat{N} =-(-f\hat{N})\cdot\hat{N}=f$ Thus $$f=\tau$$. Putting the values of $$b=\kappa$$, $$c=0$$ and $$f=\tau$$ into (1), (2) and (3) gives the final formulas as: \begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\,\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\,\tau \hat{N} \end{aligned}

Since $$\dfrac{d}{dt}=\dfrac{ds}{dt}\dfrac{d}{ds}=|\vec v|\dfrac{d}{ds}$$ we conclude:

The derivatives of $$\hat{T}$$, $$\hat{N}$$ and $$\hat{B}$$ satisfy: \begin{aligned} \dfrac{d\hat{T}}{dt}&=\qquad \quad |\vec v|\kappa \hat{N} \\ \dfrac{d\hat{N}}{dt}&=-\,|\vec v|\kappa \hat{T} \quad + \quad |\vec v|\tau \hat{B} \\ \dfrac{d\hat{B}}{dt}&=\qquad -\,|\vec v|\tau \hat{N} \end{aligned}

Supported in part by NSF Grant #1123255 