Curvature in Terms of r', r'', and Their Magnitude

 This article is intended to provide a simpler formula for the curvature in terms of easy-to-solve $r'$, $r''$, $| r' |$, and $|r'' |$. 

Let curve $C$ be defined by $r(t)$ such that

$$r:[a,b] \subseteq \mathbb{R}  \rightarrow \mathbb{R}^n$$

Then we can define Tagent, Normal, and Binormal Unit Vectors to be $$\begin{align*} T &= \frac{{r}'}{|{r}'|} \\  \\ T' &= \left(  \frac{r '}{|r'|} \right)'  \\ &= \frac{|r'|r'' + | r '|' r'}{| r '|^2} \\  |T'| &= \frac{\left |  | r '| r'' + | r' |' r' \right |}{| {r}'|^2} \\ \\ N &= \frac{T'}{| T' |} \\ &= T' \frac{ 1 }{ |T'| } \\ &=  \frac{ | {r}'|  {r}'' + | r' |'  r' }{ | {r}'|^2 }  \frac{ | {r}' |^2 }{ \left |  | {r}'| {r}'' + | {r}'|' {r}' \right | } \\ &= \frac{| {r}'| {r}'' + | {r}'|' {r}'}{\left |  | {r}'|  {r}'' + | {r}'|' {r}' \right |} \\ \\ B &= T\times N \\  &= \frac{ {r}'}{| {r}'|} \times \frac{| {r}'|r'' + | {r}'|'r'}{\left |  | {r}'| r'' + | {r}'|' r' \right |} \\ &= \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( r' \times (|r'| r'' - |r'|' r') \right), \text{ Factor our Scalars.} \\ &= \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( |r'| \, r' \times r'' - |r''| \, r' \times r' \right), \text{Distribute out cross product. }\\ &=  \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( |r'| \, r' \times r'' \right) \\ &= \frac{ r' \times r'' }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \end{align*}$$

Notice that $$|T| = 1 \\ \therefore |T|^2 = T\cdot T = 1$$

$$\begin{align*} \because T \cdot T &= 1 \\ \therefore (T\cdot T)'  &= 0\\ T'\cdot T + T\cdot T' &= 0\\ 2T\cdot T'  &= 0\\ T\cdot \frac{T'}{| T' |} &= 0\\ \therefore T\cdot N &= 0  \end{align*}$$ This shows that $T$ and $N$ are orthogonal. 

$$\therefore |B| = \sin(\theta_{T,N})|T||N| = |T||N|= 1$$

$$\begin{align*} \because |B| &= \frac{ \left |  r' \times r''  \right | }{ \left | |r'| r'' - |r'|' r' \right | } \\ \\ \therefore \left | |r'| r'' - |r'|' r' \right | &=  \left |  r' \times r''  \right | \\ &= \sin(\theta_{r', r''})|r'||r''| \end{align*}$$

By definition of cross product, where $\theta_{r', r''}$ is angle between the vectors $r'$ and $r''$.

$$\begin{align*} \therefore \sin(\theta_{r', r''})|r''| &= \left |  r'' - |r'|' \frac{r'}{|r'|} \right | \\ &=  \left |  r'' - |r'|' T \right | \\ \end{align*}$$

Arclength is defined as $$s(t) = \int_{a}^t | {r}'| dt \\ \therefore s' = | {r}' |$$

Curvature is originally defined as $$\begin{align*}  \kappa &= \left | \frac{dT}{ds} \right | \\ &= \left | \frac{dT}{dt} \frac{dt}{ds}  \right | \\ &= \left | \frac{T'}{s'} \right | \\ \therefore \kappa &= \frac{ \left | T' \right | }{ \left | {r}' \right | } \end{align*}$$

We can define the second derivatives of the curve using the terms above.

$$\begin{align*} r'' &= (r')' \\ &= \left( |r'|T \right)' \\ &= |r'|' T + |r'| T' \\ &= |r'|' T + |r'| |T'| \frac{T'}{|T'|} \\ &= |r'|' T + |r'| |T'| N \\ &= |r'|' T + |r'| ^2\frac{|T'|}{|r'|} N \\ r'' &= |r'|' T + |r'| ^2\kappa N \\ \\ \end{align*}$$

$$\therefore |r'| ^2\kappa N = r'' - |r'|' T$$

Remember that $$\begin{align*} \sin(\theta_{r', r''}) |r''| &= \left |  r'' - |r'|' T \right | \\ &= \left | |r'|^2 \kappa N \right | \\ &= |r'|^2 \kappa \end{align*}$$

So we can redefine Curvature as

$$\kappa = \frac{ \sin(\theta_{r' , r''} ) |r''| }{ |r'|^2 }$$

Where $\theta_{r',r''} \in [0,\pi)$ is the angle between the two vectors $r'$ and $r''$.

Notice that because $\theta_{r',r''} \in [0,\pi)$, 
$$\sin(\theta_{r',r''}) \geq 0$$   We also know from properites of dot product and trig identity that $$\begin{align*} \because \cos(\theta_{r',r''}) &= \frac{r'\cdot r''}{|r'||r''|} \\ \because \sin^2(\theta) + \cos^2(\theta) &= 1 \\ \\ \therefore \sin^2(\theta_{r',r''}) &= 1 - \cos^2(\theta_{r',r''}) \\ &= 1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2 \end{align*}$$

Notice that because $\sin(\theta_{r',r''}) \geq 0$, we can solve for $\sin(\theta_{r',r''})$ by rooting both sides. $$\sin(\theta_{r',r''})  = \sqrt{1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2}$$

Finally, we find that

$$\begin{align*} \kappa &= \frac{ |r''| }{ |r'|^2 } \sqrt{1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2} \\ \\ &= \frac{1}{|r'|^3} \sqrt{ (|r'||r''|)^2 - (r' \cdot r'')^2 }  \end{align*}$$ QED