We have seen that any regular curve can be reparametrized by arclength. Of course, the process involves computing an integral and then describing the inverse of some increasing function. A little bit of Calculus tells us that it is theoretically possible, but it doesn't make it computationally feasible.
The first place we feel the pinch of this trouble is in computing curvature and torsion. We have convenient formulae for finding these functions for a unit speed curve, but if some guy from a dark alley steps out and opens his trenchcoat to reveal... lots of beautiful geometric space curves, and says, "Hey, buddy. Wanna buy a curve?", the chance that you get a curve which is already unit speed is pretty small. I mean, its probably not a real swiss timepiece, even if the guy claims its a Bernoulli Brachistochrone.
So, how can we deal with such a thing?
Now, we can get by for any single curve with the chain rule, but then we have to compute the whole Frenet-Serret apparatus, too. It would be much nicer to have compact and easy to compute formulae for the general case. That is our goal. (Note: I don't recall learning this when I first encountered the subject. I take the below from Struik [1].)
To keep things straight, let us set out our notation very carefully. We will have a curve described by a vector-valued function
Now, we let the algebra machine run its game, and we end up with
\[ \kappa^2 = \langle \dot{x} , \dot{x} \rangle^{-3} \left\{ \langle \dot{x}, \dot{x} \rangle \langle \ddot{x}, \ddot{x} \rangle - \langle\dot{x}, \ddot{x} \rangle \langle \ddot{x}, \dot{x} \rangle \right\} . \]
This puts us in a position to apply the Cauchy-Binet identity, so we get
\[ \kappa^2 = \dfrac{ \langle \dot{x} \times \ddot{x}, \dot{x}\times \ddot{x} \rangle }{\langle \dot{x} , \dot{x} \rangle^{3} } . \]
At some point, I will have to figure out what the geometry is behind that thing. I have seen a proof, but it hasn't really resonated deeply with me, yet. Anyone have a lead?
\[ \tau = \dfrac{(x'\, x''\, x''')}{\langle x'' , x'' \rangle} \]
the denominator is something we have already computed. So we only need to handle the numerator. Since this is a triple product, it is really a determinant. This will allow us to use determinant tricks, like adding a multiple of one row to another without changing the value of the determinant!
But we have seen in equations (1), (2), and (3):
\[ x' = \dfrac{ \dot{x} }{ \langle \dot{x} , \dot{x} \rangle^{1/2} } , \qquad x'' = \dfrac{ \ddot{x} }{ \langle \dot{x}, \dot{x}\rangle } + C \dot{x}, \qquad x''' = \dfrac{\dddot{x} }{\langle \dot{x}, \dot{x}\rangle^{3/2}} + D \ddot{x} + E\dot{x} \]
for some functions $C, D, E$. Hence, by the row operation trick, we have
\[ (x'\, x''\, x''') = \det \begin{pmatrix} \dfrac{\dot{x}}{\langle \dot{x} , \dot{x} \rangle^{1/2} } \\ \dfrac{\ddot{x}}{\langle \dot{x}, \dot{x}\rangle} + C \dot{x} \\ \dfrac{ \dddot{x} }{ \langle \dot{x}, \dot{x}\rangle^{3/2} } + D \ddot{x} + E \dot{x} \end{pmatrix} = \det \begin{pmatrix} \dfrac{ \dot{x}}{\langle \dot{x} , \dot{x} \rangle^{1/2} } \\ \dfrac{\ddot{x}}{\langle \dot{x}, \dot{x}\rangle} \\ \dfrac{\dddot{x} }{\langle \dot{x}, \dot{x}\rangle^{3/2}} \end{pmatrix} = \langle \dot{x}, \dot{x}\rangle^{-3} ( \dot{x} \, \ddot{x} \, \dddot{x} ) . \]
Putting this together with what we found for the curvature, we obtain
\[ \tau = \dfrac{ ( \dot{x} \, \ddot{x} \, \dddot{x} )}{\langle \dot{x} \times \ddot{x}, \dot{x}\times \ddot{x} \rangle } . \]
And that is it. These are now reasonable formulae for our important functions $\kappa$ and $\tau$ that can be computed for any regular curve by taking derivatives with respect to the given parameter.
The first place we feel the pinch of this trouble is in computing curvature and torsion. We have convenient formulae for finding these functions for a unit speed curve, but if some guy from a dark alley steps out and opens his trenchcoat to reveal... lots of beautiful geometric space curves, and says, "Hey, buddy. Wanna buy a curve?", the chance that you get a curve which is already unit speed is pretty small. I mean, its probably not a real swiss timepiece, even if the guy claims its a Bernoulli Brachistochrone.
So, how can we deal with such a thing?
THE CHAIN RULE!
Now, we can get by for any single curve with the chain rule, but then we have to compute the whole Frenet-Serret apparatus, too. It would be much nicer to have compact and easy to compute formulae for the general case. That is our goal. (Note: I don't recall learning this when I first encountered the subject. I take the below from Struik [1].)
To keep things straight, let us set out our notation very carefully. We will have a curve described by a vector-valued function
\[ x: (a, b) \rightarrow \mathbb{R}^3 .\]
We will use $t$ for the parameter of the curve, and we will use the notation $\dot{x}$ to mean the derivative of $x$ with respect to $t$.
Of course, the standard notation for the arclength parameter is $s$, and we will stick with the notation $x'$ for derivatives with respect to arclength so as to match our previous work.
If we take $x$ as if it is arclength parametrized, then we have previously found:
\[ \kappa^2 = \langle x'', x'' \rangle \tag{curvature} \]
\[ \tau = \dfrac{(x'\, x''\, x''')}{\langle x'' , x'' \rangle} . \tag{torsion} \]
We will replace all of the derivatives with respect to $s$ with some taken with respect to $t$. To do so, we just apply the chain rule carefully. To make this easier, we will write $t = u(s) $ for the inverse function that comes up in reparametrizing the curve by arclength.
Watch the first equation carefully, as it has all of the idea in it.
\[ x' = \dfrac{dx}{ds} = \dfrac{dx}{dt}\dfrac{dt}{ds} = \dot{x} u' = \dot{x} (\langle \dot{x}, \dot{x} \rangle )^{-1/2} \tag{1}\]
\[ x'' = \ddot{x} (u')^2 + \dot{x}u'' = \left[\langle \dot{x}, \dot{x}\rangle \ddot{x} - \langle \dot{x}, \ddot{x}\rangle \dot{x} \right] \langle \dot{x}, \dot{x}\rangle^{-2} \tag{2}, \]
and finally,
\[ x''' = \dddot{x} (u')^3 + 3\ddot{x}u'u'' + \dot{x} u''' . \tag{3} \]
Hang on to your hat: the curvature computation
What happens next is that we will take the formulas (curvature) and (torsion) and substitute in equations (1), (2), and (3), and then power through to simplify. Our goal is to get to formulae for $\kappa$ and $\tau$ which do not have any derivatives with respect to $s$ left (the primes), and instead have only derivatives with respect to $t$ (the dots).
\[ \kappa^2 = \langle x'' , x'' \rangle = \langle \dot{x} , \dot{x} \rangle^{-4} \left\langle \left( \langle \dot{x}, \dot{x}\rangle \ddot{x} - \langle \dot{x} , \ddot{x}\rangle \dot{x}\right) , \left( \langle \dot{x}, \dot{x}\rangle \ddot{x} - \langle \dot{x}, \ddot{x}\rangle \dot{x}\right) \right\rangle \]
Now, we let the algebra machine run its game, and we end up with
\[ \kappa^2 = \langle \dot{x} , \dot{x} \rangle^{-3} \left\{ \langle \dot{x}, \dot{x} \rangle \langle \ddot{x}, \ddot{x} \rangle - \langle\dot{x}, \ddot{x} \rangle \langle \ddot{x}, \dot{x} \rangle \right\} . \]
This puts us in a position to apply the Cauchy-Binet identity, so we get
\[ \kappa^2 = \dfrac{ \langle \dot{x} \times \ddot{x}, \dot{x}\times \ddot{x} \rangle }{\langle \dot{x} , \dot{x} \rangle^{3} } . \]
At some point, I will have to figure out what the geometry is behind that thing. I have seen a proof, but it hasn't really resonated deeply with me, yet. Anyone have a lead?
The torsion computation
In our formula for torsion\[ \tau = \dfrac{(x'\, x''\, x''')}{\langle x'' , x'' \rangle} \]
the denominator is something we have already computed. So we only need to handle the numerator. Since this is a triple product, it is really a determinant. This will allow us to use determinant tricks, like adding a multiple of one row to another without changing the value of the determinant!
But we have seen in equations (1), (2), and (3):
\[ x' = \dfrac{ \dot{x} }{ \langle \dot{x} , \dot{x} \rangle^{1/2} } , \qquad x'' = \dfrac{ \ddot{x} }{ \langle \dot{x}, \dot{x}\rangle } + C \dot{x}, \qquad x''' = \dfrac{\dddot{x} }{\langle \dot{x}, \dot{x}\rangle^{3/2}} + D \ddot{x} + E\dot{x} \]
for some functions $C, D, E$. Hence, by the row operation trick, we have
\[ (x'\, x''\, x''') = \det \begin{pmatrix} \dfrac{\dot{x}}{\langle \dot{x} , \dot{x} \rangle^{1/2} } \\ \dfrac{\ddot{x}}{\langle \dot{x}, \dot{x}\rangle} + C \dot{x} \\ \dfrac{ \dddot{x} }{ \langle \dot{x}, \dot{x}\rangle^{3/2} } + D \ddot{x} + E \dot{x} \end{pmatrix} = \det \begin{pmatrix} \dfrac{ \dot{x}}{\langle \dot{x} , \dot{x} \rangle^{1/2} } \\ \dfrac{\ddot{x}}{\langle \dot{x}, \dot{x}\rangle} \\ \dfrac{\dddot{x} }{\langle \dot{x}, \dot{x}\rangle^{3/2}} \end{pmatrix} = \langle \dot{x}, \dot{x}\rangle^{-3} ( \dot{x} \, \ddot{x} \, \dddot{x} ) . \]
Putting this together with what we found for the curvature, we obtain
\[ \tau = \dfrac{ ( \dot{x} \, \ddot{x} \, \dddot{x} )}{\langle \dot{x} \times \ddot{x}, \dot{x}\times \ddot{x} \rangle } . \]
And that is it. These are now reasonable formulae for our important functions $\kappa$ and $\tau$ that can be computed for any regular curve by taking derivatives with respect to the given parameter.
Nice formulas!
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