Proving that $k(t)=\frac{|\alpha'\wedge\alpha''|}{|\alpha'|^3}.$

Question

I understand this question has already been asked, but I don't exactly follow that answer. I saw I had a similar process but I got lost in their explanation.

Following do Carmo, Section 1.5-Exercise 12

Let $α:I→R^3$ be a regular parametrized curve (not necessarily by arc length) and let $β:J→R^3$ be a reparametrization of $α(I)$ by the arc length $s=s(t)$, measured from $t_0∈I$. Let $t=t(s)$ be the inverse function of $s$ and set $dα/dt=α′$, $d^2α/dt^2=α′′$, etc.

Prove that the curvature of $\alpha$ at $t\in I$ is

$$k(t)=\frac{|\alpha'\wedge\alpha''|}{|\alpha'|^3}.$$

So, my first thought it to start with what we already know

$$s=s(t)=\int_{t_0}^t|\alpha'(t)|dt\implies s'(t)=|\alpha'(t)|.$$

If we rewrite this is a differentiable form, we get

$$s'(t)=\frac{\alpha'(t)\cdot\alpha'(t)}{|\alpha'(t)|}.$$

Differentiating both sides of the equation we get

$$s''(t)=\frac{2\alpha'(t)\cdot\alpha''(t)}{|\alpha'(t)|}\frac{dt}{ds}\implies s'''(t)=2\frac{\alpha'(t)\cdot\alpha''(t)}{|\alpha'(t)|}.$$

But now, I have no idea what I'm supposed to do? I don't know how I'm supposed to get the $k(t)$ from this equation.

Answer 1

I'm not sure if this will be entirely helpful to you just yet, but I will first try and explain the answer given here as you had mentioned in the OP.

We start with as you say $s'(t) = \lvert \alpha'(t) \rvert $. Since the latter is a scalar, we can scale other vectors by $s'$, and in particular by $\frac{1}{s'}$. In particular we can do this to the tangent vector $\alpha'$ to obtain $\frac{1}{s'} \cdot \alpha'$ (I write it in this way to emphasise the nature of $1/s'$), and differentiate this entire vector to obtain (a multiple of) the normal vector, $$\left[\frac{1}{s'}\cdot \alpha'\right]'.$$ Finally I multiply this once more by $1/s'$, which gives me the left-hand side of equation (1) in the answer. Choose your favourite method of computing the derivative of $1/s' \cdot \alpha'$. I choose the quotient rule, so we get $$ \frac{1}{s'} \left[ \frac{1}{s'} \cdot \alpha'\right]' = \frac{\alpha''s' - \alpha's''}{(s')^3}.$$

Note here: it appears you may be confused on the notation. The denominator here is $s'^3$, the cubic power of $s'$. The convention for $s'''$ is the third derivative of $s$, sometimes also written as $s^(3)$, and these two things are (as you would know) usually not the same thing.

Then, since $\frac{1}{s'} \alpha' = \frac{1}{\vert \alpha' \vert} \alpha'$ is a unit vector, its derivative is perpendicular (note we don't actually need it to be a unit vector, just a vector of fixed length). By the linearity of the wedge (cross) product and its distributivity over vector addition (and subtraction), we have (you should do this calculation yourself, it is very straightforward) $$\frac{1}{s'} \alpha' \times \frac{1}{s'^3} (\alpha''s' - \alpha's'') = \frac{1}{s'^3} (\alpha' \times \alpha'').$$

In this setup, we have a formula for the unit tangent vector (the first term in the product above) and the normal vector (second term). Since the length of the cross product of a unit vector $u$ and any other vector $v$ is the length of $v$ itself (this you should know), we know the length of the normal vector is the length of the vector given in the right hand side of the above formula. This is precisely the curvature.

Answer 2

The selected answer is good, however, for those that would like a more "algebraic" derivation, I have thought of including it here.
To simplify notation, let the derivatives with respect to the t parameter be represented with dots and the derivatives with respect to arclength be represented with primes.
First, the desired identity expresses curvature in terms of the t derivatives of our curve. Hence, to obtain it we should begin to relate those derivatives with the arclength derivatives from which curvature is more naturally defined. Let us then compute the second derivative w/respect to t. $$\frac{d^2 \alpha}{{dt}^2}=\frac{d}{dt}\left(\frac{d\alpha}{ds}\frac{ds}{dt}\right)=\frac{d^2\alpha}{dt\,ds}\frac{ds}{dt}+\frac{d\alpha}{ds}\frac{d^2s}{{dt}^2}=\left(\frac{ds}{dt}\right)^2\frac{d^2\alpha}{{ds}^2}+\frac{d\alpha}{ds}\frac{d^2s}{{dt}^2}=(\dot{s})^2\alpha''+\alpha'\ddot{s}$$ Because the curvature is the magnitude of $\alpha''$, we solve for it. $$\alpha''=\frac{\ddot{\alpha}-\ddot{s}\alpha'}{{\dot{s}}^2}$$ Now, $\alpha'={\dot{s}}^{-1}\dot{\alpha}$. We multiply and divide $\alpha''$ by $\dot{s}^2$ to obtain $$\alpha''=\frac{\dot{s}^2\ddot{\alpha}-\ddot{s}\dot{s}\dot{\alpha}}{\dot{s}^4}$$ We're almost done. Now we pursue a known cross product identity that states $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$. We can see that our equation almost matches that identity. All that's left is to somehow express the scalars as dot products of vectors. We know that $\dot{\alpha}\cdot \dot{\alpha}=\dot{s}^2$. The second scalar can be rewritten by noting that $|\dot{\alpha}|=|\alpha'|\dot{s}=\dot{s}$. By squaring both sides and differentiating with respect to t we obtain $\dot{\alpha} \cdot \ddot{\alpha}=\dot{s}\ddot{s}$. Our equation then becomes, and by the cross product identity, $$\alpha''=\frac{(\dot{\alpha}\cdot \dot{\alpha})\ddot{\alpha}-(\dot{\alpha} \cdot \ddot{\alpha})\dot{\alpha}}{\dot{s}^4}=\frac{\dot{\alpha}\times (\ddot{\alpha}\times \dot{\alpha})}{\dot{s}^4}$$ And we're pretty much done. $$k(t)=|\alpha''|=\frac{|\dot{\alpha}||\ddot{\alpha}\times \dot{\alpha}|}{\dot{s}^4}=\frac{|\ddot{\alpha}\times \dot{\alpha}|}{\dot{s}^3}$$