5

I'm reading the classical Manfredo's differential geometry book and I couldn't prove formally the following statement written on page 16:

Why does the tangent vector measure the rate of change of the angle which neighboring tangent make with the tangent at $s$?

user42912
  • 23,582
  • You can adapt the argument here to answer your question: https://math.stackexchange.com/a/2080466/565 – AnonymousCoward Jun 26 '17 at 14:12
  • @FaraadArmwood In $\mathbb{R}^3$, the tangent space to a curve is a line, not a plane. I do not know what you mean by "tangent plane" to a curve in $\mathbb{R}^3$. – AnonymousCoward Jun 26 '17 at 14:30
  • Maybe you are thinking of the osculating plane? – AnonymousCoward Jun 26 '17 at 14:40
  • @AnonymousCoward. Yes. I made a few errors when writing. Here $\langle \alpha', N(s) \rangle = P$ is a plane which contains $\alpha'$. But the example in which I was trying to illustrate assumes $\alpha$ is a plane curve. I was just making the remark that $\alpha'$ is always contained in a plane, but the curve just necessarily isn't. – Faraad Armwood Jun 26 '17 at 14:44
  • The statement "$\alpha'(s)$ is contained in a plane" is demonstrably false, unless you mean something else. – AnonymousCoward Jun 26 '17 at 14:51
  • I am fixing $s$, so then it is contained in the that plane $P$ above. Anyway, I'll just leave it alone. – Faraad Armwood Jun 26 '17 at 14:56
  • The problem is that if you fix $s$ in $\alpha'(s)$, then how can you say anything about $\alpha''(s)$. – AnonymousCoward Jun 26 '17 at 20:02

1 Answers1

7

For simplicity, lets explain this at the point where $s=0$.

Let's try to do this the naive way first, and we will see where we get stuck.

Attempt 1 The angle formed by the tangent vector to the curve at time $0$ and the tangent vector to the curve at time $s$ is the angle $\theta(s)$ formed by the two vectors $\alpha'(0)$ and $\alpha'(s)$. Since these are unit vectors, this angle is given by the formula $$\cos(\theta(s)) = \alpha'(0)\cdot \alpha'(s)$$ Note that when $s=0$, the RHS is $|\alpha'(0)|^2 =1$, so $\theta(0) = 0$.

If we evaluate the derivative of both sides of this equation at time $s=0$, we get $$-\sin(\theta(0)) \theta'(0) = \frac{d}{ds}\bigg\vert_{s=0}\alpha'(0)\cdot \alpha'(s) = \alpha'(0)\cdot \left(\frac{d}{ds}\bigg\vert_{s=0}\alpha'(s)\right) = \alpha'(0)\cdot \alpha''(0)$$ (since dot-product is linear in the second argument). Unfortunately, this equation tells us nothing about $\theta'(0)$ since $\sin(0)= 0$ (indeed, this tells us that $\alpha'(0)\cdot \alpha''(0)=0 $ always, which is why the equation won't tell us $\theta'(0)$).

Ok, so the naive solution doesn't work, but the problem was differentiating $\cos$ to $\sin$, so we have an idea.

Attempt 2 Let's use a different formula relating $\theta$ and the unit vectors $\alpha'(0)$ and $\alpha'(s)$:

$$\sin(\theta(s)) = |\alpha'(0)\times \alpha'(s)|$$

If we differentiate both sides of this equation away from $s=0$, we get $$\cos(\theta(s)) \theta'(s) = \frac{d}{ds}|\alpha'(0)\times \alpha'(s)| = \frac{\alpha'(0)\times \alpha''(s)}{2|\alpha'(0)\times \alpha'(s)|}$$ which is the same as $$2\sin(\theta(s))\cos(\theta(s)) \theta'(s) = \alpha'(0)\times \alpha''(s)$$ and I'm still stuck. Huh.

Attempt 3 Ok, let's finish this answer incorporating the argument I linked in the comments. Differentiating the formula from Attempt 1 a second time, we get that

$$-\cos(\theta(s)) (\theta'(s))^2 -\sin(\theta(s))\theta''(s) = \alpha'(0)\cdot \alpha'''(s)$$ Plugging in $s=0$ yields $$\theta'(0)^2 = - \alpha'(0)\cdot \alpha'''(0).$$ But we also know that $$0 = \frac{d}{ds}\alpha'(s)\cdot \alpha''(s) = \alpha''(s)\cdot \alpha''(s)+\alpha'(s)\cdot \alpha'''(s) $$ so $$\theta'(0)^2 = \alpha''(0)\cdot \alpha''(0) = |\alpha''(0)|^2. $$

AnonymousCoward
  • 5,548
  • A solution can be found here: https://math.stackexchange.com/a/2080466/565. The trick is to differentiate the formula from Attempt 1 once more wrt s. – AnonymousCoward Jun 26 '17 at 14:11