# How to prove that $\alpha''(s)$ goes towards the inside of the curve $\alpha(s)$

Let $\alpha:[a,b]\longrightarrow \mathbb{R}^2$ be a plane curve parametrized by arc length by $\alpha(s)$.

$T(s):$ unit tangent vector

Note that $||T(s)||=1\Longrightarrow T'(s)\perp T(s)$

How to prove that $T'(s)$ goes towards the '' inside of the curve ''.

'' inside of the curve '' $:$ in sense to the convex region that encloses the curve (locally at the point $\alpha(s)$)

Any hints would be appreciated.

• Hint: What can you say between the curve near $\alpha(s)$ and the circle centered at $\alpha(s) + \frac{T'(s)}{|T'(s)|^2}$ with radius $\frac{1}{|T'(s)|}$. – achille hui Dec 28 '13 at 19:28

Let $\phi$ is the angle between $T$ and the $x$-axis, then

$$\frac{dT}{ds}=\frac{dT}{d\phi}\frac{d\phi}{ds}.$$

Since length of $T$ is constant, $\frac{dT}{ds}$ and $\frac{dT}{d\phi}$ both are perpendicular to $T$. They have same direction if $\frac{d\phi}{ds}$ be positive and opposite direction if $\frac{d\phi}{ds}$ be negative. We can write $$T=(\cos \phi) \mathbf{i} +(\sin\phi)\mathbf{j}$$ so $$\frac{dT}{d\phi}=-(\sin \phi)\mathbf{i}+(\cos \phi)\mathbf{j}=\cos(\phi+\frac{\pi}{2})\mathbf{i}+\sin(\phi+\frac{\pi}{2})\mathbf{j}$$ is a unit vector which obtain by rotating $T$ counterclockwise $\frac{\pi}{2}$ radians. Thus if you stand at the point $\alpha(s)$ and in direction of $T$, the vector $$\frac{dT}{ds}=\frac{dT}{d\phi}\frac{d\phi}{ds}$$ is to the left if $\frac{d\phi}{ds}$ be positive and, is to the right if $\frac{d\phi}{ds}$ be negative. On the other words $\frac{dT}{ds}$ goes toward concavity of the curve. (See following figures)

• as you choose the sense of $T'(s)$? – felipeuni Dec 28 '13 at 19:38
• Sorry, I edited the answer. – Sepideh Bakhoda Dec 28 '13 at 20:57

You may assume $\alpha(0)=(0,0)$ and $\alpha'(0)=(1,0)$. Since your curve $\gamma$ is parametrized by arc length we know that $\alpha''(0)\perp\alpha'(0)$, so $\alpha''(0)=(0,\kappa)$ for some $\kappa\in{\mathbb R}$. It follows that for $|t|\ll1$ we have $$\alpha:\quad t\mapsto\cases{x(t)&=t+\ ?t^3\cr y(t)&={\displaystyle{\kappa\over2}}t^2+\ ?t^3\cr}\quad .$$ Eliminating $t$ we see that in the neighborhood of $(0,0)$ the curve $\gamma$ can be viewed as graph $$y={\kappa\over2} x^2+\ ?x^3\ .$$ When $\kappa>0$ this is in first approximation a parabola bended upwards, as indicated by the arrow $\alpha''(0)=(0,\kappa)$; and similarly, when $\kappa<0$ we have a parabola bended downwards.

I suppose it really depends on how you define the words “inside of the curve”. Note that in your picture, in the point on the central arrow, it's not clear (or at least, it isn't to me) which is the inside and the outside of the curve.

To follow OP's definition: assume that $\gamma$ is locally the graph of a function. That is $\gamma (t)=(t,f(t)),$ in a neighborood of $t=0$. Since $\gamma ''(t)=(0,f''(t))$ if $f''(t)>0$ ($<0$), the acceleration vector points up (down), the function is locally convex (concave) and the graph lies entirely above (under) the tangent line.

• The inside may not be clear at that point, but $T'\approx0$ there, so that, for twice-differentiable curves, any definition of "inside direction" could be undefined when that happens, and be just fine. In other words, if you ignore felipeuni's question, then you could take the the direction of $T'$ when it's nonzero as the definition of the inside direction. – Mark S. Dec 29 '13 at 19:22
• Yes, it is exactly what I wrote in the unedited answer. :-) However, seems to me (basing the observation on my sketch of proof above) that the two definitions are equivalent. – pppqqq Dec 29 '13 at 20:42