User:InfoTheorist/Differentiable curves

In this short note, I'll prove an identity regarding differentiable curves. Let $\gamma :[a,b]\rightarrow \mathbb {R} ^{n}$ be a differentiable curve. For every $t\in [a,b]$, the following identity holds

$\gamma '(t)\cdot \gamma (t)=|\gamma (t)|'|\gamma (t)|.$

The left hand side equals

$\gamma (t)\cdot \gamma '(t)=\sum _{k=1}^{n}\gamma _{k}(t)\gamma '_{k}(t).$

On the other hand, we have

$|\gamma (t)|=\left(\sum _{k=1}^{n}\gamma _{k}^{2}(t)\right)^{\frac {1}{2}},$

thus

$|\gamma (t)|'={\frac {1}{2|\gamma (t)|}}\sum _{k=1}^{n}2\gamma _{k}(t)\gamma '_{k}(t).$

Therefore,

$\gamma '(t)\cdot \gamma (t)=|\gamma (t)|'|\gamma (t)|.$

This result has several implications. First note that by the Cauchy-Schwarz inequality,

$\gamma (t)\cdot \gamma '(t)\leq |\gamma (t)||\gamma '(t)|.$

Thus the above identity implies

$|\gamma (t)|'\leq |\gamma '(t)|.$

If we apply the above inequality to ${\tilde {\gamma }}(t)=\gamma (t)-\gamma (a)$ and integrate both sides over $[a,b]$ we get

$|\gamma (b)-\gamma (a)|\leq {\text{length}}(\gamma ).$

In words, a straight line has the shortest distance among all curves connecting two points.

Second, if $\gamma (t)$ represents the position of a particle at time $t$, then its velocity and acceleration vectors at time $t$ are given by $v(t)=\gamma '(t)$ and $a(t)=\gamma ''(t)$, respectively. If we apply the identity to the velocity vector we get

$v(t)\cdot a(t)=|v(t)||v(t)|'.$

From this identity we deduce the following:

1) the velocity of a particle is perpendicular to its acceleration vector if and only if speed is constant (e.g., a particle in uniform circular motion), and

2) at any given moment, speed is increasing (decreasing) if and only if $v(t)\cdot a(t)$ is positive (negative) at that moment.

InfoTheorist (talk) 06:30, 28 September 2014 (UTC)