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)