Show that a curve has constant speed if and only if its acceleration is everywhere orthogonal to its velocity.

Show that a curve has constant speed if and only if its acceleration is everywhere orthogonal to its velocity.
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Cheyanne Charles
Let the curve be r(t). Then, the velocity of the curve is v(r) =r'(t).
Use the deivative of the product,
$\frac{d}{dt}\left(\nu \left(t\right),\nu \left(t\right)\right)=\nu ‵\left(t\right)\nu \left(t\right)+\nu \left(t\right)\nu ‵\left(t\right)\phantom{\rule{0ex}{0ex}}=2\nu \left(t\right)\nu ‵\left(t\right)\phantom{\rule{0ex}{0ex}}=2\nu \left(t\right)a\left(t\right)$
Necessary part:
Let the curve has a constant speed. That is, $\parallel \nu \left(t\right)\parallel =c\phantom{\rule{0ex}{0ex}}$. To prove, acceleration is everywhere
orthogonal to its velocity.
$\parallel \nu \left(t\right)\parallel =c\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(r\left(t\right){\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(\nu \left(t\right),\nu \left(t\right)\right)=0\phantom{\rule{0ex}{0ex}}2\nu \left(t\right)a\left(t\right)=0\phantom{\rule{0ex}{0ex}}\nu \left(t\right)a\left(t\right)=0$Copyright ©2011-2012 CUI WEI. All Rights Reserved.
Hence, acceleration is everywhere orthogonal to its velocity.
Sufficient part:
$\nu \left(t\right)a\left(t\right)=0\phantom{\rule{0ex}{0ex}}2\nu \left(t\right)a\left(t\right)=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(\nu \left(t\right)\nu \left(t\right)\right)=0\phantom{\rule{0ex}{0ex}}\parallel \nu \left(t\right)\parallel =\sqrt{\nu \left(t\right),\nu \left(t\right)}$constant
Hence the proof.