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

Shannon Andrews
2022-07-25
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Cheyanne Charles

Answered 2022-07-26
Author has **13** answers

Let the curve be r(t). Then, the velocity of the curve is v(r) =r'(t).

Use the deivative of the product,

$\frac{d}{dt}(\nu (t),\nu (t))=\nu \u2035(t)\nu (t)+\nu (t)\nu \u2035(t)\phantom{\rule{0ex}{0ex}}=2\nu (t)\nu \u2035(t)\phantom{\rule{0ex}{0ex}}=2\nu (t)a(t)$

Necessary part:

Let the curve has a constant speed. That is, $\parallel \nu (t)\parallel =c\phantom{\rule{0ex}{0ex}}$. To prove, acceleration is everywhere

orthogonal to its velocity.

$\parallel \nu (t)\parallel =c\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(r(t){)}^{2}=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(\nu (t),\nu (t))=0\phantom{\rule{0ex}{0ex}}2\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}\nu (t)a(t)=0$Copyright ©2011-2012 CUI WEI. All Rights Reserved.

Hence, acceleration is everywhere orthogonal to its velocity.

Sufficient part:

$\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}2\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(\nu (t)\nu (t))=0\phantom{\rule{0ex}{0ex}}\parallel \nu (t)\parallel =\sqrt{\nu (t),\nu (t)}$constant

Hence the proof.

Use the deivative of the product,

$\frac{d}{dt}(\nu (t),\nu (t))=\nu \u2035(t)\nu (t)+\nu (t)\nu \u2035(t)\phantom{\rule{0ex}{0ex}}=2\nu (t)\nu \u2035(t)\phantom{\rule{0ex}{0ex}}=2\nu (t)a(t)$

Necessary part:

Let the curve has a constant speed. That is, $\parallel \nu (t)\parallel =c\phantom{\rule{0ex}{0ex}}$. To prove, acceleration is everywhere

orthogonal to its velocity.

$\parallel \nu (t)\parallel =c\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(r(t){)}^{2}=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(\nu (t),\nu (t))=0\phantom{\rule{0ex}{0ex}}2\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}\nu (t)a(t)=0$Copyright ©2011-2012 CUI WEI. All Rights Reserved.

Hence, acceleration is everywhere orthogonal to its velocity.

Sufficient part:

$\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}2\nu (t)a(t)=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}(\nu (t)\nu (t))=0\phantom{\rule{0ex}{0ex}}\parallel \nu (t)\parallel =\sqrt{\nu (t),\nu (t)}$constant

Hence the proof.

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