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I know the derivative of position vector function $r(t)$ is velocity, and its unit vector is tangent vector, therefore, the derivative of tangent vector is the unit vector of acceleration and normal vector. Why isn't that the case?

tc216
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user3908838
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    Imagine a particle moving in a straight line. Is acceleration normal to the line? – Bobbie D Mar 04 '17 at 00:35
  • What causes the difference? Based on the formula, I can't tell how that happens – user3908838 Mar 04 '17 at 00:38
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    The derivatives in the Frenet-Serret frame are with respect to arclength, not time. Also I think it may be undefined for a particle moving in a straight line (just based on symmetry), but I'm not sure. – Bobbie D Mar 04 '17 at 00:41

2 Answers2

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If speed $(\|r'(t)\|)$ is constant, then the acceleration must be normal to the direction of travel.

But if speed is variable then acceleration can be broken into a component parallel to the direction of travel and a component that is perpendicular to the direction of travel.

Doug M
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  • The formula shows the differences. But I'm still confused, why isn't the derivative of unit tangent vector parallel with acceleration vector. – user3908838 Mar 04 '17 at 01:31
  • If we choose a space curve in the xyz 3-dimensional coordinate space, for two different unit tangent vectors, the difference between them is only the direction. Is that right? – user3908838 Mar 04 '17 at 01:35
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    $N$ is the second derivative of $r$ with respect to arc length $= \frac {d^2r}{dS^2}, a$ is the second derivative of $r$ with respect to time.

    When we unitize the tangent vector, then we are dividing by $| r'(t)|$ (speed). It is the direction of the particle, moving at constant speed. $a = \frac {d^2r}{dt^2} = \frac {d^2S}{dt^2} T + (\frac {dS}{dt})^2 N$

     – Doug M Mar 04 '17 at 01:59
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By definition, acceleration is the variation of velocity in time, and thus the derivative of the velocity function.

Write $x(t)$ for the position, $v(t):=\dot x(t)$ for the velocity, and $a(t):=\dot v(t)$ for the acceleration. The direction vector is $$d(t):=\frac{v(t)}{\|v(t)\|}\ .$$ If we differentiate it, we obtain \begin{align} \frac{d}{dt}d(t) =&\ \frac{\dot v(t)}{\|v(t)\|} -\frac{\langle\dot v(t),v(t)\rangle}{\|v(t)\|^3}v(t)\\ =&\ \frac{a(t)}{\|v(t)\|} -\frac{\langle a(t),v(t)\rangle}{\|v(t)\|^3}v(t)\ . \end{align} This is of course different from the normal vector in general.

Daniel Robert-Nicoud
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  • Is this what is called the principal normal? and this not orthogonal to the tangent plane which the common normal is? – user123124 Jul 30 '19 at 08:36