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I have always wondered why considering paths, $\gamma(t)$, there is inevitably a condition that $\gamma'(t) \neq 0$ associated within the same sentence.

Can someone please give me a motivational reasoning behind this; whether by examples or goals etc.

I have intentionally left this vague as I have seen this in Calculus, real/Complex analysis, geometry and so forth.

Any comments will be appreciated.

Harch
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  • I don't see $\gamma'(t) \neq 0$ as such a central condition. Do you mean that the derivative should exist? – Thomas Oct 08 '13 at 05:59

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I think it's often because we're mostly interested in considering smooth curves (i.e. ones that don't have sharp corners in them). If $\gamma'(t_0) = 0$, then there might be a sharp corner at $t=t_0$ even though the function $\gamma$ is infinitely differentiable. For example, consider the curve $\gamma(t) = (t^2, t^3)$ at $t=0$.

Another way of saying the same thing ... at places where $\gamma'(t) = 0$, the naive computation of the unit tangent (by unitizing the first derivative vector) fails, so, without some extra fuss, you can't even get started doing simple differential geometry. Best to exclude such places.

bubba
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  • Does this also mean that this condition does not allow us to travel backwards on the same path? – Harch Oct 09 '13 at 06:26
  • The condition doesn't prevent you from traversing the entire path in the reverse direction. But it does prevent you from doing a u-turn in the middle of the path. To do a u-turn, the derivative vector has to flip directions. To do this, by a continuity argument, there must be some point in the flipping process at which its length is zero. – bubba Oct 10 '13 at 00:13