Let $\gamma : I \rightarrow \mathbb{R}^3$ be a parametretrized smooth curve with unit speed. Assume there exist a fixed vector $q$ such that $\gamma ''(s)=q, \ \forall s \in I$. Show that $\gamma$ is a straight line

How would one approach this?

I've thougt that since $\gamma$ have unit speed, then $||\gamma'(t)||=1,\ \forall t$ and then the arc length between any $t_1,t_2\in I$ would be $\int_{t_1}^{t_2} 1 \ dt=t_2-t_1$. But could I conclude it's a straight line from this, since the distance between two points, is just found by subtracting the to points with eachother


closed as off-topic by T. Bongers, Shailesh, Leucippus, Daniel W. Farlow, user91500 Jun 22 '16 at 4:50

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$\gamma''(s) = q$ gives $\gamma'(s) = qs + p$. Geometrically, $\gamma'$ is a straight line and every point on this straight line is at the same distance $1$ from the origin, so $\gamma'$ must be reduced to a point, hence $q = 0$ and $\gamma' = p$, with $\|p \|= 1$. This gives $\gamma = ps + c$. Another approach is to plug in several values of $s$ in the equation $\|qs + p\| = 1$, to get $q = 0$.

  • $\begingroup$ Is it because $\gamma$ have unit speed, that every point on $\gamma'$ have distance 1 from origin? $\endgroup$ – jta Jun 21 '16 at 21:51
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    $\begingroup$ @Jta yes, because a point on the image of $\gamma'$ takes the form $\gamma'(s)$, and $\|\gamma'(s)\| = 1$ $\endgroup$ – user258700 Jun 21 '16 at 21:52
  • $\begingroup$ Thanks! But I don't see why it has to have unit speed? Wouldnt it work as long $||\gamma'(t)||=c,\ \forall t$ for some constant $c$? $\endgroup$ – jta Jun 21 '16 at 22:10
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    $\begingroup$ @Jta of course it would, by the same reasoning. The reason why the question supposes that $\gamma$ is unit speed is that, in general, one is only interested in unit-speed curves because any (regular) curve with constant speed can be made into a unit-speed curve. $\endgroup$ – user258700 Jun 21 '16 at 22:17

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