# Show that $\gamma$ is a straight line [closed]

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, user91500Jun 22 '16 at 4:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Shailesh, Leucippus, Daniel W. Farlow, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

$\gamma''(s) = q$ 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 $q = 0$.
• Is it because $\gamma$ have unit speed, that every point on $\gamma'$ have distance 1 from origin? – jta Jun 21 '16 at 21:51
• @Jta yes, because a point on the image of $\gamma'$ takes the form $\gamma'(s)$, and $\|\gamma'(s)\| = 1$ – user258700 Jun 21 '16 at 21:52
• 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 $c$? – jta Jun 21 '16 at 22:10
• @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. – user258700 Jun 21 '16 at 22:17