Differential geometry for curves that are linearly dependent

Question

Show that a curve is a straight line if $X’$ and $X’’$ are linearly dependent for all $t$.

Who knows how to solve this question?

Answer 1

Let us assume that the curve is regular ($X'(t)$ is not zero), otherwise the problem is not well-stated. Then the assumption that $X''$ and $X'$ are linearly dependent means that there exists a real function $a(t)$ such that $$ X''(t)= a(t) X'(t). $$ If you write $X(t)=\left(X_1(t),X_2(t),X_3(t)\right)$, then for every $i=1,2,3$ we can integrate $X''_i(t)=a(t)X_i'(t)$ twice to obtain an expression for $X_i(t)$: $$ X_i(t) = b_i \int\left( \exp \left( \int a(t)\,dt\right) \right)\,dt + c_i,$$ where $b_i,c_i$ are constants. This expression is not particularly nice, but you have now obtained a parametrisation of a line.

I didn't fill in every tiny detail since I don't want to give away the answer completely. I hope this will be helpful.