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I'm trying to solve this question in the classical Do Carmo's differential geometry book (page 23):

  1. A regular parametrized curve $\alpha$ has the property that all its tangent lines pass through a fixed point. Prove that the trace of $\alpha$ is a (segment of a) a straight line.

My attempt

Following the statement of the question, we have $\alpha(t)+\lambda(s)\alpha'(s)=const$.

Taking the derivative of both sides we have $\alpha'(s)+\lambda'(s)\alpha'(s)+\lambda(s)\alpha''(s)=0$ which is equal to $(1+\lambda'(s))\alpha'(s)+\lambda(s)\alpha''(s)=0$.

Since $\alpha'(s)$ and $\alpha''(s)$ are linearly independent, we have $\lambda'(s)=-1$ and $\lambda(s)=0$ for every $s$ which I found strange, since the derivative of the zero function is zero.

I need a clarification at this point and a hand to finish my attempt of solution.

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    $\begingroup$ Why do you think that the velocity and acceleration vectors are linearly independent? For a particle accelerating along a line, the velocity and acceleration vectors are parallel. $\endgroup$ – symplectomorphic Oct 12 '17 at 3:09
  • $\begingroup$ @symplectomorphic because they are perpendicular with each other $\endgroup$ – user42912 Oct 12 '17 at 3:18
  • $\begingroup$ They are perpendicular only if you are parametrizing by arclength, which you did not say. Take the curve $r(t)=\langle t^2, 0\rangle$. The velocity is $\langle 2t, 0\rangle$ and the acceleration is $\langle 2, 0\rangle$. Obviously these vectors aren't perpendicular. $\endgroup$ – symplectomorphic Oct 12 '17 at 3:22
  • $\begingroup$ @symplectomorphic I'm sorry, it's parametrized by arc length. $\endgroup$ – user42912 Oct 12 '17 at 3:23
  • $\begingroup$ @symplectomorphic one question. Do you think I can take the derivative of $\lambda$? I don't know in the $\lambda$ is differentiable or not. $\endgroup$ – user42912 Oct 12 '17 at 13:49
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I guess you assumed that $\alpha$ is parametrized by arc length (or constant length), so $|\alpha'(s)|=1$ and differentiating gives $\langle \alpha' , \alpha''\rangle = 0$. Thus, if $\alpha''(s)\neq \vec 0$, then $\alpha'(s), \alpha''(s)$ are linearly independent.

So like you said, you find $\lambda (s) = 0$ and $\lambda'(s) = -1$ whenever $\alpha''(s)\neq 0$.

The set $\{ s : \alpha''(s)\neq \vec 0\}$ is an open set. If it is nonempty, it contains some intervals $I$. But your assertion on $\lambda$ cannot be true on an interval. Thus

$$\{ s : \alpha''(s)\neq \vec 0\}$$

is empty. So $\alpha''\equiv \vec 0$ and $\alpha$ defines a straight line.

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  • $\begingroup$ Thank you for your answer. Why my assertion on $\lambda$ can't be true on an interval? $\endgroup$ – user42912 Oct 12 '17 at 13:45
  • $\begingroup$ @user42912 if $\lambda$ is a constant function on an interval then it's derivatives is 0 $\endgroup$ – user99914 Oct 12 '17 at 17:49
  • $\begingroup$ Thank you again. Out of curiosity, If $\lambda$ is constant on a different domain, could the function be different than zero? Do you have some example of such function? $\endgroup$ – user42912 Oct 12 '17 at 17:54

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