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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.

user42912
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    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. – symplectomorphic Oct 12 '17 at 03:09
  • @symplectomorphic because they are perpendicular with each other – user42912 Oct 12 '17 at 03:18
  • 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. – symplectomorphic Oct 12 '17 at 03:22
  • @symplectomorphic I'm sorry, it's parametrized by arc length. – user42912 Oct 12 '17 at 03:23
  • @symplectomorphic one question. Do you think I can take the derivative of $\lambda$? I don't know in the $\lambda$ is differentiable or not. – user42912 Oct 12 '17 at 13:49

2 Answers2

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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.

  • Thank you for your answer. Why my assertion on $\lambda$ can't be true on an interval? – user42912 Oct 12 '17 at 13:45
  • @user42912 if $\lambda$ is a constant function on an interval then it's derivatives is 0 –  Oct 12 '17 at 17:49
  • 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? – user42912 Oct 12 '17 at 17:54
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Let's assume that the tangents pass trough the point $c$ in $\mathbb{R}^2$. Then we the vectors $\alpha(s) - c$ and $\alpha'(s)$ are proportional, that is $(\alpha(s)-c) + \lambda(s) \alpha'(s)=0$ for some scalar function $\lambda(s)$. Replacing $\alpha(s)$ with $\alpha(s) - c$, we may assume that $c=0$, that is $$\alpha(s) + \lambda(s) \alpha'(s)=0$$

The set $\{s \ | \ \lambda(s) \ne 0\}$ is open and dense in the domain (since the curve is regular). On this domain we can write $$\alpha'(s) = \mu(s) \alpha(s)$$ and so on each connected component $$\alpha(s) = e^{M(s)} \cdot \mathbb{a}$$ where $M(s)$ is an antiderivative of $\mu(s)$. It is now easy to conclude (since $\gamma$ is regular) that the vector constants $\mathbb{a}$ are the same for all the connected components. Therefore, $\gamma$ describes a (part of ) the line of direction $\mathbb{a}$ through the origin.

orangeskid
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