Can I take this derivative?

Question

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

5. 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.

Following the statement of the question, we have $\alpha(s)+\lambda(s)\alpha'(s)=const$.

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

So before I continue my attempt of solution I realized that maybe I can't take this derivative above because I don't know if $\lambda$ is differentiable.

Can I take the derivative above?

Answer 1

Note that you were only given a regular parametrized curve, not an arclength-parametrized curve, although of course you could reparametrize by arclength. But what regularity tells you is that $\alpha'(t)\cdot\alpha'(t) > 0$ for all $t$.

So let's take your equation $$\alpha(t)+\lambda(t)\alpha'(t) = P,$$ where $P$ is a constant vector, and subtract $\alpha(t)$ from both sides. Now dot with $\alpha'(t)$ and we get $$\lambda(t) = \frac{(P-\alpha(t))\cdot\alpha'(t)}{\|\alpha'(t)\|^2}.$$ The denominator is never $0$, and both numerator and denominator are smooth functions of $t$, so $\lambda$ is likewise a smooth function of $t$.

Answer 2

Because $\|\alpha'(s)\|\not=0$, you can scalar multiply by $\alpha'(s)$ and rearrange for $\lambda(s)$:

$$\lambda(s)=\frac{\mathrm{const}-\langle\alpha(s),\alpha'(s)\rangle}{\|\alpha'(s)\|^2}.$$

As everything on the right is differentiable (assuming $\alpha\in C^2$), so is $\lambda(s)$.