1
$\begingroup$

Here is the Ex.2.4.6 in Do Carmo's Differential geometry of curves and surfaces.

Let $\alpha:I\rightarrow \mathbb R^3$ be a regular parametrized curve with everywhere nonzero curvature. Consider the tangent surface of $\alpha$, with the parametrization defined by $$X(t,v)=\alpha(t)+v\alpha'(t), t\in I, v\neq 0.$$

Show that the tangent planes along the curve $x(t, const)$ are all equal.

I think the proof is straightforward. Let $c$ denote that constant. Then the tangent planes along the curve $x(t, const)$ can be written as

$$[X_t\wedge X_v]\cdot(x-x(t,c))=0$$ Since $$X_t(t,c)=\alpha'(t)+c\alpha''(t), X_v(t,c)=\alpha'(t)$$ then $$[\alpha'(t)\wedge\alpha''(t)]\cdot(x-\alpha(t))=0$$

This is all I can do for this problem. But intuitively, I am afraid the tangent planes $[\alpha'(t)\wedge\alpha''(t)](x-\alpha(t))=0$ are not equal to each other, since $\alpha'(t)\wedge\alpha''(t)$ does not necessarily have a constant direction. What's wrong with it?

$\endgroup$
2
$\begingroup$

You want to be looking at the tangent plane along the rulings, i.e., fixing $t$ and varying $v$. Now your calculation works great (if you fiddle with what should be $c$).

$\endgroup$
  • $\begingroup$ So you believe that the author wants us to show that the tangent planes along the curve x(t,c), where t is fixed and c is variable are all equal? My English is poor, but it seems to me that he wants the constant fixed and t vary~~~ $\endgroup$ – No One Oct 9 '16 at 17:17
  • $\begingroup$ Did you copy the problem exactly as it was written? There's no hope moving along the curve, as the planes contain varying tangent lines to the curve, and those will usually be skew lines. This is a classic result, and, yes, it's about the rulings — i.e., moving along the lines. $\endgroup$ – Ted Shifrin Oct 9 '16 at 17:18
  • $\begingroup$ I finally went back and looked at a copy of the book. There is, indeed, a typo in that problem. But I guess you just won't listen to reason. $\endgroup$ – Ted Shifrin Oct 10 '16 at 4:56
  • $\begingroup$ Thank you so much! Now I am sure it is really a typo. It should be x(const, v). $\endgroup$ – No One Oct 11 '16 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.