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Below is an example from Do Carmo's Differential Geometry page 139 "The Geometry of the Gauss Map".

Let us analyse the point $p=(0,0,0)$ of the hyperbolic paraboloid $z=y^2-x^2$. For this, we consider a parametrisation $\textbf{x}(u,v)$ given by $$\textbf{x}(u,v)=(u,v,v^2-u^2),$$

and compute the normal vector $N(u,v)$. We obtain successively

$\textbf{x}_u=(1,0,-2u),$

$\textbf{x}_v=(0,1,2v),$

$N=\Big(\frac{u}{\sqrt{u^2+v^2+\frac{1}{4}}},\frac{-v}{\sqrt{u^2+v^2+\frac{1}{4}}},\frac{1}{2\sqrt{u^2+v^2+\frac{1}{4}}}\Big)$.

Notice that at $p=(0,0,0)$ $\textbf{x}_u$ and $\textbf{x}_v$ agree with the unit vectors along the $x$ and $y$ axes, respectively. Therefore, the tangent vector at $p$ to the curve $\alpha(t)=\textbf{x}(u(t),v(t))$, with $\alpha(0)=p$, has, in $\mathbb{R}^3$, coordinates $(u'(0),v'(0),0)$.

I understand up until this point.

Now my question is what follows:

How can Do Carmo get the following:

Restricting $N(u,v)$ to this curve and computing $N'(0)$, we obtain

$N'(0)=(2u'(0),-2v'(0),0)$

I have little clue on how can he get $2u'(0)$ and $2v'(0)$?

Could somebody please help clarify this confusion? Thanks.

user338393
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  • Does $'$ denote the derivative with respect to arc length? I don't know the answer to your question, but explaining which derivative is meant might help you get an answer – Chill2Macht Sep 30 '16 at 08:07
  • @William It is with respect to $t$. For example we can write $N(t)=(x(t),y(t),z(t))$, so $N'(t)=(x'(t),y'(t),z'(t))$. So in the example $N'(0)$ is $N'(t)$ at $t=0$. Hope that helps. – user338393 Sep 30 '16 at 08:10

1 Answers1

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You need to differentiate $N(u(t), v(t))$, given that $u(0) = v(0) = 0$. To calculate the first component of $N$, which is $\frac{u(t)}{\sqrt{u(t)^2 + v(t)^2+ 1/4}}$, we have

$$\frac{d}{dt} \frac{u(t)}{\sqrt{u(t)^2 + v(t)^2+ 1/4}} = \frac{u'(t)}{\sqrt{u(t)^2 + v(t)^2+ 1/4}} - \frac{u(t)(u(t)u'(t) + v(t)v'(t))}{(u(t)^2 + v(t)^2+ 1/4)^{3/2}}$$

so setting $t=0$ gives

$$\frac{u'(0)}{\sqrt{0^2 + 0^2+ 1/4}} = 2u'(0)$$

similar for the other components.