# Calculating the normal vector of a surface.

Let $\alpha: I\rightarrow \mathbb{R}^3$ be a parametrized curve with non-zero curvature every where and parametrized by arc length. Let $$x(s,v)=\alpha(s)+ r(n(s)\cos v + b(s)\sin v), r\not =0, s\in I$$ a parametrized surface, where $n$ is the normal vector of $\alpha$ and $b$ is the vector binormal of $\alpha$. Show that when $x$ is regular then the normal vector of the surface is $$N(s,v)=-(n(s) \cos v+b(s)\sin v)$$

I arrive anywhere trying to calculate $$x_s\times x_v$$

I think this is the correct way, but I don't know how to get the expression above.

I have seen a "prove" that check the properties of the normal vector on $N(s,v)$, but I'm not sure that it is well done, because of the direction.

Thanks!

• Can you show us your computation of $x_s$? I assume that $n(s)$ and $b(s)$ are the normal and binormal vectors for the curve $\alpha$, so you'll want to use the Frenet formulas to simplify your expression once you've found it. – John Hughes Sep 22 '14 at 13:44
• I have edited the post. – YTS Sep 22 '14 at 14:12

## 1 Answer


Thus your starting point, \begin{align} x_s&=\al'(s)+r(n'(s)\cos v+b'(s)\sin v) \\ x_v &=r(-n(s)\sin v+b(s)cos v) \end{align} becomes \begin{align} x_s&=\al'+\cv\NN'+\sv\BB' \\ x_v &=-\sv\NN+\cv \BB \end{align} And since $\al' = \TT$, we have \begin{align} x_s\times x_v&=(\TT+\cv\NN'+\sv\BB')\times (-\sv\NN+\cv \BB) \end{align} At this point in your work, $\alpha'(s) = \TT(s)$ seems to have gotten lost; I'll correct your work continuing from here: \begin{align} x_s\times x_v &=(\TT+\cv\NN'+\sv\BB')\times (-\sv\NN+\cv \BB)\\ &=\left(\TT \times (-\sv\NN+\cv \BB)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right)\\ &=\left(-\sv\BB-\cv \NN)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right), \end{align} where I've used the fact that $\TT \times \NN = \BB$ to simplify the first term. Now I'll simplify by using the Frent-Serret formulas to write $\NN' = -\ka \TT + \ta \BB$ and $\BB' = -\ta \NN$: \begin{align} x_s\times x_v &=\left(-\sv\BB-\cv \NN)\right)+\cv\left(\NN'\times (-\sv\NN+\cv \BB)\right)+\sv\left(\BB'\times (-\sv\NN+\cv \BB)\right)\\ &=\left(-\sv\BB-\cv \NN)\right)+\cv\left( (-\ka \TT + \ta \BB) \times (-\sv\NN+\cv \BB)\right)+\sv\left((-\ta \NN)\times (-\sv\NN+\cv \BB)\right) \end{align}

The first term is almost exactly the answer we want. Let's simplify the 2nd and 3rd terms one at a time: \begin{align} \text{2nd term} &=\cv\left( (-\ka \TT + \ta \BB) \times (-\sv\NN+\cv \BB)\right)\\ &=\cv\left( (-\ka \TT \times (-\sv\NN+\cv \BB)+ \ta \BB \times (-\sv\NN+\cv \BB)\right)\\ &=\cv\left( (\ka \sv\BB + \ka\cv \NN)+ \ta \sv\TT\right)\\ \end{align}

and \begin{align} \text{3rd term} &=\sv\left((-\ta \NN)\times (-\sv\NN+\cv \BB)\right)\\ &=-\ta\sv\cv \TT\\ \end{align}

The third term exactly cancels the last bit of the second term, so

\begin{align} x_s\times x_v &=\left(-\sv\BB-\cv \NN)\right)+\cv( (\ka \sv\BB + \ka\cv \NN))+ \ta \sv\TT) -\ta\sv\cv \TT \\ &=(-\sv\BB-\cv \NN))+\cv( \ka \sv\BB + \ka\cv \NN) \\ &=(-\sv\BB-\cv \NN))+\ka\cv( \sv\BB + \cv \NN) \\ &=(1-\ka) (-\sv\BB-\cv \NN)) \end{align} This final vector is just $(1-\ka)$ times a unit vector, so normalized, we get that the surface normal is \begin{align} -\sv\BB-\cv \NN. \end{align} When you redo this computation including $r$, that final factor will be $(1 - r\ka)$, I believe, showing that if $r$ is less than the radius of curvature, everything works fine. If not, then there's a sign change ... but in that case ($r$ greater than radius of curvature), the offset surface is singular, so we're not in the "regular surface" case any more.

• Thanks! Typo: you missed the constant "r" in many parts. – YTS Sep 25 '14 at 3:41
• Nope: you missed the fourth sentence: "Finally, I'm going to set r=1 so that it disappears from the equations." The idea was that you could fill it back in for yourself. :) – John Hughes Sep 25 '14 at 11:02