Why is $\frac{\partial n}{\partial v}$ in the tangent plane?

65 Views Asked by At

I got a ruled surface $f(u,v) = c(u)+v\cdot X(u)$.

I first calculated the partial derivatives...

\begin{align*} f_u = c'(u)+vX'(u),\quad f_v = X(u),\quad f_{uu} = c''(u)+vX''(u),\quad f_{uv} = X'(u),\quad f_{vv} = 0 \end{align*}

According to the solutions, I need to look at the normal vector $n=\frac{f_u \times f_v}{\| f_u \times f_v\|}$ and notice following...

\begin{align*} \frac{\partial}{\partial v}\langle n,f_u \rangle=\langle \frac{\partial n}{\partial v},f_u \rangle + \langle n, X'\rangle =0\Rightarrow -\langle \frac{\partial n}{\partial v},f_u \rangle = \langle n, X'\rangle \end{align*}

The solutions then say that for $\frac{\partial n}{\partial v}\neq 0$ we see that $\frac{\partial n}{\partial v}$, $f_u$ and $f_v$ lie in the tangent plane and are linear dependent.

But I don't really see how that's obvious. Can someone explain?

EDIT for Context:

enter image description here