fluid free surface water The free surface of a fluid moving in two dimensions is given parametrically by $r(x, t) =(x, \eta(x, t))$
Show that a unit normal to the surface is $\dfrac{1}{\sqrt{1+ (\eta_x)^2}}(-\eta_x,1)$.
I think for this I should take the gradient but how do I get $(-\eta_x,1)$?
 A: OK, for any and every fixed $t$, the tangent to
$r(x, t)$ is $(1, \partial \eta / \partial x)$.
It is easy to see that 
$ (1, \partial \eta / \partial x ) \cdot (-\partial \eta / \partial x, 1) = 0; \tag{1}$
$1 / \sqrt{1 + (\partial \eta / \partial x )^2}$ is simply a 
normalizing factor.
Note Added in Edit, Saturday 6 May 2017 10:50 AM PST:  This in response to stedmoaoa's comment:  for fixed $t$, the free surface is given by the curve $r(x, t) = (x, \eta(x, t))$.  This curve $r(x, t)$ is  parameterized by $x$, much as the graph of any function $y = f(x)$ may be thought of as the $x$-parameterized curve $(x, f(x))$.  The tangent vector to such a curve is given by taking the derivative of each component with respect to the curve parameter, $x$.  Thus the tangent vector is
$(\partial x / \partial x, \partial \eta (x, t) / \partial x) = (1, \partial \eta (x, t) / \partial x); \tag{2}$
we obtain a normal vector to the curve $r(x, t)$ by rotating its tangent through an angle of $\pi / 2$; such a rotation is given by the matrix
$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{3}$
indeed, we have
$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} 1 \\  \partial \eta (x, t) / \partial x \end{pmatrix} = \begin{pmatrix} -\partial \eta (x, t) / \partial x  \\ 1 \end{pmatrix}; \tag{4}$
furthermore, 
$\Vert (-\partial \eta (x, t) / \partial x,  1) \Vert = \sqrt{1^2 + (\partial \eta / \partial x )^2} = \sqrt{1 + (\partial \eta / \partial x )^2}, \tag{5}$
so that 
$\dfrac {1} {\Vert (-\partial \eta (x, t) / \partial x,  1) \Vert} (-\partial \eta (x, t) / \partial x,  1) = \dfrac {1} {\sqrt{1 + (\partial \eta / \partial x )^2}}(-\partial \eta (x, t) / \partial x,  1);  \tag{6}$
thus $\dfrac {1} {\sqrt{1 + (\partial \eta / \partial x )^2}}(-\partial \eta (x, t) / \partial x,  1)$ is in fact a unit normal vector to the curve $r(x, t) = (x, \eta(x, t))$.
Observe that the normal vector $(-\partial \eta (x, t) / \partial x,  1)$ may also be obtained from $(1, \partial \eta (x, t) / \partial x)$ by simple inspection, which is indeed what I did in the above.
Finally, I think it is worth pointing out that we don't really use the gradient operator $\nabla$ here; rather, we take the derivative of the curve $r(x, t)$ with respect to the parameter $x$, component-wise.
