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Not too sure about this.

A surface is described as $y=\phi(x,t)$ Find a unit vector orthogonal to the surface.

I was thinking of a new function $f(x,y,t) = y - \phi(x,t) = 0$ and taking the gradient over the magnitude of the gradient.

Any help appreciated, thanks.

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Let $F(x,y,z)=0$ be the equation of a surface ,say $S$, such that it is continuously differentiable. According to the definition at any point $P(x_0,y_0,z_0)\in S$, the vector $$\nabla F|_P$$ is the normal vector you are searching for.

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  • $\begingroup$ I think what confuses me is the surface is defined in terms of time. I'm not sure if that is 2D or 3D. $\endgroup$ – Mike Miller Sep 29 '13 at 17:20
  • $\begingroup$ @BritMiller: I meant that your assumption of taking $f$ is right. $\endgroup$ – mrs Sep 29 '13 at 17:23
  • $\begingroup$ I think I understand. So: $\frac{(-\frac{\partial \phi}{\partial x}, 1)}{\sqrt{(\frac{\partial \phi}{\partial x})^2+1}}$? $\endgroup$ – Mike Miller Sep 29 '13 at 17:41
  • $\begingroup$ @BritMiller: It would be $$\frac{(-\frac{\partial \phi}{\partial x}, 1,-\frac{\partial \phi}{\partial t})}{\sqrt{(\frac{\partial \phi}{\partial x})^2+1+(\frac{\partial \phi}{\partial t})^2}}$$ $\endgroup$ – mrs Sep 29 '13 at 17:55
  • $\begingroup$ @ Babak S. That's what I initially thought, but the answer is given as the one I stated - I just checked. I actually took the partial w.r.t. z rather than t, but I was guessing. $\endgroup$ – Mike Miller Sep 29 '13 at 18:15

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