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This is a problem I haven't thought about or encountered in many years but has popped up again incidentally, so please correct anything ahead if something is incorrect. If we are given the condition that $\{f(x,y,z) = 0\}$ for some differentiable function $f$, then the normal vector to the surface governed by said equation is given by

$\nabla f(x,y,z)=(f_1(x,y,z), f_2(x,y,z),f_3(x,y,z))$

where $f_i := \frac{\partial f}{\partial x_i}$ is the $i^{\operatorname{th}}$ partial derivative of $f$.

Good and all, but say that instead I am given $f$ parametrically in the form $f(u,v) = (x(u,v), y(u,v), z(u,v))$. How would I now go about determining the normal vector at arbitrary $u,v \in \mathbb{R}$?

user93334
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  • Note that in the case $f:\mathbb{R}^n\to\mathbb{R}$, it's important that $\nabla f$ is never zero (the zero vector) where $f(\underline{x})=0$. – Jonathan Y. Sep 05 '13 at 21:58

3 Answers3

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One way is to take the tangents $f_u$,$f_v$ and take the cross product to get an orthogonal vector.

Evan
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As your surface is given parametrically, it is easy to obtain normal vector simply by using the vector equation. Thus if your vector equation of the surface is:

$\vec{r}(t)=\left \langle x(u,v);y(u,v);z(u,v) \right \rangle$

You find normal vector as:

$\hat{n} = \frac{\vec{r}_u\times\vec{r}_v}{\left | \vec{r}_u\times\vec{r}_v \right |}$

where $\vec{r}_u$ and $\vec{r}_v$ designate partial derivatives of $\vec{r}(t)$ with respect to $u$ and $v$. Note that this vector is normalized.

Jefeter_7
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It's not the same situation. In the first case, you assume $f$ to be scalar valued, whereas in the second case, you don't.

Étienne Bézout
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