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I'd like some help with the following question:

Find the equation of the tangent plane in $(x_0,y_0,z_0)$ to a regular surface given by $f^{-1}(0)$, where $0$ is a regular value.

I tried to find local parametrization but It didn't work.

Thanks.

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    The [surface normal](http://en.wikipedia.org/wiki/Surface_normal) of an implicitly defined surface $f(\vec{x})=0$ at the point $\vec{x}=\vec{a}$ is $\vec{\nabla} f(\vec{a}).$ The tangent plane is perpendicular to this and intersects the point $\vec{x}$ (obviously), so the equation is $$(\vec{x}-\vec{a})\cdot\vec{\nabla}f(\vec{a})=0.$$ Your point is $\vec{a}=\vec{0}$. Does this make sense to you?2011-09-17
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    @anon: I think that $\overrightarrow{a}=(x_{0},y_{0},z_{0})$.2011-09-17
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    @Americo: Right, I wasn't thinking straight.2011-09-17

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If a surface is defined by a regular function $z=g(x,y)$, the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ in terms of the partial derivatives of $g$ is

$$z=z_{0}+\left. \frac{\partial g}{\partial x}\right\vert _{(x_{0},y_{0})}(x-x_{0})+\left. \frac{\partial g}{\partial y}\right\vert _{(x_{0},y_{0})}(y-y_{0}).$$

enter image description here

If the surface is defined implicitly by $f(x,y,z)=0$, then $z=g(x,y)=f^{-1}(0)$ (i.e. $f(x,y,g(x,y))\equiv 0$). Since

$$\frac{\partial g}{\partial x}=-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}$$

and

$$\frac{\partial g}{\partial y}=-\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z},$$

the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ is given by

$$z=z_{0}-\left(\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(x-x_{0})-\left(\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(y-y_{0})$$

or

$$(x-x_{0})\left(\frac{\partial f}{\partial x}\right)_{(x_{0},y_{0},z_{0})}+(y-y_{0})\left(\frac{\partial f}{ \partial y}\right)_{(x_{0},y_{0},z_{0})}+(z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}=0.$$

In compact notation, $\mathbf{x}=\left( x,y,z\right) ,\mathbf{x}_{0}=\left( x_0,y_0,z_0\right) $, we get

$$\left( \mathbf{x}-\mathbf{x}_{0}\right) \cdot \mathbf{\nabla }f(\mathbf{x}_{0})=0.$$