plane tangent to 3d surface mmn14 3a

**transgalactic** · June 7th 2011, 12:56 PM

i have a suface $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}$

find the formula of a plane which is tangent to it in (x0,y0,z0)

i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

?

**FernandoRevilla** · June 7th 2011, 08:46 PM

If $S\equiv f(x,y,z)=0$ then, the tangent plane to $S$ at $P_0(x_0,y_0,z_0)$ is

$\dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$

**transgalactic** · June 7th 2011, 08:51 PM

i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

?

**FernandoRevilla** · June 7th 2011, 10:30 PM

Originally Posted by transgalactic

i need to find the vector whic is vertical to this surface
the formula of it is $<f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form

Sincerely, it is rather difficult to understand your question. Perhaps you mean that if the surface is written in explicit form $z=f(x,y)$ then, the equation of the tangent plane at $P_0(x_0.y_0,z_0)$ is $\pi\equiv <(f_x(P_0),f_y(P_0),-1),(x-x_0,y-y_0,z-z_0)>=0$ . For the general case, see my previous post.

**transgalactic** · June 7th 2011, 11:32 PM

Originally Posted by FernandoRevilla

If $S\equiv f(x,y,z)=0$ then, the tangent plane to $S$ at $P_0(x_0,y_0,z_0)$ is

$\dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$

i was told that the norm vector <df/dx,df/dy,-1>

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