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Given a surface $z=f(x,y)$ is $\nabla (f(a,b))$ always perpendicular to the surface at the point $(a,b,f(a,b))$.

I see many problems using this gradient function to find a normal vector to a surface so it is most likely true I guess but I can't seem to come up with an intuitive reason or better still a proof to show me why this is the case.

Any help?

Marlo P
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    It is not perpendicular to the surface. It is perpendicular to the level line $f(x,y)=const=f(a,b)$ – Svetoslav Oct 05 '15 at 21:25
  • Note that $\nabla f$ is a vector in $\mathbb{R}^2$ but your surface is a subset of $\mathbb{R}^3$. Your question is ill-posed as it stands. – Giuseppe Negro Oct 05 '15 at 21:30

2 Answers2

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The tangent plane to your surface in the point $(a,b,f(a,b))$ is generated by $(1,0,\frac{\partial f}{\partial x}(a,b))$ and $(0,1,\frac{\partial f}{\partial y}(a,b))$, so a normal vector to the surface in $(a,b,f(a,b))$ is $(\frac{\partial f}{\partial x}(a,b)),\frac{\partial f}{\partial y}(a,b)),-1)$, which is not the gradient of your $f$, $(\frac{\partial f}{\partial x}(a,b)),\frac{\partial f}{\partial y}(a,b))$

Laz
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Your problem can be fixed.

That means that one should take instead the function $F(x,y,z)=f(x,y)-z$ and tell that $${\rm grad}F=({\rm grad f},-1)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},-1)$$ This triplete is perpendicular to the tangents $(1,0,\frac{\partial f}{\partial x})$ and $(0,1,\frac{\partial f}{\partial y})$ of the level surface $F(x,y,z)={\rm constant}$, i.e. normal to them.

janmarqz
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