Gradient and normal

1. Feb 23, 2010

azay

It states in course notes:

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If y = f(x) defines a surface in (n+1) dimensional space then the normal is defined as the (n+1)-dimensional vector:
$$(\frac{\partial f(x)}{\partial x1},(\frac{\partial f(x)}{\partial x2},...,(\frac{\partial f(x)}{\partial n1},-1)$$

This implies that the projection of this vector to the plane y = 0 is the gradient. The gradient is perpendicular to the tangent line of the contourline at the point, which is the projection of P onto the plane y = 0.

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I think I understand why the gradient is 'normal' or 'perpendicular' to the level set of a function. After all, moving along, say the contour line (if the number of variables is 2), implies no change at all since f is constant on that line. So if I'm correct it's kind of intuitive to say the direction in which change is maximal is 'perpendicular' to that line. This gives however rise to the question: doesn't the projection of the normal to any plane (not just y = 0 I mean but also y = 1 etc.) give the gradient?

And my main question: why is the normal vector, defined like above, perpendicular to the tangent hyperplane at a certain point (intuitively, not mathematically)? Where does this '-1' come to play?

2. Feb 23, 2010

slider142

They are looking at the surface as the graph of the solutions of the equation h(x, y) = f(x) - y = 0. In that case, the gradient vector of h(x, y) is normal to the level set h(x, y) = 0 = f(x) - y, and the last component of the gradient vector is dh/dy = -1.
Down to R2+1 dimensional space, we have z = f(x, y), for example x2 + y2, defining a surface in R3, which is a level set of the function h(x, y, z) = f(x, y) - z for h(x, y, z) = 0.

3. Mar 5, 2010

azay

Thanks

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