love your visually intuitive explanation, very helpful. I look on maths as based on the physical reality, and not the other way around. What you have done is uncover this basis to this particular problem, and then the maths follows easily as it is clear what is going on.

A simple way of rephrasing what you have illuminated above, is to say that the shortest ie the steepest, route between contours or level surfaces, is by definition, in a direction perpendicular to the plane of the contour, since this is how 2D contours are constructed ie in a plane perpendicular to the third dimension axis.

This perpendicular direction, parallel to the 3rd dimensional axis, when projected onto the 2D contour, will necessarily be perpendicular to the 2D tangent (or gradient of the 2D function to any point on it.

Thus by definition, ‘grad f’ or the gradient function of a 3D shape is perpendicular to the gradient of the 2D shape. thanks again Robyn

Proof: Suppose that (a,b) is any vector that is tangent to the level curve of f through (x0,y0). So the derivative of f along the vector (a,b) is zero [some call this the directional derivative, but the directional derivative is along a unit vector]. But this means that dot(gradf(x0,y0), (a,b))=0 since dot(gradf(x0,y0), (a,b)) is the derivative of f along the vector (a,b). But this means that gradf(x0,y0) is perpendicular to the vector (a,b), which is what we wanted to prove.

This generalizes when you are talking about level surfaces and the gradient there is perpendicular to the surface instead of the level curve.

Let F(x, y, z) =c define a family of (level) surfaces.
F(x1 , y1 ,z1 )=c1 and F(x2 , y2 ,z2 )=c2 are two adjacent surfaces where x2 = x1 + Δx, y2 = y1 + Δy, and
z2 = z1 + Δz.
Then in a small neighborhood ΔF = F(x2 , y2 ,z2 ) – F(x1 , y1 ,z1 )
= F(x1 , y1 ,z1)+(dx F) Δx + (dyF) Δy +(dz F) Δz – F(x1 , y1 ,z1 )
= grad(F) • Δr = c2 – c1
If we restrict to one surface, so that the vector Δr lies within that surface (technically tangent to it) and c2 – c1= 0, then the vector grad(F) must be perpendicular to the surface.

Seems to me you plotted F(1,1) =

The rigorous proof that the gradient produces a vector normal to a given level surface is evidently pretty complex, since it is omitted from my vector analysis text. There is, however, an informal explanation.

Imagine a function f(x,y,z) defined everywhere in a box. Pick a point (x0,y0,z0), and let’s say that f(x0,y0,z0)=C. Then it isn’t hard to believe that a smooth surface could exist where f(x,y,z)=C, which passes through (x0,y0,z0) and is continuous throughout the box.

If you took a directional derivative while staying on the surface, it is 0, since f(x,y,z) is constant on the surface.

Notice that the directional derivative is in a direction tangent to the surface, and its magnitude is zero. df/du = grad(f).u implies that this direction, tangent to the surface, is perpindicular to the direction of greatest change. This means that the direction of greatest change (aka the gradient vector) is normal to the level surface.

Jonathan

In fact, you’re both heading in the same direction as the book. You do need to take partial derivatives. To refresh, is the slope of the function if you hold y constant and vary x. Similarly, is the slope of the function if you hold x constant and vary y.

The tangent plane to a surface at a point is .

However, when I read this, it didn’t make clear what was going on intuitively/concretely. In some sense, the first term of the three terms makes sure that that point is on the plane, and the second and third terms describes what’s happening locally when you move slightly to the right or left. This sounds like what you are talking about Jonathan.

But I don’t think that explanation is totally correct — or makes great sense to me. And then the book goes on to say that another way to write the tangent plane is , for any function . Well, that was a real challenge for me — where the heck this comes from.

Which led to this post.

Ah, yes, I see the point now. I had made the erroneous assumption that there existed a technique to do ‘differentiation in 3D’ to get the gradient of the plane – without knowing whether such a technique actually existed!

I guess that’s the difference between a mathematician and and engineer (me): a mathematician would demand to see the proof that ‘differentiation in 3D’ existed before using it, but an engineer would just try extrapolating from 2D and then try to verify empirically the results were correct (or at least close enough for the problem at hand). Which would not work in this case…

Come to think of it (now I’m really beyond what I recall from calculus) isn’t there a notion of partial derivatives where you can get the rate of change of z with respect to x AND y? Wouldn’t that be the ‘gradient’ of the plane that is tangential to our surface? Or can partial derivatives only be done ‘one axis at a time’?