gradient of level curves

[PG1995]

Hi

Could you please help me with this query? Thanks a lot.

Regards
PG

Useful Link(s):
1: https://tutorial.math.lamar.edu/Classes/CalcIII/MultiVrbleFcns.aspx

 

[steveB]

First, isnt that first function a sphere and not a plane?

Also, what is the definition of a "level curve". It's not clear what this means so it is hard to answer.

Basically, the curve shapes do not matter. All that matters is the directions of the normals at the point specified. The angle between the normals of each function, at that point, defines the angle.

EDIT: I looked up what a "level curve" is. It seems to me that the technique specified does not require a level curve, but probably a potential function is needed.

EDIT AGAIN: Looking further into this, it seems that probably "potential functions" that I'm used to in field theory are basically the same as what mathematicians call "level sets" where f(x1,x2, ... xr)=c, which is a generalization of the level curve f(x,y)=c.

The problem you presented is in 3 dimensions, so you would want to have a case where f(x,y,z)=constant. This is essentially a potential function because we can assign meaning to the value of the constant (for example potential energy), and all points on that surface would have the same value. We then get a set of surfaces as we vary the constant, and you can imagine that the maximum change must occur in the direction that is orthogonal to that surface, because there is no change in the direction along the surface.

 

[PG1995]

Thank you.

We then get a set of surfaces as we vary the constant, and you can imagine that the maximum change must occur in the direction that is orthogonal to that surface, because there is no change in the direction along the surface.

Could you please help me with this follow-on this query? This is the reference I have used. Thanks.

Regards
PG

 

[steveB]

This seems to be a "terminology" question. Basically, a vector that is normal to a curve forms a right angle with the tangent line to the curve, at the point of interest on that curve. The angle is not referenced to any particular coordinate axis.

In this particular example, the gradient of sqrt(x^2+y^2) is (x i + y j)/sqrt(x^2+y^2) which is a unit vector in the radial direction outward, hence this vector always points outward from the origin of the xy-coordinate system. The level curves are circles, and the tangent to a circle at any point is always orthoganal to the position vector of that point. Oh, and you should note that x i +y j is the postition vector, which is obviously parallel to the unit vector (x i + y j)/sqrt(x^2+y^2) found by taking the gradient.

 

[PG1995]

Thank you.

So, I take the arrow "3" represents the grad's direction. But I'm little confused about the quoted text below. There doesn't occur any change, let alone maximum change, in the direction of of grad, i.e. the arrow "3". In other words, could you please elaborate a bit on the quote below in the context of the example at hand where we have both surface and its level curves? Thanks

We then get a set of surfaces as we vary the constant, and you can imagine that the maximum change must occur in the direction that is orthogonal to that surface, because there is no change in the direction along the surface.

Regards
PG

 

[steveB]

Yes, arrow 3 indicates the direction, but I would prefer to see that arrow projected into the xy-plane to be consistent with what we are saying. The arrow 3 is orthogonal to the level curves, but it is not orthogonal to the cone's surface.

This may be why you are confused about the wording. If we restrict ourselves to the xy-plane, then the function f(x,y) is a scalar function that we can take the gradient of. The value of that function sees maximum change in the direction of the gradient which is an outward direction from the origin in this case.

This concept is central to the potential theory we use in many branches of physics. In particular, we see this in electro-statics where the electric field is the negative gradient of the voltage potential. We often draw constant potential surfaces (3d) or constant potential curves (2d) and the constant potential contours are shown to be orthoganal to the electric field lines, which is given by the (negative) gradient of potential.

 

[PG1995]

Thank you.

Please don't mind my asking. Think that the level curve k=3 has been projected onto the x-y plane, then I don't know why I think it's the arrow "2" which gives direction of maximum rate of change. When both the level curve and arrow "2" are projected onto the x-y plane, then the arrow does seem to be orthogonal to the curve as far as x-y plane matters. Kindly have a look here; perhaps it might help you to see what I'm trying to say. Please guide me with this. Thanks.

Regards
PG

 

[steveB]

There are two separate issues to consider here.

First, a path along the direction of the vector 2 has to traverse a greater distance than that for vector 3 to affect the same change in the value k. Hence, the vector 2 direction could not possibly have greater rate of change. Then, you could consider if there is another vector in the 3d space that defines a direction of greater rate of change than that of vector 3. It doesn't seem that there is.

The second issue is much more important. That is, it is a mistake to even be thinking of 3 dimensions when the 2D gradient is being considered. A gradient is an operation performed on a scalar function, and the dimensionality of the domain of that function determines whether a 1D, 2D or 3D gradient is being considered. A 2d gradient in xy-system does not know anything about the direction z.

However, there is another important point here. In a sense, there is a mapping from the 2D xy-plane to the 3D surface, but the surface, which abstractly exists in the 3D space, is still a 2D structure. Note that, because of this mapping, as you travel along the direction of the vector 3, you are also traversing along the direction of the vector 2, due to this mapping. So, in some sense your visualization is not wrong. However, rate of change, or gradient, or any of the other vector space notions are things to consider only in the xy-plane.

It's important to understand this, so let's try to put a physical example in place. In a 2D electrostatics problem, you might determine the potential voltage V as a function of the coordinates x and y. Hence V(x,y) is something we can take the gradient of to determine the electric field. That electric field will have direction as a function of position (x,y) and will define directions only in the xy-plane.

One can extend the example to a 3D problem. Then, we have V(x,y,z), which will define electric fields that have directions in 3D space. In this case you would not even be able to visualize a fourth dimension to create a vector analogous to your vector 2.

 

[PG1995]

Thanks a lot, Steve.

But I'm still struggling. I have found something which is quite, at least it looks like that, in line with what I said in my previous post. Please have a look here and kindly help me. Thank you.

Regards
PG

 

[steveB]

So, yes I do think that arrow is like the arrow 2 from your other drawing. As I hinted at, there is nothing wrong with your visualization and thinking in terms of that arrow. However, notice how they project that arrow down into the xy-plane in order to talk about the gradient. So direction has meaning in the xy-plane only, and when we talk about rate of change or derivatives, it is relative to the xy-coordinate system. When dealing with a purely mathematical problem, you might not see this so clearly, which is why I mentioned the physical example with electric fields and potential voltage, above. If the third dimension is potential voltage and we are dealing with a 2D xy-plane electro-statics problem, what meaning is there to a vector 2i+4j+6k? The i and j components are coordinate distances, and the k component is a voltage. These are not compatible for making a meaningful vector with the proper transformation properties needed in vector calculus. In this case, the third dimension is a scalar number as a function of the coordinates x and y. In a 3 dimensional problem, you would have a scalar number as a function of the coordinates x, y and z (i.e. w=V(x,y,z). Do not confuse this latter z with the z from the 2 dimensional case (i.e. z=f(x,y)). They are different, and "z" is just a label that says nothing about the nature of the objects involved.