Can a Cross Product Determine a Gradient Vector for a Non-Function Surface?

[fk378]

This is a general question. If we have a parametric equation r(u,v) and we take r_u and r_v, then take their cross product, does it give us the gradient vector? Or just a vector parallel to the gradient vector?

 

[Dick]

If r(u,v) has a gradient vector, then it's a scalar function. If r_x and r_y are vectors then r is a vector function. Which is it? I don't get you.

 

[fk378]

I just edited my original post; I meant r_u X r_v would give a vector orthogonal to the tangent plane. And since the gradient is also (r_u, r_v), is this vector the same vector?

 

[Defennder]

I just edited my original post; I meant r_u X r_v would give a vector orthogonal to the tangent plane. And since the gradient is also (r_u, r_v), is this vector the same vector?

You mean to say the gradient of the surface described by a scalar function. Well there are two possible vectors for r_u X r_v, you might end up getting the inward pointing normal vector instead.

 

[Dick]

If r is a scalar function, r_u and r_v are scalars. There is no cross product. If r(u,v) is a vector function then r_u x r_v is a normal to the surface defined by r(u,v), but then there is no gradient. Someone is confused here, and I don't think it's me.

 

[HallsofIvy]

You are using the terminology incorrectly. A function has a gradient, not a surface.

If a surface is given by f(x,y,z)= constant, the grad f is perpendicular to the surface.
If that surface is written as a position vector $\vec{r}(u,v)$, then $\vec{r}_u\times\vec{r}_v$ is also perpendicular to the surface. Your question is whether those two vector must also have the same length.

The answer is "no" because the gradient of f is a specific vector while different parameterizations of the surface f(x,y,z)= constant will give vectors of different length.

 

[fk378]

If r is a scalar function, r_u and r_v are scalars. There is no cross product. If r(u,v) is a vector function then r_u x r_v is a normal to the surface defined by r(u,v), but then there is no gradient. Someone is confused here, and I don't think it's me.

Why would a surface not have a gradient?

And yes I am confused, which is why I'm asking questions, but I appreciate your input.

 

[Dick]

Why would a surface not have a gradient?

And yes I am confused, which is why I'm asking questions, but I appreciate your input.

The big confusion here is that you have two different representations of the surface, as a position vector $\vec{r}(u,v)$ and as a level surface by R(x,y,z)=C. Those two r's are completely different objects. You are confusing me (at least) by labeling them the same and using them interchangeably. As Halls said, if r and R do happen to represent the same surface, then, yes, the cross product and the gradient are parallel, not necessarily equal.

 

[D H]

Why would a surface not have a gradient?

Because a surface isn't a function. Think of it this way: The unit 2-sphere is the locus of points in R³ that satisfy $x^2+y^2+z^2-1=0$. Note well: $2x^2+2y^2+2z^2-2=0$ also describes the unit 2-sphere. The gradients of the functions $x^2+y^2+z^2-1$ and $2x^2+2y^2+2z^2-2$ are obviously different.