Question

Basic multivariable question.

div(f(x,y,z))=∇⋅f(x,y,z), for f(x,y,z)=f(x)i+f(y)j+f(z)k is ∇⋅f(x,y,z)=((∂/∂x)f(x)i+(∂/∂y)f(y)j+(∂/∂z)f(z)k)⋅(f(x)i+f(y)j+f(z)k)

∇⋅f(x,y,z)=∇f(x,y,z)cos(φ)=((((∂/∂x)f(x))^2+((∂/∂y)f(y))^2+((∂/∂z)f(z)k)^2)^(1/2))((f(x)^2+f(y)^2+f(z)^2)^(1/2))cos(φ)

But ∇⋅f(x,y,z)=(∂/∂x)f(x)i+(∂/∂y)f(y)j+(∂/∂z)f(z)k according to textbook...

I'm confused...

Answer 1

∇⋅f(x,y,z)=(∂/∂x)f(x)i+(∂/∂y)f(y)j+(∂/∂z)f(z)k
is how I know it
where did you see the other one?

Answer 2

Isn't it according to the rules of dot product? a⋅b=abcos(φ)

Answer 3

actually I know it as\[\nabla \cdot f=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\]remember that del is not a function, it is an /operator/ like d/dx

Answer 4

Oh, it's an operator. Gah, why do I keep forgetting stuff like this. This is why I need to review. X(

But isn't the gradient the vector sum, not just the sum of magnitudes?

Answer 5

Oh, right, the derivative of the function f will include the unit vectors.

Answer 6

\[\nabla\neq\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\]but\[\nabla \cdot f=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\]but this is not the gradient
keep in mind operators have to be operating on something to be set equal to anything

d/dx=k means nothing

Answer 7

also remember that\[\nabla f\neq\nabla \cdot f\neq\nabla\times f\]and all that
the middle is a scalar, the others are vectors

Answer 8

What, what's the difference between ∇ƒ and ∇⋅ƒ

Answer 9

And how can you take the cross product of the gradient operator and a function?

Answer 10

a lot of difference\[\nabla f=<f_x,f_y,f_z>\]which is the gradiant, the vector perpendicular to the level curve of the function, as opposed to\[\nabla \cdot f=f_x+f_y+f_y\]which is a scalar
you are not actually taking a dot product. that is impossible for the reason you stated.
instead the notation of divergence is made similar to that of the dot product because of the obvious mathematical similarity
it is a bit misleading though

Answer 11

Ah, okay! Thanks!

Answer 12

If we again think of F as the velocity field of a flowing fluid then div(F) represents the net rate of change of the mass of the fluid flowing from the point (x,y,z) per unit volume.

Answer 13

welcome!