• ericm1234
In summary, on pages 14-15, the authors derive the normal vector to a surface by cutting the surface with a plane and then using the curve 'c' in the xz plane. They then want to use the components of 'u' but they have a problem because 'u sub x' in the x direction is not a vector but a function that is equal to 'partial f/partial x times U sub X'.
ericm1234
on pages 14-15, in deriving the normal vector to a surface, they use a plane to cut the surface (the plane is parallel to the xz plane) then use the curve 'c' in the xz plane (this curve being where the plane intersects the surface), draw a tangent vector 'u' and want to use the components of 'u'. Now, they use an arbitrary length called 'u sub x' in the x direction (this again is in the xz plane) but WHY is the component in the z direction designated 'partial f/partial x times u sub x'? My question is NOT "why is the derivative of z with respect to x replaced with the partial derivative of f with respect to x", that I understand. But why in the first place is the change in the z direction a derivative times change in x?

ericm1234 said:
on pages 14-15, in deriving the normal vector to a surface, they use a plane to cut the surface (the plane is parallel to the xz plane) then use the curve 'c' in the xz plane (this curve being where the plane intersects the surface), draw a tangent vector 'u' and want to use the components of 'u'. Now, they use an arbitrary length called 'u sub x' in the x direction (this again is in the xz plane) but WHY is the component in the z direction designated 'partial f/partial x times u sub x'? My question is NOT "why is the derivative of z with respect to x replaced with the partial derivative of f with respect to x", that I understand. But why in the first place is the change in the z direction a derivative times change in x?
"on pages 14-15" of what?

If ux is a vector in the (positive) x direction, then "the component in the z direction" cannot be "partial f/partial x time ux" because that is a vector in the x direction, not the z direction. Are you sure you are quoting correctly?

Pgaes 14-15 of 'div grad curl and all that', second edition however.
Let me rephrase:
there is a surface S in 3-d (x, y, z). This is cut by a plan parallel to the xz plane. This intersection of the plane and surface creates a curve C.
Now there is a picture showing the x and z axes with this curve C, and a tangent vector called U. U is decomposed into U sub X and U sub Z HOWEVER, U sub Z is 'equal' here to as 'partial f/ partial x' times U sub X.
So my question is why is U sub Z equal to partial f/partial x time U sub X.

Oh, and f refers to the 3-d surface: z=f(x,y). The second picture I referred to is a 2-d picture with the x and z axes, just to clarify.

## What is the main concept of "Div Grad Curl and All That"?

The main concept of "Div Grad Curl and All That" is to introduce the fundamental ideas of vector calculus, specifically the concepts of divergence, gradient, and curl, and how they relate to physical phenomena.

## What are the applications of div, grad, and curl in real-world problems?

Div, grad, and curl have many applications in fields such as physics, engineering, and mathematics. They are used to describe the behavior of fluids, electric and magnetic fields, and the motion of particles, among other things.

## Can you provide an example of how div, grad, and curl are used in a real-world scenario?

One example is the use of curl to describe the rotation of a fluid in a vortex. Another example is the use of gradient to determine the path of steepest descent in a topographical map.

## What is the relationship between div, grad, and curl?

The relationship between div, grad, and curl is described by the fundamental theorem of calculus for vector fields. It states that the curl of a gradient is always zero, and the divergence of a curl is always zero.

## How can I improve my understanding of div, grad, and curl?

To improve your understanding of div, grad, and curl, it is important to practice solving problems and applying these concepts in different scenarios. Reading additional materials and seeking help from a tutor or teacher can also be helpful.

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