Question from reading Div Grad Curl and All That

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The discussion focuses on the derivation of the normal vector to a surface as presented in "Div Grad Curl and All That." It examines the relationship between the tangent vector components in the xz plane, specifically questioning why the z component is expressed as 'partial f/partial x times u sub x.' The clarification emphasizes that u sub z is not simply a vector in the z direction but is derived from the surface's behavior in relation to changes in x. The inquiry seeks to understand the mathematical reasoning behind this representation, particularly in the context of the surface defined by z = f(x, y). This highlights the connection between partial derivatives and the geometric interpretation of tangent vectors on surfaces.
ericm1234
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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?
 
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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.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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