Deriving the surface area equation

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The discussion centers on the derivation of the surface area equation, specifically questioning the treatment of vectors a and b as forming a rectangle instead of a parallelogram. It highlights that a and b are tangent vectors on a curved surface, which do not create a rectangular area due to the surface's curvature. The calculations attempted to simplify the area to a rectangle resulted in an additional term, suggesting that this term must equal zero, though the reasoning behind this is unclear. The conclusion emphasizes the need to project the vectors onto the xy-plane for accurate area calculations, where they may not remain perpendicular. Understanding the geometric properties of the surface is crucial for deriving the correct surface area formula.
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consider the following image

4IrsvlR.png


(the red is the surface area element and the green is the differential element that I'm integrating over)

when we derived this in class, we treated the area formed by vectors a and b, as the area of a parallelogram. the thing is, a and b should be at right angles implying that rest of the angles should be right angles as well. if that's the case, then why can't i just treat the area formed from a and b, as a rectangle?

well, that's what i attempted to try but failed. i get all the appropriate terms as they do in the surface area formula, except i get an extra term as shown in the link below, implying, perhaps.. that that extra term must go to zero. but i don't know why it should.

if I'm incorrect in treating the shape formed from a and b as a rectangle, why is this incorrect?

calculations:

http://i.imgur.com/6MFO4xI.jpg
(image was too large to put here)
 
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It's very difficult to read that- the colors do not show up on black very well! But to answer your question, the "a" and "b" on the surface are NOT vector IN the surface because it is not flat- they are tangent vectors to a curved surface and so do NOT form a rectangle in the surface. In order to calculate and area we have to project down to the xy- plane and the vectors are no longer necessarily perpendicular there.
 

Thread 'Proving that convexity implies second order derivative being positive'

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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