High School How to Visualize 3D Surfaces for a PVT Diagram

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    Calc 3 Surfaces
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To visualize 3D surfaces for a PVT diagram, start by folding a piece of paper in half both lengthwise and widthwise. After unfolding, manipulate the paper by aligning the left horizontal crease with the bottom vertical crease and creating a -45 degree bend. This technique helps illustrate the plane represented by the equation 2x + 2y + z = 6 by drawing traces in the coordinate planes. The discussion emphasizes the importance of clear instructions for effective visualization. Overall, the method provides a practical approach to understanding complex 3D surfaces.
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TL;DR
How to visualize 3-d surfaces in a test friendly manor
1. Fold in half the long way
8C3B63F5-0397-4C14-B408-AF0C7B8C5B47.jpeg

2. Fold in half the short way

53BE26D0-C33C-4FEE-86DD-29C59E96FA10.jpeg

3. Unfold paper, grasping the left side horizontal crease, bend the paper as shown by making the left crease flush with the bottom vertical crease and crease along the -45 degree bend (not shown)

46A766D2-4969-4B6C-A39C-9150D956FC83.jpeg


4. This is the visualization for the plane
2x+2y+z=6, made by drawing traces of the given surface in the coordinate-planes. For example, in the xy-plane set z=0 to get 2x+2y=6.
DFD72ACE-D5AC-4430-A283-778E68A88464.jpeg
 
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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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