SOLUTION: Stanford University Polar Coordinates & ?Orientations Diffeomorphism Proof

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π‘₯ 3 βˆ’ 3π‘₯𝑦 2
𝑓(π‘₯, 𝑦) = π‘Ÿ cos(3πœƒ) = 2
, 𝑓(0,0) = 0.
π‘₯ + 𝑦2
Proof. It is simpler to use polar coordinates short. For (π‘₯1 , 𝑦1 ) = (0,0) we have
|𝑓(π‘₯1 , 𝑦1 ) βˆ’ 𝑓(π‘₯2 , 𝑦2 )| = |𝑓(π‘₯2 , 𝑦2 )| = |π‘Ÿ cos(3πœƒ)| ≀ π‘Ÿ.
For (π‘Ÿ1 , πœƒ1 ) not at the derivation and any (π‘Ÿ2 , πœƒ2 ) deliberate a track from (π‘Ÿ2 , πœƒ2 ) to (π‘Ÿ1 , πœƒ1 ) consisting of two
parts: π‘Ÿ = π‘Ÿ2 until we get the similar πœƒ and then πœƒ = πœƒ1 until we get the similar π‘Ÿ:

We see |𝑓(π‘Ÿ1 , πœƒ1 ) βˆ’ 𝑓(π‘Ÿ2 , πœƒ2 )| ≀ |𝑓(π‘Ÿ2 , πœƒ1 ) βˆ’ 𝑓(π‘Ÿ2 , πœƒ2 )| + |𝑓(π‘Ÿ2 , πœƒ1 ) βˆ’ 𝑓(π‘Ÿ1 , πœƒ1 )|.

Estimate the pristine summand:
|𝑓(π‘Ÿ2 , πœƒ1 ) βˆ’ 𝑓(π‘Ÿ2 , πœƒ2 )| = π‘Ÿ2 |cos(3πœƒ1 ) βˆ’ cos(3πœƒ1 )| ≀ 3π‘Ÿ2 |πœƒ2 βˆ’ πœƒ1 |.
Note that it is 3 times the extension of the pristine multiply of the track.
Estimate the promote summand:
|𝑓(π‘Ÿ2 , πœƒ1 ) βˆ’ 𝑓(π‘Ÿ1 , πœƒ1 )| = |(π‘Ÿ2 βˆ’ π‘Ÿ1 ) cos(3πœƒ1 )| ≀ |π‘Ÿ2 βˆ’ π‘Ÿ1 |.
It is simply the extension of the promote multiply of the track.
Now we deficiency to appreciate the sum by 𝐾dist((π‘Ÿ1 , πœƒ1 ), (π‘Ÿ2 , πœƒ2 )) for some trustworthy 𝐾. Deliberate the bold
blue triangle. The angle betwixt pristine two sides is frequently oblite (we can frequently choose
πœƒ1 , πœƒ2 : |πœƒ2 βˆ’ πœƒ1 | ≀ πœ‹). Therefore
dist((π‘Ÿ1 , πœƒ1 ), (π‘Ÿ2 , πœƒ2 )) = π‘Ž2 + 𝑏 2 βˆ’ 2π‘Žπ‘ cos βˆ π‘Žπ‘ β‰₯ π‘Ž2 + 𝑏 2 ,
dist((π‘Ÿ1 , πœƒ1 ), (π‘Ÿ2 , πœƒ2 )) β‰₯ βˆšπ‘Ž2 + 𝑏 2 .
πœ‹

One further being to do is to appreciate the pristine incurvation extension 𝐿1 by π‘Ž. It is further or near plain that 𝐿1 ≀ 2 π‘Ž
(the whack contingency when ...


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