# SOLUTION: Stanford University Polar Coordinates & ?Orientations Diffeomorphism Proof

π₯ 3 β 3π₯π¦ 2
π(π₯, π¦) = π cos(3π) = 2
, π(0,0) = 0.
π₯ + π¦2
Proof. It is simpler to use polar coordinates short. For (π₯1 , π¦1 ) = (0,0) we have
|π(π₯1 , π¦1 ) β π(π₯2 , π¦2 )| = |π(π₯2 , π¦2 )| = |π cos(3π)| β€ π.
For (π1 , π1 ) not at the derivation and any (π2 , π2 ) deliberate a track from (π2 , π2 ) to (π1 , π1 ) consisting of two
parts: π = π2 until we get the similar π and then π = π1 until we get the similar π:

We see |π(π1 , π1 ) β π(π2 , π2 )| β€ |π(π2 , π1 ) β π(π2 , π2 )| + |π(π2 , π1 ) β π(π1 , π1 )|.

Estimate the pristine summand:
|π(π2 , π1 ) β π(π2 , π2 )| = π2 |cos(3π1 ) β cos(3π1 )| β€ 3π2 |π2 β π1 |.
Note that it is 3 times the extension of the pristine multiply of the track.
Estimate the promote summand:
|π(π2 , π1 ) β π(π1 , π1 )| = |(π2 β π1 ) cos(3π1 )| β€ |π2 β π1 |.
It is simply the extension of the promote multiply of the track.
Now we deficiency to appreciate the sum by πΎdist((π1 , π1 ), (π2 , π2 )) for some trustworthy πΎ. Deliberate the bold
blue triangle. The angle betwixt pristine two sides is frequently oblite (we can frequently choose
π1 , π2 : |π2 β π1 | β€ π). Therefore
dist((π1 , π1 ), (π2 , π2 )) = π2 + π 2 β 2ππ cos β ππ β₯ π2 + π 2 ,
dist((π1 , π1 ), (π2 , π2 )) β₯ βπ2 + π 2 .
π

One further being to do is to appreciate the pristine incurvation extension πΏ1 by π. It is further or near plain that πΏ1 β€ 2 π
(the whack contingency when ...