Exercise 9.17

Exercise 17: Let f ˇ = ( f 1 , f 2 ) be the mapping of 2 into 2 given by

f 1 ( x , y ) = e x cos y , f 2 ( x , y ) = e x sin y .

(a) What is the range of f ˇ ?

(b) Show that the Jacobian of f ˇ is not zero at any point of 2 . Thus every point of 2 has a neighborhood in which f ˇ is one-to-one. Nevertheless, f ˇ is not one-to-one on 2 .

(c) Put ǎ = ( 0 , π 3 ) , b ˇ = f ˇ ( ǎ ) , let ǧ be the continuous inverse of f ˇ , defined in a neighborhood of b ˇ , such that ǧ ( b ˇ ) = ǎ . Find an explicit formula for ǧ , compute f ˇ ( ǎ ) and ǧ ( b ˇ ) , and verity the formula (52).

(d) What are the images under f ˇ of lines parallel to the coordinate axes?

Answers

(d) Fix y = y 0 . Then f ˇ maps ( x , y 0 ) , x R , to the ray from (but not including) the origin through the point ( cos y 0 , sin y 0 ) on the unit circle. The negative values of x map to points on the ray inside the circle, and the positive values of x map to points outside the unit circle.

Fix x = x 0 . Then f ˇ maps ( x 0 , y ) , v R , to the circle of radius e x . Note that f ˇ ( x 0 , y ) = f ˇ ( x 0 , y + 2 ) for all integers n .

(a) The results of (d) show that the range of f ˇ is 2 minus the origin.

(b) Since

D 1 f 1 ( x , y ) = e x cos y D 1 f 2 ( x , y ) = e x sin y D 2 f 1 ( x , y ) = e x sin y D 1 f 2 ( x , y ) = e x cos y

the Jacobian of f ˇ at ( x , y ) is e 2 x ( cos 2 y + sin 2 y ) = e 2 x 0 for all x . From (d) we have f ˇ ( x , y ) = f ˇ ( x , y + 2 ) for all integers n , so f ˇ is not one-to-one on 2 .

(c) Let w = f 1 ( x , y ) = e x cos y , z = f 2 ( x , y ) = e x sin y . Then, in a neighborhood of b ˇ = ( 1 2 , 3 2 ) where w 0 , we have w 2 + z 2 = e 2 x , z w = tan y , so

ǧ ( w , z ) = ( log w 2 + z 2 , arctan ( z w ) ) f ˇ ( x , y ) = ( e x cos y e x sin y e x sin y e x cos y ) ǧ ( w , z ) = ( w ( w 2 + z 2 ) z ( w 2 + z 2 ) z ( w 2 + z 2 ) w ( w 2 + z 2 ) ) f ˇ ( ǎ ) ǧ ( b ˇ ) = ( 1 2 3 2 3 2 1 2 ) ( 1 2 3 2 3 2 1 2 ) = ( 1 0 0 1 )

User profile picture
2023-08-07 00:00
Comments