EDITED: The first attempt didn't really solve anything, so I deleted it. Please take a look at the below post.
I would be greatful for your comments.
A function is defined by
(a) Find the set of critical points of .
Roughly sketch the u and v contours in the -plane together with the critical points of . Inlcude and label the contours u=0 and v=0.
(b) Write down the derivative of the function and the derivative of the local inverse at a non-critical point .
Deduce an expression for and evaluate it at the point =(2,1).
(c) Find, if possible, an appropriate formula for the local inverse function at each of the points =(2,1) and (1,1) justifying your answers. Carefully state the region in the -plane on which the local inverse is defined and its image on the -plane.
On separate diagrams, sketch the and contours through each of the given points. Comment on the relationship between the contours in each diagram.
(d) Find directly from the local inverse function you found in part (c).
Evaluate it at the image point of =(2,1) in the -plane and verify that this agrees with the result in part (b).
(e) Write down the equations of the tangent flat to at the point =(2,1).
I'll post my work in a separate post.
Attempt at solution # 2.
(a) critical points and sketch of u, v contours on (x,y)-plane.
The contour of u is a straight line of the type y=x-u, and u can take any value on the real line -> it is a series of straight lines with slope of 1.
Contour are hyperbolas with different distance from the origin (varies with v).
The critical point of f are the points where the determinant of Jacobian is zero:
If Jacobian is
then |J|=-2y+2x=2(x-y)=0 iff x=y. So the critical point is an infinite set represented by a line y=x.
The derivative of a function is a matrix (Jacobian) calculated as follows:
The derivative of the local inverse is the inverse Jacobian:
At point (2,1) the value of this is
(c) formula for local inverse
I solve for x and y the system of equations
and I get
At (2,1) it is
We want x>0, y>0
Solving the system of equations, I got a pair:
u=1, v=3 so (1,3) is the value of at the point (2,1). For the local inverse to exist, each (x,y) must have only one pair (u,v). Therefore, I think the relevant area is (b):
At the point (1,1) the local inverse does not exist because at this point the |J|=0 and the Jacobian matrix is not invertible.
as in (b)
(e) tangent flat equation at (2,1)
therefore, via some matrix multiplication we get the equation
I would really appreciate your comments! Thank you!