# Thread: f: R^2 to R^2 function, derivatives and inverses

1. ## f: R^2 to R^2 function, derivatives and inverses

Question.

A function $f:R^2->R^2$ is defined by

$f(x,y)=(u,v)^T=(x-y,x^2-y^2)$.

(a) Find the set of critical points of $f$.

Roughly sketch the u and v contours in the $(x,y)$-plane together with the critical points of $f$. Inlcude and label the contours u=0 and v=0.

(b) Write down the derivative of the function $f$ and the derivative of the local inverse at a non-critical point $(x,y)$.

Deduce an expression for $\frac{\partial{x}}{\partial{u}}$ and evaluate it at the point $(x,y)$=(2,1).

(c) Find, if possible, an appropriate formula for the local inverse function at each of the points $(x,y)$=(2,1) and (1,1) justifying your answers. Carefully state the region in the $(u,v)$-plane on which the local inverse is defined and its image on the $(x,y)$-plane.

On separate diagrams, sketch the $u$ and $v$ contours through each of the given points. Comment on the relationship between the contours in each diagram.

(d) Find $\frac{\partial{x}}{\partial{u}}$ directly from the local inverse function you found in part (c).

Evaluate it at the image point of $(x,y)$=(2,1) in the $(u,v)$-plane and verify that this agrees with the result in part (b).

(e) Write down the equations of the tangent flat to $f$ at the point $(x,y)$=(2,1).

I'll post my work in a separate post.

2. EDITED: The first attempt didn't really solve anything, so I deleted it. Please take a look at the below post.

3. 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 $v=x^2-y^2$ 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

|1 -1|
|2x -2y|

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.

(b) Derivative

The derivative of a $R^2->R^2$ function is a matrix (Jacobian) calculated as follows:

$|\frac{\partial{u}}{\partial{x}}\frac{\partial{u}} {\partial{y}}|$
$|\frac{\partial{v}}{\partial{x}}\frac{\partial{v}} {\partial{y}}|$

which is

|1 -1|
|2x -2y|

The derivative of the local inverse is the inverse Jacobian:

$|\frac{-2y}{2(x-y)} \frac{1}{2(x-y)}|$
$|\frac{-2x}{2(x-y)} \frac{1}{2(x-y)}|$

At point (2,1) the value of this $J^{-1}$ is $\frac{-2*1}{2(2-1)}=-1$

(c) formula for local inverse

I solve for x and y the system of equations

$u=x-y$
$v=x^2-y^2$

and I get $x=\frac{u^2+v}{2u}, y=\frac{v-u^2}{2u}$

At (2,1) it is $2=\frac{u^2+v}{2u}, 1=\frac{v-u^2}{2u}$

We want x>0, y>0
(a) $u^2+v<0, u<0, v-u^2<0$
(b) $u^2+v>0, u>0, v-u^2>0$

Solving the system of equations, I got a pair:
u=1, v=3 so (1,3) is the value of $f$ 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): $v>u^2 (v>0), u>0$

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.

(d)

$\frac{\partial{x}}{\partial{u}}=\frac{1}{2}\frac{( u^2+v)'-(u^2+v)u'}{u^2}=\frac{1}{2}\frac{u^2-v}{u^2}=\frac{-2}{2}=-1$ as in (b)

(e) tangent flat equation at (2,1)

$(u,v)^T=f(2,1)+f'(2,1)(x-2, y-1)^T$

$f'(2,1)$:

|1 -1|
|4 2|

$f(2,1)=(1,3)^T$

therefore, via some matrix multiplication we get the equation

$u=1+(x-2)-(y-1)$
$v=3+4(x-2)+2(y-1)$