Part 2.: problems. solutions, short test solutions

Part 3.: problems, solutions, homework solutions.

### Tangent lines and planes

Notice that an equation describes a curve on the plane. Imagining the graph of function we see that it is its level line. The direction of this level line in the given point such that is the direction in which the directional derivative is zero (thus the level is constant). Since the directional derivative at in direction equals , it means that the equation

where and , describes the line intersecting and parallel to the tangent line to the considered curve at . To get the equation of the actual tangent line we obviously have to change it:

where .

In the three-dimensional case an equation describes a surface, and analogously the equation of the plane tangent to it at point such that , is

where , , and .

### Diffeomorphisms

A diffeomorphism between open sets and is a function such that is its image (it is ,,onto” ), and is one-to-one and (differentiable, and with continuous first order partial derivatives), and its inverse function is of class as well.

Intuitively, you can think of a diffeomorphism as of a transformation of an open such which may stretch it but not reap it and squeeze it but not glue any points together (continuity), not making and ,,folds” (continuous derivative).

A standard example of a diffeomorphism is the radial mapping mapping the inside of the rectangle with vertices , , and onto the inside of a unit circle with centre in without the radius from to by the formula

### Theorem of inverse function

If , where is open and is of class, and for , , then there exists and an open set , such that , where is a bijection (if then and for every there exists , such that ) and

- the mapping is of class on ,
- , where , and .

In other words if the matrix of the derivative is invertible at a given point, the function itself has an inverse on a neighborhood of this point. Yet, in other words, the function restricted to this neighborhood is a diffeomorphism.

Indeed, to understand the theorem consider function . Generally there is no inverse function — it is not possible to get out of uniquely , because even though can be determined uniquely, we do not know the sign of . We can write that , but we cannot determine the sign.

But in a neighborhood of a given point it may be possible. E.g. consider the point . In its neighborhood , so for , we get — and it works. Looking at the Theorem, indeed,

so

and . Moreover, indeed

is the same as

On the other hand at (for any ) there is not inverse function since

and so .

### Theorem of implicit function

This theorem describes when an equation of the form defines some variables as function of the other variables.

Let assume that a function takes arguments and gives values. We consider an equation (actually a system of euqations!) and we ask, whether knowing , are uniquely determined. The theorem describes when it is possible. Notice that is a matrix. We denote by the matrix consisting of right columns of and by the matrix consisting of left columns of .

Let be open and let . Let be of class and let and also . Then there exist open and , such that and , and crucially a function of class , which determines from in a neighborhood of , meaning for any and we have if and only if . Moreover, .

E.g. let us consider the following system of equations

which corresponds to , where . The question is whether we can determine from in a neighborhood of this point.

Let us consider neighborhood of and try to solve the problem manually. We have that . We also know that (neighborhood of ) so . We know that is positive (neighborhood of ), thus . Then .

We check whether it is consistent with the theorem. We have

so

and

thus

and the determinant equals , so the assumptions are met. Indeed, exists. Moreover,

(because since to ), which is exactly the same as

Meanwhile at

has zero determinant and manually we also will not be able to determine from , since may be equal to zero.