Lecture. Differentiable functions of two variables
This lecture will discuss differentiable functions of two variables.
If we remember the one variable function, the concepts "existence of a finite derivative at the point" and "differential existence" ("differentiability") considered.
It turns out that in the case of a function of two variables, these concepts are different.
Let's deal with it.
Let the function of two variables set in some surrounding of point M0 with coordinates x0 and y0.
Let us give a variable increment Δx, Δy, and function as a result will receive a total increment.
The definition of differentiability.
The function is called differentiable at point M0, if increment at this point can be represented as (see video).
Remember that Δx, Δy are variables in this case, (x0, y0) are fixed points, and the resulting expression is A * Δx + B * Δy, where A and B are linear expression.
Another two summands are infinitely small quantities at Δx, Δy tending to zero.
The main linear part of the function increment A * Δx + B * Δy is called a function f differential at point (x0, y0).
Let's write down that.
The differential is of the form according to the definition (see video).
To calculate the differential, it is better to deduce its formulas, to know what numbersf A and B are equal to.
Let's try to do one by one.
We write the complete conversion, denote it by an asterisk, and now we assume that Δy - the increments variable y - is zero, then Δf becomes a special increment of variable x, changing only coordinate x.
The asterisk equality takes the form (see video).
Try to calculate the limit of rise over run by variable x.
Substituting this expression for the private increment, we very simply calculate the limit, we get number A, but the calculated limit is nothing else but partial derivatives of f with respect to x at point (x0 y0).
Value A is obtained.
Then in the equation the asterisk assume that zero Δx.
That is, only variable y changes, then the increment becomes a special increments by variable y and it has the form (see video).
Again calculate the limit of rise over run by variable y to increment Δy and this value is equal to B.
Number B is also found.
The next stage.
We can write down the function differential a little differently.
But now, our task is to change Δx and Δy for the formula unification.
Let us consider the function of two variables, equal to x.
The differential of this function at any point is equal to Δx.
The function differential (i.e., differential x) is Δx.
We argue quite similar for the function of two variables which takes value y, we get dy equal to Δy.
As a result the differential formula acquired the final form (see video).
In practice, we sometimes write differently (see video).
Look, you can write dz = z'x dx + z'ydy, two summands: the first is related to variable x, we call it the private differential of variable x.
The second summand is related to variable y, which we call the partial differential z of variable y.
It is easy to verify that the function differential of two variables have the properties we already know (see video).
Try to prove the validity of these formulas.
And now let's try to find the function differential, using only properties.
Begin.
We find differential sin (5x + x * y).
Sin is a function of one variable, it has a property of differential invariant, so the derivative of sin is equal to cos at the same point, multiplied by the differential of the inner function.
Then we calculate the differential, the differential of the sum of two functions.
In the first summand 5 can be taken off the differential sign as a constant factor.
In the second summand we use the formula of differential product.
Look what we get (see video).
Then we will open brackets, group the summands containing dx, and summands containing dy.
Look what we get as a result (see video).
The differential is obtained.
Let us remember the formula that we originally deduced, the differential formula.
Look, without finding partial derivatives with respect to variable x, variable y, we can write down their values.
These are factors in front of dx and dy before. Of course, in reality, we will find the differential according to this equation.
We calculate the partial derivatives with respect to x and y.
Substitute into the differential formula, but look, reverse operation is possible.
Find the differential, and then, knowing the differential, write out partial derivatives.
Differentials of the second and higher order are determined by induction, i.e. the second order differential is the differential of the differential and so on.
What's the problem when calculating the differential of the differential?
There is a variable dx, dy, and in addition, z ' of x and z' of y are functions of variables x and y.
We calculate the differential, considering that dx and dy - increments of variables x and y - adopt the same values, that is they are constants.
Therefore, using differential linearity property, we take off dx and dy as factors, and then instead of differentials write according to formulas those expressions, which they are: differential z ' of x, differential z' of y.
Further we open brackets and group summands, assuming that mixed derivatives z '' of x, y, and z '' of y, x are equal.
This occurs when mixed derivatives are continuous, we often have to deal, of course, with such functions.
The formula of the second order differential is obtained (see video).
There is another issue which we also consider today.
This is the concept of the tangent plane and surface normal.
Again, let the function of two variables set in some surrounding of point M0, and P0 is the corresponding point of the graph.
So, what is the tangent plane?
This is such a plane α, here it is built in the figure (see video), where all tangents to all possible curves drawn on the surface through point P0 are located.
Normal also determined in the same way, as in the case of a tangent normal to a one variable function.
This is a line drawn perpendicular in this case to the tangent plane through the tangency point.
How do you get the equation of the tangent plane?
Suppose that the function is differentiable.
By definition the function increment can be represented as (see. For video), where A and B, we remember, are the values of partial derivatives with respect to x, y at point (x0, y0).
Let's write increments Δx, Δy, Δz as differences x - x0, y - y0, z - z0, where (x, y, z) are arbitrary points on the surface.
Then we get the following equation (see video).
Look, if we remove the last summand, we get the plane equation.
With Δx, Δy tending to zero, surface points and plane points become arbitrarily close.
This is the equation of the plane tangent to the given surface (see video).
So how do you describe a normal?
We are a little ahead of our mathematics course, so let's take the following information as reference.
The plane is defined by the following equation (see video), where A, B, C are not simultaneously 0.
The vector which coordinates are of the form (A, B, C), where A, B, C are coefficients of the variables, is a normal vector, i.e. perpendicular to this plane.
In this case, the coefficients of x, y, z, if we bring the equation of the tangent plane to the mentioned form,are the coordinates of the normal vector.
Look what they are (see video).
Make sure that this is so.
And then, by analogy with the equation for the normal of one variable function, the equation of the normal to the surface at the tangency point is the three-dimensional space generalization.
What do we pay attention to?
Of course we can use this equation, if the values partial derivatives are not zero.
What to do if suddenly the denominator is zero?
In this case, the easiest way is to go to the equation in a parametric form (see video).
We write down that all these fractions are equal, and take value t.
Introduce parameter t.
In this case, from three equations (each part equals t), we can easily get x, y, z, which depend on parameter t.
In this case partial derivatives can be equal to zero.