Математика. Материалы курса

Interesting concepts: the directional derivative and the gradient. You have probably heard in the speeches of our politicians such phrases as “you need to specify the gradient of development.” What does it mean? What should we specify? Let’s get this straight. And you, too, will be able to parade the knowledge of such a word.

So, let the function of two variables z = f (x, y) be defined in some neighborhood of point М₀. The direction, we are going to analyze the concept of the derivative in the direction, we will set the ray l, leaving point М₀. Let M be an arbitrary point of this ray, and we will denote the current point by x and y coordinates. So, what is called the derivative in the direction l of the function f at the point М₀?

So, this is the limit of the ratio of the difference between the values of the function f at points M and М₀, M lies on the ray l, to the distance М₀М when point M tends to point М₀. So, again, we find the limit of the ratio of the function increment to the argument increment. The function increment occurs when moving from point M to point М₀. Our task now is to figure out how to calculate this value. It is clear that when calculating this limit, we will have problems. Is it possible to find a formula for calculating the direction derivative?

What will we do? The direction l can be set as a unit vector. Let’s build it. The beginning is at the point O, its length is equal to 1, it is co-directed with l. Let’s denote the angles that this vector forms with the coordinate axes Ox and Oy – a and b. Then, look, from the resulting right triangles, it is easy to find that the length of the vector is 1, the first x coordinate of this vector is cos a, and the second coordinate is cos b. Frequently, cos a and cos b  are called cosine alpha and cosine beta – coordinates of the unit vector – guide cosines.

So, we have the formula that defines what the directional derivative is at a point. We will try to derive the formula. So, М₀ has coordinates x0 and y0. The current point M belongs to ray l and has x and y coordinates. The vector М₀М is co-directed to the unit vector e, so it is the product of the vector e by some positive number t. So, the coordinates of the vector М₀М are (х-х₀, у-у₀), and the coordinates of the vector e are (cos a, cos b). So we can write a system of equalities. From here, expressing x and y, we get the coordinates of the point M (x, y). Well. So, let’s see what is there in the limit. In fact, we can find the numerator. We will write the value at point M, and we will write the value at point М₀. What is the denominator – length М₀М? So, the formula is the distance between points, and the difference between the coordinates x and x0 is t* cos a, t*cos b. But the sum of the squares of cosines is the length of the unit vector, that is, 1. So, we substitute the numerator and denominator in the formula.

At this point, of course, we should think a little. Because according to the definition of the derivative, it turns out the value of the derivative of a function of one variable at the point 0. And what is this function? Look, f is a function of two variables, each of which turns out to be a function of the variable t. So, we have come again to the derivative of a function of one real variable.

Let's see how to calculate this derivative. We have come again to a complex function. So, let’s figure it out. So, this is a complex function, and the diagram for it looks like this (see on the screen). Let’s record the transitions.

Note that z depends on two variables x and y, so we write round letters d – these are partial derivatives with respect to z, z with respect to x, and z with respect to dy. And here is the transition to the variable t, we write the direct letters d, which we use to denote the derivative of a function of one variable: dx by dt, dy by dt. Then the value of the derivative function of one variable t is calculated using the formula (see on the screen). Let’s see, well, what is the derivative of the function x with respect to the variable t? After all, x is х₀ + t cos a, and the derivative of t is это cos a, and accordingly the derivative of y is t - cos b  . What do we see? We got the formula. The derivative in the direction l at point М₀ is calculated using the formula (see on the screen).

So, what do we need to know to apply this formula? The value of the partial derivatives of the functions z with respect to x and y at the point М₀ and the coordinates of the unit vector indicating the direction of differentiation – (cos a, cos b). So, here we met with such a concept as a vector. Look at two numbers-partial derivatives with respect to x at point М₀ and partial derivatives with respect to y at point М₀. These are two real numbers. We can assume that they are coordinates of some vector. This vector is called the gradient of the function f at point М₀ (x₀, y₀). So, the gradient is a vector. Let us return to the calculation of the derivative in the direction. (cos a, cos b)  are the coordinates of the unit vector e. Then, if we recall the scalar product formula, where the coordinates of the vectors are used, we see that the derivative in the direction at the point М₀ is the scalar product of two vectors: the gradient and the unit vector. So, this is a scalar product of vectors.

But generally speaking, the scalar product of vectors is also calculated differently. We need to find it out. Let’s figure it out. If two vectors a and b change. In what way? Their lengths are the same, but the angle between them may be different. In which case does this scalar product take the largest value? When the cosine φ is equal to 1. If we say that the length is the same, and we change the direction of the vector only. It means φ=0.

Therefore, the scalar product of two vectors that have a constant length takes the greatest value when the angle between them is zero, that is, they are co-directed.

This means, look, the gradient is also the value of the derivative at the point, the length of the vector is always the same, it is generally a constant vector. But we know that the vector e has a length of 1. See when this scalar product takes the largest value? Apparently, when the direction of differentiation (vector e) is co-directed to the gradient, that is, the direction of differentiation must coincide with the direction of the gradient.