Practical lesson 1. The directional derivative, the gradient
The topic of the lesson is “The directional derivative, the gradient.” First, let’s think these concepts over, they appeared for a function of several variables only. For a function of a single variable, we did not meet such a concept. We have already had a derivative in the direction when we considered partial derivatives in x, in y. It turns out that the derivative by x is the directional derivative of the Ox axis, and the partial derivative with respect to variable y is the derivative in the direction of the Oy axis.
So, let’s move on to solving problems.
The task is to find the derivative of a function at a point in the direction of another point. At first, let’s build these points, what they are. Let’s take the point with coordinates (1, 2), let’s call it M zero, and this is point M. So, M zero has the coordinates (1, 2), and the point M has the coordinates (2, -5). We connect the vector, this is direction l. We need to find the value of the derivative at the point M zero in direction l. How is dz by dl at the point m zero calculated? The formula is as follows: (see the video). We need 4 numbers, what is what? Of course, we will find the value of partial derivatives at the point M zero. What is cosine alpha, cosine beta? These are the coordinates of a unit vector that is co-directed with the direction of differentiation, with the vector M zero M. So, let’s find all of this in turn. So, z’ by x at any point 2x-3y, z’ by y = 3x+2y, z’ by x at the point M zero -4, z’ by y at the point M zero is 7.
We have already found two numbers, so let’s turn to the unit vector. First, the vector m zero M has the coordinates (1, -7). Let’s see if it is a unit one. The length of the vector M zero M is equal to the square root of the sum of the coordinates, the root of 50, five roots of two. Obviously, the vector is not a unit one. How do we get a unit vector? Obviously, you need to divide the coordinates of the vector M zero M by the length. So, the first coordinate is one divided by five roots of two, the second coordinate is minus seven divided by five roots of two. This is the cosine alpha, cosine beta, the coordinates of the unit vector.
We return to the calculation of the direction derivative. So, dz by dl at the point (1, 2) (see the video). The problem is solved.
The task is to find the gradient of the function at a point. Note that the problem sounds a little different because u is a function of three variables, the gradient of a function of three variables here. We discussed the concept of the gradient of two variables at the lecture. It turns out that we can generalize this to any finite number of variables greater than or equal to two. And the gradient is a vector that has three coordinates: u’ by x, u’ by y, u’ by z in this case. The gradient is at the point, so we will find the values of partial derivatives at the point. Let’s find u’ by x, which is 2x+yz, u’ by y = xz, and, finally, u’ by z = xy. The derivative u’ by x at the point (1, 2, 3) is calculated by substituting, we get 8, u’ by y at the point (1, 2, 3) is the product of x and z, it is 3, u’ by z at the point (1, 2, 3) is 2. And that’s it, the answer is: the gradient of the function u at the point (1, 2, 3) is a vector with the coordinates (8, 3, 2). This vector indicates the direction of the greatest increase in the function.
Sometimes the direction is indicated in the most unexpected way, sometimes it is indicated in the direction of a straight line, sometimes, as here, in the the direction of the bisector of the first coordinate angle.
Let’s draw it so that we make this situation clear. So, at the point M zero with coordinates (0, 1) in the direction of the bisector. This means that we will calculate in the direction indicated by the bisector of the first coordinate angle. You can find the coordinates of a unit vector in different ways. I will assume that this is the unit vector. The unit vector means that the point lies on the unit circle, hence, we can find the cosines of angles or just the cosines, the angles =π/4, it is the square root of 2 divided by 2 root 2 divided by 2 – the coordinates of this point.
So, to find the derivative in the direction, we need to find only partial derivatives at the point (0, 1). Let’s write z’ by x at the point (0, 1), this is 2 for x=0 and y=1, it is clear that this is 2. So, z’ by y at the point (0, 1), I also introduce the notation how you can calculate, this is the derivative of sin2y again at the point (0, 1), we get sin2. Then, how do we calculate the derivative in the direction dz with respect to dl at the point (0, 1)? (see the video). The problem is solved.
Here, it may be clear what a gradient is. The task is to specify the gradient of the largest increase in the function at the point (2, 1). The direction of the greatest increase is indicated by the gradient, the vector, its coordinates are equal to the value of the partial derivatives.
So, we find z’ by x at the point (2, 1) = 1+y, for x=2, y=1 it is 2, z’ by y at the point (2,1) = -2+x+10y, for x=2, y=1 is equal to 10.
So, the next task is to find the derivative in this direction. Look, the gradient vector and the unit vector coincide. Then, let’s see what is the scalar product of the two vectors a and e, this is the length of the vector a on the vector e, its length is 1, and the cosine of the angle between them, cos0=1, we got the length of the vector a. So, the derivative in the direction of the gradient is equal to the length of the gradient vector. The z gradient at the point (2,1) is the vector with the coordinates (2, 10). We’ve obtained the result. The derivative of the direction indicated by the gradient at the point (2,1) is the length of this vector. The value of the derivative in the direction of the gradient is equal to two roots of 26 (see the video). The problem is solved.