Directional derivative in the previous section I have seen the concept of and its largest and smallest values account of how we have learned that we will. Now, as usual, our approach these problems resolved To reinforce the concepts and calculations We want to increase our ability. As the first example. Yet it because you're getting a simple surface to see easy, or on accounts review the main concepts stuck We can miss. This is a paraboloid. x and y, z is equal to zero, nine, means a paraboloid in the top nine. Hence the downward pointing minus signs a paraboloid. When we gave the paraboloid z zero xy-plane x squared plus y the radius of which is equal to the square of nine three The circumferential cuts. Here in the xy plane view. As a perspective view where the We see the circle. We stayed here for one or two points in the xy plane we choose. One or two points here, in the xy plane. Also in perspective view in the xy plane the point is here, but in the third dimension We know that we come up here When we find the point on z. To this point, the height x of a two year a and Five out of nine to four that I remove We find means of four From this point the height of the horizontal plane In the four. We first question, one by one The derivative in accordance with the direction finding we want. However, if we look at the plane that this is a vector a point in one or two places an in the direction of a vector. Considering again a plane in three dimensions In one direction, it does not pass through the center This vector. This is when we rise above such base a three-dimensionally vector such that it is projected that such that vector. This surface and also with the point find the largest and the smallest derivatives we want. This foreboding can easily reply question. Our paraboloid such a paraboloid our. Now up on the paraboloid Heading largest slope occurs. On the way down to the smallest slope occurs. See, when this maximum slope direction If we take this inward projection an outgoing vector. Thus, the components will be minus minus. However, the smallest slope minus sign will be going down. In this direction at this point and direction vector away from its center toward the projection will be a vector. Foreboding them easily we can find. Now let's move on to the calculations, the calculations easy. We have such a simple formula, gradient We will calculate. We'll take the inner product of the vectors given but u independence from the neck of the vector to ensure that we divide the length of the where the unit vector so that it occurs. Very easy gradient, where the f function derivatives with respect to x and y by will take. Minus two x minus two years. Gradient in a two-point value When we look at the will be minus two and minus four. We're putting a minus instead of x, y instead We put four minus two. And that's just the foreboding in the same direction. Easy going. The highest slope with my fingers in it I showed up towards the gradient. This will counter the vector The x's will be too inward minus sign comes from there. it When the value of the account dividing this by the length of the inner product will account. This is a very easy thing. but the units one by one to find the length of the vector We divides the square root. Gradient for this one minus two minus four We're putting. From multiplying minus two minus four minus six as we find. The largest and smallest slopes in just as I said, in accordance with our internal önsezgi. The largest value and the smallest value of the gradient the absolute value of one plus sign One of the minus sign. The largest and the smallest value is formed negative aspects which formed the greatest value The negative components, it's just that I showed will be this kind of as a vector. This projection minus minus its components. Vector that goes down to the smallest value thereof for the a vector projection toward the outside, therefore, components plus plus is happening. So these are pretty easy calculations. However, if we consider the way again plane with a two-coordinate here we are. Here we say that we found inside Tmax towards vector, so the maximum, the largest slope This content occurs towards the direction. We have also found this foreboding. This is just like I showed up at a vector towards the gradient is brings. The gradient direction of the vector is zero a vector which is perpendicular. See here for two and if we take one minus Two times two is four, minus one times four minus four, zero, this front There is the importance of plus or minus. Now comes zero anyhow. It also shows the direction of the line. This occurs at the zero slope in the direction large and the smallest pitch in the direction in which the formation of steep. Says that the general theory. The second example we take the same surface, paraboloid facing down. And you're getting the same point, this on the plane a couple of paraboloid Select the item above point we find. But as there is a difference this time, we call In a previous post u were directly We are moving on a parabola. This is the equation of the parabola y equals two times the square root of x. If we take the square of this year is equal to the square of X, a parabola to be x-axis. However, the y-axis, which we normally We are used because there are more parabolas y equals x squared function was taking but nothing has changed here. On this parabola progresses, more precisely in space, while moving surface We will stay on this parabola izdüşümü so when you scroll in a two-point We want to find the direction of the slope. Again, an easy calculation, the formula is the same. Unit gradient vector multiply. Where does this unit vector, this curve through. Let's do an inspection before it's a wonder two points on this curve. Indeed, a data y to x is equal to two we find. So the right spot, this parabola We are working on a dot. At this point we need to find T vector. You can find a very simple way We know that because the slope of derivatives. If we account derivative, derivative of the square root of x one-half times the square root of x is happening. But one or two in the front there is this dilemma When x is equal to one another in simplification We find a slope that. Slope of course, but a number of us, a vector You have to have a tangent vector. This tangent vector, this means horizontal When we go up a slope above We're going. So we're going above a, those of us immediately gives the vector. If there was something more complicated, perhaps in the plane we saw in the third section slope of the tangent directly in vector, or even finding the unit tangent vectors We used formulas but a much simpler way wherein able solutions. We need the unit tangent vector. His take on the length of the u We divide. It is already in a previous problem We tried to strike. Therefore, the result is always the same turns. Here we can provide as a. I wonder if this value is the value in line with the is the greatest and the smallest claim that we are staying within the value I really The maximum value of the length of the gradient We know that. Gradient that is minus two minus four length of first plus second squared components of the square and the square root of four plus one six, twenty, twenty square root. Is approximately four and a half. We have seen that before. It is the smallest value of the minus sign. Therefore, we do not need an account. We can now test the following monitoring, we can provide. Really in line with slope derivatives, directional derivatives, these two values Does staying in between, it's true. Because this value is smaller than the maximum value, is bigger than the smallest value. Therefore, as a result, we found to internalize We're going to have a checksum. So far, we've worked with two arguments, three variables to generalize in this section we want. But now take a break, that you We saw a little internalization of topics. Problems by yourself again maybe Earn thoroughly solving skills. These three variables in three dimensions then we learned chain derivatives, directional derivatives and We will generalize the concept of gradient.
