Visualizing the gradient vector Another example here. We'll look at a picture, and we will see that the gradient of a levelcurve is normal to the level curves. So the function we have been working with since early is x squared plus y squared. If I take the derivative of its x- and y-components, I get 2x and 2y. Therefore, its gradient is(2x, comma 2y). The gradient at a point depends on the coordinates of the point. It can be written as aformula in terms of them. Let us just write down a few of those formulas for you. If you take the gradient at (1,0) and plug it into the function, you get (2,0). If you take the gradient of the point (1,1) and plug that into the equation, you get (2,2). If we take the gradient at (0,1) and plug it into our equation, we get (0,2). Good. And let us draw a picture of the level curves. We are moving to visualizing these gradients. The level curves of the function x2+y2 are pictured below.
We know that there are circles. For example, we can think about the point (1, comma 0). And the gradient of f at (1, 0) is 2;0. So I want you to imagine in your mind's eye a vectorstarting at (1, 0). The vector is (2, comma 0). So what would this do? Yes, it would go straight to the right and off the screen. You can notice that this would be perpendicular to the level curve at that point. In order to illustrate the trend, we generated a number of data points. Maybe let’s alsocheck point (1, 1).
So, let's say we draw a line with a gradient of of (1, comma 1). We got (2, comma 2).That's the vector that goes diagonally up and to the right. Imagine drawing the vector (2, comma 2) starting at that point. The vector would beperpendicular to the circle tangent to the level curve at that point. We drew a variety of these figures and then converted the images into vectors. The resultsare shown below.
Pay attention: we rescaled the vectors so they would fit in the picture. The real vector (2,comma 2) would go way out of the picture. So we divided them all by 10 anyway, so that they look kind of nice, but that's what it lookslike. You can see that this vector always points perpendicular to the level curves, which are thecircles.