Visualizing the gradient vector
Another example here. We’ll look at a picture, and we will see that the gradient of a level curve is normal to the level curves. So the function we have been working with since early is × 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). f(x,y)=x^2 +y^2 F_x=2x F_x=2y Δf=(2x,2y) The gradient at a point depends on the coordinates of the point. It can be written as a formula 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). ∇F(0,1)=(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 ×2+2 are pictured below.
