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A small exploration of the connection between the gradient vector and level surfaces.

For a function f, the gradient vector grad f has the properties that
1. it points in the direction in which f increases the fastest, and
2. it is perpendicular to level curves or surfaces of f.

A key point to note is that in the first item above the direction that we're looking at is a direction in terms of the independent variables of the function. Therefore, if we're considering a function of two variables, f(x,y), the direction is in the x-y plane. If we have a function of three variables, f(x,y,z), the direction is in three-space.

Example:
Suppose we have consider the function f(x,y,z) = x2 - y2 + z2. Then grad f = < 2 x, -2 y, 2 z >. To think about a specific direction, let's consider the point (2,2,1).

figure 1: the level surface f(x,y,z) = 1

figure 2: the level surfaces f(x,y,z) = 9 (with grid) and f(x,y,z) = 1 (smooth).

figure 3: the level surfaces f(x,y,z) = 9 (with grid) and f(x,y,z) = 1 (smooth), with an exaggerated gradient vector (yellow) shown.

At the point (2,2,1), the gradient vector is grad f = < 4, -4, 2 >. So, to get the largest possible increase in f, we should go in this direction (as a unit vector, in the direction u = (1/6) < 4, -4, 2 >). The rate of change in f in this direction is the value of the directional derivative, Du f, which is in this direction u equal to the magnitude of the gradient: Du f = 6. Check that all of that makes sense.

To connect this to level surfaces, note that f(2,2,1) = 1. So the point (2,2,1) is on the level surface given by f(x,y,z) = 1, or x2 - y2 + z2 = 1. What does this look like? It's graphed in the picture to the right, and is a double-ended trumpet bell. Notice that this makes sense: if we were to solve this equation for z, we'd get a plus-or-minus square root of stuff. The plus in the square root gives the top half of the bell, and the negative the bottom.

If f is increasing, the value of the function increases, and we get a different level surface. For example, the value 9 gives the level surface f(x,y,z) = 9, or x2 - y2 + z2 = 9. We can graph this as well, as is shown in the second figure to the right. In this the first level surface is left in, but smoothed out so that it's clear how it's related to the surface for the larger value of f. A good way of thinking about these different surfaces is to pretend that f represents something physical like temperature. Then the temperature is increasing as we move from the first level surface to the second.

Now, how is the gradient related to this? One last picture! This is the third figure on the right, and has a(n exaggerated) gradient vector in yellow added to the graph. Notice that the point (in blue) (2,2,1) is located on the surface f(x,y,z) = 1 (check that you can see that it looks like it's just about in the right place), and that the gradient vector also looks like it points in the right direction: < 4, -4, 2 > is in the positive-x, negative-y and positive-z direction. Then, is f increasing (as fast as possible) in this direction? It certainly seems reasonable given the relative positions of the two level surfaces. Is grad f perpendicular to the level surface? It also looks like that's the case.

Second Example:
Find the equation of the tangent plane to the surface x2 - y2 + z2 = 1 at the point (2,2,1).

Notice that the point (2,2,1) does satisfy the equation, and so is on the surface. To solve this, we use our work above. We know that this surface can be thought of as the level surface of the function f(x,y,z) = x2 - y2 + z2 where f = 1. And we know that grad f gives a vector that is perpendicular to this surface. So we can use the gradient vector we found above as our normal vector, we have a point already, and the equation of the tangent plane is easy:

4(x - 2) - 4(y - 2) + 2(z - 1) = 0.