In the activity Directional Derivatives in Matlab, we investigated the derivative in an arbitrary direction, called the directional derivative. Let's repeat some of that work here.

We begin by picking an arbitrary point (a,b) at which we wish to find the directional derivative. Then we pick a unit vector vec u=langle u_1, u_2 rangle, is whose direction we wish to find the derivative. To find the equation of the line in the xy-plane that passes through (a,b) in the direction of vec u, let X(x,y) be an arbitrary point on the line. Then the equation of the line in vector form is

vec (PX)= t vec(u).

This is equivalent to the vector equation

langle x-a,y-b rangle=t langle u_1, u_2 rangle,

which leads to the parametric equations of the line.

$\begin{eqnarray} x&=&a+u_1 t\\ y&=&b+u_2 t \end{eqnarray}$

In the activity Directional Derivatives in Matlab, we looked at points on the surface having x- and y-values restricted to the above line. There we had:

z=f(x,y) \quad\text{where}\quad x=a+u_1t \quad\text{and}\quad y = b+u_2t

Using the chain rule, we calculate dz//dt.

frac(dz)(dt)=frac(partial f)(partial x) frac(dx)(dt)+frac(partial f)(partial y)frac(dy)(dt)

Because dx//dt=u_1 and dy//dt=u_2, this becomes

frac(dz)(dt)=frac(partial f)(partial x) u_1+frac(partial f)(partial y) u_2.

However, note that the right-hand side of this last result can be written as a dot product.

frac(dz)(dt)=langle frac(partial f)(partial x),frac(partial f)(partial y) rangle cdot langle u_1,u_2 rangle.

Given a function f:R^2 \to R, the gradient of f is defined to be

nabla f=langle frac(partial f)(partial x), frac(partial f)(partial y) rangle.

With this definition, the directional derivative can be written as follows:

frac(dz)(dt)=langle frac(partial f)(partial x),frac(partial f)(partial y) rangle cdot langle u_1,u_2 rangle =nabla f cdot vec(u).

Finally, if we adopt the notation D_u f to mean the derivative of f in the direction of the unit vector vec u (the directional derivative), then we have the following result.

Suppose that f:R^2\to R and vec u is a unit vector. Then the derivative of f in the direction of vec(u) is given by

D_u f=nabla f cdot vec(u).

Thus, the slope of the tangent line to the surface at the point (a,b,f(a,b)) is given by

D_u f(a,b)=nabla f(a,b) cdot vec(u).

Let's make an example calculation.

Given f(x,y)=9-x^2-y^2, calculate the gradient, then evaluate the gradient at the point (1,1).

Solution: Because f(x,y)=9-x^2-y^2, we have partial f//partial x = -2x and partial f//partial y=-2y. Thus, the gradient, evaluated at (x,y), is given by

nabla f(x,y)=langle frac(partial f)(partial y), frac(partial f)(partial y) rangle=langle -2x,-2y rangle.

Therefore the gradient, evaluated at the point (1,1), is the vector

nabla f(1,1)=langle -2,-2 rangle.

Let's build on this last example and calculate a directional derivative.

Given the function f(x,y)=9-x^2-y^2, find the directional derivative of f at the point (1,1) in the direction of the unit vector vec u=langle sqrt(2)//2,sqrt(2)//2 rangle.

$\begin{eqnarray} D_f f(1,1) &=&\nabla f(1,1) \cdot \vec u\\ &=&\langle -2, -2 \rangle \cdot \langle \sqrt2/2, \sqrt2/2 \rangle\\ &=&-\sqrt2-\sqrt2\\ &=&-2\sqrt2. \end{eqnarray}$

Readers might recognize this as the slope of the tangent line from the activity Directional Derivatives in Matlab.

In Example 1, for the function f(x,y)=9-x^2-y^2, the gradient vector at the point (1,1) was nabla f(1,1)=langle -2,-2 rangle. We've plotted this gradient vector at the point (1,1) in Figure 1. We've also calculated an additional gradient vector at the point (5,-5), getting nabla f(5,-5)=langle -10,10 rangle and including it in Figure 1. Similar calculations show that the further we get from the origin, the longer the gradient vectors become.

We could continue by hand, calculating the gradient vector at each point of the grid in Figure 1 and plotting the result. The result would be a field of vectors, or equivalently, a vector field. However, a hand plot is tedious, particularly when Matlab does an excellent job of plotting gradient fields.

First, create a grid of (x,y) pairs at which we will plot gradient vectors.

[x,y]=meshgrid(-5:1:5);


Note that -5:1:5 produces the vector [-5,-4,-3,-2,-1,0,1,2,3,4,5] and meshgrid(-5:1:5) uses this vector for both x and y. Next, calculate the gradient vectors at each point of the grid. Because we are using the function f(x,y)=9-x^2-y^2, we have partial f//partial x = -2x and partial f//partial y = -2y.

vx=-2*x;
vy=-2*y;


We use the familiar quiver command, turning off all scaling.

quiver(x,y,vx,vy,0);


Adding the usual annotations produces the result in Figure 2.

Although the gradient field of Figure 2 is completely accurate, it is very difficult to interpret the result due to overcrowding. (Indeed, it appears that the gradient vectors are pointing outward, an optical illusion contradicted by the evidence in Figure 1.) A standard practice is to scale the vectors appropriately to aid in the interpretation. This is accomplished by letting Matlab's quiver command sketch the field with scaling turned on.

quiver(x,y,vx,vy);


The resulting gradient field is shown in Figure 3.

The vectors in Figure 3 have the inccorrect length. For example, the vector at the point (1,1) should have the length shown in Figure 1. Hoever, the vector is pointing in the correct direction and it has the correct relative length, as compared to the remaining vectors in the gradient field.

The gradient possesses several important properties, one of which is its relation to the set of level curves of the function. Let's add some contours to the gradient field in Figure 4.

First create a finer mesh for more accuracy.

[x,y]=meshgrid(linspace(-5,5,50));


Recalculate the function f(x,y)=9-x^2-y^2 on this finer mesh.

z=9-x.^2-y.^2;


Hold the plot and add the level curves.

hold on
contour(x,y,z)


Equalize the axes and add a grid.

grid on
axis equal


The result is shown in Figure 4. Note that the gradient vectors all appear to be orthogonal to the level curves of the function.

One final question remains: If the gradient vectors are orthogonal to the level curves, there are two possible directions that they could point, inward or outward. Why do they point inward in our example? this is most easily vizualized by drawing a surface and adding both level curves and the gradient field to the image. We start a new image by closing all open figure windows.

close all


Create the mesh and recalculate the function values forf(x,y)=9-x^2-y^2.

[x,y]=meshgrid(linspace(-5,5,40));
z=9-x.^2-y.^2;


Now we draw the surface and contours with the meshc command. We also use Matlab's handle graphics to color the surface with magenta and set the mesh lines to black.

h1=meshc(x,y,z);
set(h1,'FaceColor','m',...
'FaceAlpha',0.5,...
'EdgeColor','k')


Add a grid, turn on the box, and rotate into the standard orientation.

grid on
box on
view([130,30])


The result is shown in Figure 5.

Now, let's add the gradient field. Because our figure is three-dimensional, we will use the quiver3 command to add the gradient field. Begin with a coarser grid where you want the gradient vectors to be plotted.

[x,y]=meshgrid(-5:5);


Calculate the positions where you want to plot the gradient vectors. In Figure 5, note that the bottom plane is set at z=-50. Thus, we want a matrix of z-values the same size as x- and y-matrices that is filled with the constant -50.

z=-50*ones(size(x));


Recall that nabla f(x,y)=langle -2x, -2y rangle. In addition, we want the z-component of the vectors to be zero, so we fill a matrix vz with zeros the same size as the matrix vx.

vx=-2*x;
vy=-2*y;
vz=zeros(size(vx));


Hold the plot and use the quiver3 command to plot the gradient field in the desired position.

hold on
quiver3(x,y,z,vx,vy,vz);


Add a grid and turn on the box, then rotate into the standard orientation.

grid on
box on
view([130,30])


xlabel('x-axis')
ylabel('y-axis')


The result is the surface, contours, and gradient field shown in Figure 6.

The question of direction now seems to be resolved. In class you will be given a more rigorous proof, but from the appearance in Figure 6, the gradient vectors seem to be pointing in the direction of greatest increase of the function. If you consider the contours to be a topographical map, then the gradient vectors are pointing in an "up-hill" direction.

### Matlab Files

Although the following file features advanced use of Matlab, we include it here for those interested in discovering how we generated the images for this activity. You can download the Matlab file at the following link. Download the file to a directory or folder on your system.

1. Consider the function f(x,y)=x^2+y^2-4. Use Matlab to sketch a gradient field superimposed on contours as in Figure 6. Submit a printout of your result and the M-code that produced your result. Submit all mathematical computations required to produce your result.
2. Consider the function f(x,y)=2-|x|-|y|. Sketch a surface, level curve, and gradient plot such as that shown in Figure 6. Here's a little hint to help with the differentiation.
$\begin{eqnarray} D_x|x| &=&D_x\sqrt{x^2}\\ &=&D_x(x^2)^{1/2}\\ &=&\frac12(x^2)^{-1/2}(2x)\\ &=&\frac{x}{(x^2)^{1/2}}\\ &=&\frac{x}{\sqrt{x^2}}\\ &=&\frac{x}{|x|} \end{eqnarray}$