The pictures above represent two views of the graph of a function of two variables,

I want to find something called a directional derivative,

where the direction vector for the line is given by

We could simply parameterize the line in the xy-coordinate plane by

with direction vector given by

We can compute the directional derivative by using a form of the chain rule if we make the parameterization of the line the arclength parameterization and differentiate with respect to this parameter. Take theta to be the angle the direction vector for the line makes with the positive x-axis. In this case, if

Thus the line can be represented in the xy-coordinate plane as

with its direction vector the unit vector

With this parameterization t will be the arclength parameter since if

then

At the point (1,2,5) on the surface we find that

or just

as long as u is a unit vector.

Maximizing the Directional Derivative Using the Gradient

We know from Theorem 10.5 in LHE that for two vectors u and v and the angle theta between them

This shows us that the directional derivative will be maximized if the cosine of theta equals one, i.e., if theta equals zero degrees.  Thus the directional derivative will be maximized when the measure of the angle between the gradient and the direction vector is zero.  From this we can conclude that

 

the directional derivative will be maximized in the direction of the gradient and this maximum value will be the magnitude of the gradient.

Example 3 Find the maximum value of the directional derivative at the point (1,1,7) for the function given below. Thus the maximum value for the directional derivative at the point (1,1,7) will be in the direction of the vector < -2, -2 > and will be Click here to see a picture that includes the plane for this problem as described above in example 2.

Click on the picture to Enlarge.

Example 4

In this example I will show how the equations of the lines, curve, and plane in example 3 were found.

The line in the xy-coordinate plane would have a direction vector of < -1, -1 > so it could be given parametrically by

x = 1 - t

y = 1 - t

z = 0

The plane would contain the line y = x in the xy-coordinate plane and be perpendicular to the xy-coordinate plane so the equation of the plane in space would be

y = x.

To find the curve of intersection of this plane and the surface given by z = 9 - x² - y² we could set x = t.  Thus to satisfy the equation of the plane we would also have y = t.  We can substitute into the equation of the surface to get z.  Thus the curve is given parametrically by

x = t

y = t

z = 9 - 2t²

We can represent this curve as a vector valued position function r(t).

At the point (1,1,7) on the surface the t value would be 1 and the parametric representation of the line tangent to the curve of intersection of the plane and the surface would be

x = 1 + t

y = 1 + t

z = 7 - 4t

Example 5:  Direction of Steepest Descent

We want to find a vector from the point (1,1,7) on the surface described in example 4 above in the x and y direction of steepest descent based on the negative gradient evaluated at (1,1) with the terminal point of the vector lying on the surface.  We want to use a vector in the direction of the negative gradient that has been normalized (unit vector).

 

Click on each picture to enlarge. Maple Worksheet

Example 6:  Comparing Vectors

Here is a Maple worksheet comparing "Directional Derivative Vectors".  The colors of the vectors correspond to the colors in the computations below.  The horizontal vectors correspond to the direction vectors r, u, v, and w and the signed (up for positive, down for negative) magnitude of the vertical vectors correspond to the value of the corresponding directional derivative.  The hypotenuse vectors are what I am calling the "Directional Derivative Vectors" corresponding to unit direction vectors.  This Maple worksheet is like the one linked to above except it focuses only on the direction vector r which would give the direction of steepest descent.

return to exam III notes and links