cp /math/calclab/MA1024/Pardiff_grad_start.mws ~

to compute the directional derivative. However, the following computation, based on the definition, is often simpler to use.

One way to think about this that can be helpful in understanding directional derivatives is to realize that is a straight line in the plane. The plane perpendicular to the plane that contains this straight line intersects the surface in a curve whose coordinate is . The derivative of at is the rate of change of at the point moving in the direction .

As described in the text, the gradient has several important properties, including the following.

- The gradient can be used to compute the directional derivative
as follows.

- The gradient points in the direction of maximum increase of the value of at .
- The gradient is perpendicular to the level curve of that passes through the point .
- The gradient can be easily generalized to apply to functions of three or more variables.

- Consider the following function.

- A)
- First, plot the graph of this function over the domain
and
using the
`plot3d`command. Then use the`contourplot`command to generate a contour plot of over the same domain having 20 contour lines. - B)
- What does the contour plot look like in the regions where the surface plot has a steep incline? What does it look like where the surface plot is almost flat?
- C)
- What can you say about the surface plot in a region whe re the contour plot looks like a series of nested circles?

- Consider again the function from the first exercise. Using
either method from the
`Getting Started`worksheet, compute the directional derivative of at the point , in the three directions below.- A)
- B)
- C)

- Using the method from the
`Getting Started`worksheet, plot the gradient field and the contours of on the same plot. Use the domain of and . Use 20 contours and a 20 by 20 gradient field. Can you use this plot to explain the values for the directional derivatives you obtained in the previous exercises? By explaining the values, I only mean can you explain what kind of surface it is and why the values were positive, negative, or zero in terms of the contours and the gradient field?

2010-02-05