Hello friends, today I will talk about directional derivative.
Here I don’t go much into the theory. Instead, I will try to explain it in a simpler way.
Few things to know about directional derivative
- Directional derivative only happens to some scalar field or say some function.
- Directional derivative has the notation with as a scalar field or function.
- is the projection of the gradient vector of on the given unit vector in the direction of .
Example of directional derivative
Note: This example does not belong to me. I have chosen this from a book. I have also given the due reference at the end of the post.
So here it is.
According to Stroud and Booth (2011)* “Find the directional derivative of at the point in the direction of .”
In this example, the given surface is
To get the directional derivative of this surface, I will follow certain steps. The first step is to get the gradient of this surface.
In one of my earlier posts on vector analysis, I have explained how to get the gradient of any surface.
Here also I will do the same.
To get the value of grad , I will start with the partial differentiations.
So will be
Now I’ll get Thus it will be
Similarly, will be
Thus the gradient of is
Now I will find out the value of the gradient vector at the point
Thus at the point the value of the gradient vector will be
Next I have to find out the unit vector in the direction of the given vector
In order to find out the unit vector , first of all I will determine the magnitude of the vector
Thus the magnitude of the vector is
Therefore the unit vector in the direction of the given vector is
As a next step, now I will get the directional derivative.
According to the formula, directional derivative of the surface is .
Here both and are vectors. Thus is the scalar product of two vectors.
Therefore in this case it will be
Hence I can conclude that the directional derivative of at the point in the direction of is
This is the answer to this example.
So here ends my example.
Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!
*Reference: K. A. Stroud and Dexter J. Booth (2011): Advanced engineering mathematics, Industrial Press, Inc.; 5th Edition (March 8, 2011), Chapter: Vector analysis 1, Further problems 22, p. 815, Q. No. 11 (Example 1).
d φ ⁄ ds = a. grad φ