Directional Derivative

Let us consider a function g ( x, y ) to represent the height of a tower at each point x = ( x, y ). If a person stand at a point x = c, the slope of the ground in front of you depends on the direction he is facing. It might slope steeply down in one direction, be relatively flat in another direction, and slope steeply up in yet another direction. The partial derivatives of ‘ g ’  will give the slope ∂g/∂x in the positive direction of ‘ x ‘ and the slope ∂g/∂y in the positive direction ‘y’. Now from the above example it can be generalized the partial derivatives to calculate the slope in any of the direction is called the directional derivative.

In taking a directional derivative, the first step is to give the direction. One way to give a direction is by a vector u = < a, b > that points to the direction in which we would like to evaluate the slope.

In general, let us assume that u is a unit vector. We now write the directional derivative of ‘g’ in the direction u at the point ‘c’ as Dug(c). We can even define it using a limit definition just like an ordinary derivative or a partial derivative

Dug(c) = $\lim_{h\rightarrow 0}[g(c+hu)-g(c)]/h$

Definition of Directional Derivative

If ‘f’ is a differential function of ‘x’ and ‘y’, then ‘f’ has a directional derivative in the direction of any unit vector u = < a, b > and Du f(x, y) = fx (x, y)a + fy (x, y)b

The concept of the directional derivative is very simple, Dug(c) is the slope of g(x, y) when standing at the point ‘c’ and facing the direction given by u. If x and y were given in centimeters, then Dug(c) would be the change in height per centimeter as the person moved in the direction given by u when he is  at the point ‘c’.

Directional Derivative Example

Below you could see directional derivative example

Second Directional Derivative

Find the second directional derivative of p(x, y, z) = xyz at (2, 3, 1) in the direction of the vector i - j - k

Solution:

Here ∇p = ( yz, xz, xy )

The given direction (1, - 1, - 1) and the unit vector will be $\frac{(1,-1,-1)}{\sqrt{3}}$

The directional derivative of ‘p’ in direction d is g'd where g=∇p

Here  g'd is a scalar function of (x, y, z) so just like ‘p’ it too has a directional derivative in any direction ‘e’.

This is known as the second order directional derivative of ‘p’ with respect to ‘d’ & ‘e’.

The gradient of g'd is Hd where H=∇²f
For this ‘p’ at (2, 3, 1), H = { {0, z, y}, {z, 0, x}, {y, x, 0} } = {0, 1, 3}, {1, 0, 2}, {3, 2, 0} }

This problem has e = d = \frac{\sqrt{3}}{(1,-1,-1)}
so we want d'Hd

Hd = $\frac{\sqrt{3}}{(-4,-1,1)}$ and d'Hd = $\left ( \frac{1}{\sqrt{3}} \right )$ $\left ( \frac{1}{\sqrt{3}} \right )${1, - 1, - 1}(- 4, - 1, 1) = -4/3