# Thread: Directional derivatives question

1. ## Directional derivatives question

Find the max rate of change of f at the given point and the direction in which it occurs.

$f(x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}, (3, 6, -2)$

Attempt:

$f_x = 0.5(x^2 + y^2 +z^2)^{-0.5}2x$
$=0.5(3^{2} + 6^{2} + 4)^{-0.5}6$
$= 0.43$

$f_y = 0.5(x^2 + y^2 +z^2)^{-0.5}2y$
$=0.5(3^{2} + 6^{2} + 4)^{-0.5}12$
$= 0.86$

Gradient vector at $(x, y) = (f_x, f_y) = (0.43, 0.86)$

How would you find the direction in which f occurs?

2. If $\nabla f \neq \mathbf 0$ at $P$, then among all possible directional derivatives of $f$ at $P$, the derivative in the direction of $\nabla f$ at $P$ has the largest value. The value of this largest directional derivative is $||\nabla f||$ at $P$.

To see why this is so, you should know the directional derivative of $f$ in the direction of $\mathbf u$ at $(x_0,y_0,z_0)$ is denoted by $\mathbf{D_u}f(x_0,y_0,z_0)$ and is defined (for our purposes) by

$\mathbf{D_u}f(x_0,y_0,z_0) = \nabla f \cdot \mathbf u$

You should also be familiar with the following definition of the dot product of two vectors $\mathbf a$ and [/tex]\mathbf b[/tex]

$\mathbf a \cdot \mathbf b = ||a|| \, ||b|| \cos \theta$

where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.

Now, we can re-write the definition of the directional derivative as

$\mathbf{D_u}f(x_0,y_0,z_0) = || \nabla f || \, || \mathbf u || \cos \theta$

Because $\mathbf u$ is a unit vector, $|| \mathbf u || = 1$ and

$\mathbf{D_u}f(x_0,y_0,z_0) = || \nabla f || \cos \theta$

From the above equation, $\mathbf{D_u}$ is maximum when $\theta = 0$, which means $\nabla f$ and $\mathbf u$ are parallel.

Returning to your problem, first find $\nabla f$ at the given point $(3,6,-2)$.

$\nabla f = f_x(3,6,-2) \mathbf i + f_y(3,6,-2) \mathbf j + f_z(3,6,-2) \mathbf k = \dfrac{3}{7} \mathbf i + \dfrac{6}{7} \mathbf j - \dfrac{2}{7} \mathbf k$

To find the maximum rate of change of $f$, find $||\nabla f||$.

$||\nabla f|| = \sqrt{\left(\dfrac{3}{7}\right)^2 + \left(\dfrac{6}{7}\right)^2 + \left(-\dfrac{2}{7}\right)^2} = 1$

To find the direction in which the maximum rate of change of $f$ occurs, find the unit vector $\mathbf u$ in the direction of $\nabla f$.

In your case, $\nabla f$ is already a unit vector because $||\nabla f|| = 1$. In other words,

$\mathbf u = \nabla f = \dfrac{3}{7} \mathbf i + \dfrac{6}{7} \mathbf j - \dfrac{2}{7} \mathbf k$

and your problem is solved.