# Directional derivatives

• May 1st 2008, 11:53 PM
al1850
Directional derivatives
Find a vector u for which Duf(1,1) =0

Thank you.
• May 2nd 2008, 01:24 AM
CaptainBlack
Quote:

Originally Posted by al1850
Find a vector u for which Duf(1,1) =0

Thank you.

You ned to find a unit vector such that ${\rm{u}}.[(\nabla f (1,1)]=0$

RonL
• May 2nd 2008, 01:28 AM
al1850
Quote:

Originally Posted by CaptainBlack
You ned to find a unit vector such that ${\rm{u}}.[(\nabla f (1,1)]=0$

RonL

But how?
• May 2nd 2008, 01:53 AM
well first for the definition of directional derivative:

directional derivative of a scalar function, $f=f(x,y)$ in the direction of a unit vector, $\overrightarrow{u}=(u_1 , u_2)$ at a point $(x_0,y_0)$is given by:
$D \overrightarrow {u} f(x_0,y_0) = \bigtriangledown f(x_0 , y_0) \cdot \overrightarrow {u} = u_1 \frac {\partial f}{\partial x} + u_2\frac {\partial f}{\partial y}
$

$
f(x,y) = \frac{1}{x^2+y^2+1}
$

$
\bigtriangledown f(x , y) = \frac {\partial f}{\partial x} +\frac {\partial f}{\partial y}
$

$
= -\frac{2x}{(x^2+y^2+1)^2}\overrightarrow{i}-\frac{2y}{(x^2+y^2+1)^2}\overrightarrow{j}
$

at f(1,1):
$
= -\frac{2}{(1^2+1^2+1)^2}\overrightarrow{i}-\frac{2}{(1^2+1^2+1)^2}\overrightarrow{j}
$

$
= -\frac{2}{9}\overrightarrow{i}--\frac{2}{9}\overrightarrow{j}
$

therefore from;
$
D \overrightarrow {u} f(x,y)
$

$
D \overrightarrow {u} f(1,1)
$

$
= (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (\overrightarrow{u}) = 0
$

$
= (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (u_1\overrightarrow{i}+u_2\overrightarrow{j}) = 0
$

$
-\frac{2}{9}u_1-\frac{2}{9} u_2 = 0
$

now to obtain a zero $u_1$ and $u_2$ must be equal in magnitude but opposite in sign:
$
\overrightarrow{u} = (n,-n)
$

n can take any value.

Unit vector of $\overrightarrow{u}$
= $\frac{\overrightarrow{u}}{|\overrightarrow{u}|}$
= $\frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{|\sqrt{{ u_1}^2+{u_2}^2}|}$
= $\frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{|\sqrt{n ^2+(-n)^2}|}$
= $\frac{n \overrightarrow{i}-n\overrightarrow{j}}{|\sqrt{2}n|}$
= $\frac{n }{|\sqrt{2}n|}\overrightarrow{i}-\frac{n }{|\sqrt{2}n|}\overrightarrow{j}$