# Directional derivative

• Feb 4th 2011, 04:07 PM
Killer
Directional derivative
Find $a,b,c$ constants so that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ on $(1,2,-1)$ has a maximum value of $64$ on a parallel direction to the $z$ axis.

I think we can calculate the directional derivative by using $\langle\nabla f(x_0),x_0\rangle$ where $x_0=(1,2,-1),$ but a maximum value is asked which I don't get, and I don't either get when it says "on a parallel direction to the $z.$"
• Feb 4th 2011, 09:57 PM
FernandoRevilla
Use:

$\nabla f(1,2,-1)=k(0,0,1)$

so,

$\left\|{\nabla f(1,2,-1)}\right\|=|k|=64$

Fernando Revilla
• Feb 5th 2011, 10:33 AM
Killer
Quote:

Originally Posted by FernandoRevilla
Use:

$\nabla f(1,2,-1)=k(0,0,1)$

so,

$\left\|{\nabla f(1,2,-1)}\right\|=|k|=64$

Okay but, that would imply that $k=64$ ?

Then how to get $a,b,c$?

Thanks!
• Feb 5th 2011, 12:13 PM
zzzoak
The directional derivative to the direction to the http://www.mathhelpforum.com/math-he...b808451dd7.png axis is

$
\nabla f \cdot (0,0,1)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})\cdot (0,0,1)=\frac{\partial f}{\partial z} \; | \; _{at \; point \; (1,2,-1)}=64.
$
• Feb 6th 2011, 04:02 AM
HallsofIvy
The deriviative in the direction of the z axis is simply the partial derivative with respect to z, $by+ 2czx^4$, evaluated at (1, 2, -1). That will give you as single equation for b and c. a can be anything.