Results 1 to 5 of 5

Thread: Directional derivative

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    30

    Directional derivative

    Find $\displaystyle a,b,c$ constants so that the directional derivative of $\displaystyle f(x,y,z)=axy^2+byz+cz^2x^3$ on $\displaystyle (1,2,-1)$ has a maximum value of $\displaystyle 64$ on a parallel direction to the $\displaystyle z$ axis.

    I think we can calculate the directional derivative by using $\displaystyle \langle\nabla f(x_0),x_0\rangle$ where $\displaystyle x_0=(1,2,-1),$ 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 $\displaystyle z.$"
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    Use:

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

    so,

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



    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2011
    Posts
    30
    Quote Originally Posted by FernandoRevilla View Post
    Use:

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

    so,

    $\displaystyle \left\|{\nabla f(1,2,-1)}\right\|=|k|=64$
    Okay but, that would imply that $\displaystyle k=64$ ?

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

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    The directional derivative to the direction to the axis is

    $\displaystyle
    \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.
    $
    Last edited by zzzoak; Feb 5th 2011 at 01:14 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,730
    Thanks
    3011
    The deriviative in the direction of the z axis is simply the partial derivative with respect to z, $\displaystyle by+ 2czx^4$, evaluated at (1, 2, -1). That will give you as single equation for b and c. a can be anything.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Directional derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Jul 20th 2011, 04:37 AM
  2. Directional Derivative
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Nov 4th 2009, 05:23 PM
  3. Directional derivative
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Mar 15th 2009, 02:42 PM
  4. Directional derivative...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 18th 2008, 12:20 AM
  5. Total Derivative vs. Directional Derivative
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: May 30th 2008, 08:42 AM

Search Tags


/mathhelpforum @mathhelpforum