Results 1 to 2 of 2

Math Help - directional derivatives

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    129

    directional derivatives

    Suppose that you are climbing a hill whose shape is given by , and that you are at the point (90, 40, 300).
    1) In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? Answer : <-12.6/ \sqrt{174.76},-4/ \sqrt{174.76}>

    2) If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?

    How to do 2) ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    First, find your derivatives:

    f_{x}=-.14x, \;\ f_{y}=-.10y

    Sub in your given coordinates.

    \nabla f(90,40)=-12.6i-4j

    The max value of the directional derivative is ||{\nabla}f(90,40)||=\sqrt{(\frac{-63}{5})^{2}+(-4)^{2}}=\frac{\sqrt{4369}}{5}\approx 13.22

    A unit vector in this direction is:

    \frac{-12.6}{13.22}i-\frac{4}{13.22}j

    This is in decimal form, but is equivalent to your answer.

    \sqrt{174.76}=13.22
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Directional Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 13th 2010, 02:23 PM
  2. Directional derivatives
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 18th 2009, 04:18 PM
  3. Directional Derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 8th 2009, 12:45 AM
  4. Directional derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 2nd 2008, 02:53 AM
  5. Directional derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2008, 03:51 PM

Search Tags


/mathhelpforum @mathhelpforum