Results 1 to 2 of 2

Math Help - gradients/directional derivatives

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    13

    gradients/directional derivatives

    This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.
    Attached Thumbnails Attached Thumbnails gradients/directional derivatives-q2.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by asw-88 View Post
    This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.
    One approach of many:

    Polar coordinates: x = r \cos \theta and y = r \sin \theta.

    Therefore:

    r^2 = x^2 + y^2 .... (1)

    \tan \theta = \frac{y}{x} .... (2)


    Results that will be needed:

    From (1):

    2 r \frac{\partial r}{\partial x} = 2x \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \theta .... (A)

    2 r \frac{\partial r}{\partial y} = 2y \Rightarrow \frac{\partial r}{\partial y} = \frac{y}{r} = \sin \theta .... (B)

    From (2):

    sec^2 \theta \, \frac{\partial \theta}{\partial x} = -\frac{y}{x^2} = -\frac{r \sin \theta}{r^2 \cos^2 \theta} \Rightarrow \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r} .... (C)

    sec^2 \theta \, \frac{\partial \theta}{\partial y} = \frac{1}{x} = \frac{1}{r \cos \theta} \Rightarrow \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r} .... (D)

    Then:

     \frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial x} = ..... .... (E)


     \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial y} = ..... .... (F)


    where the results (A) - (D) are to be substituted into (E) and (F).

    Then substitute (E) - (F) into \nabla f, do a little bit of re-arranging and then the job's done.
    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: October 22nd 2009, 09:18 AM
  2. directional derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 5th 2008, 06:08 PM
  3. Gradients and Directional Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 7th 2008, 09:59 AM
  4. Directional Derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 26th 2008, 12:05 PM
  5. Directional derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2008, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum