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Directional Derivative

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the directional derivative of ##f## at ##P## in the direction of ##a##.

    ## f(x,y) = 2x^3y^3 ; P(3,4) ; a = 3i - 4j ##

    2. Relevant equations

    ## D_u f(x_0, y_0, z_0) = f_x(x_0, y_0, z_0)u_1 + f_y(x_0, y_0, z_0)u_2 ##

    3. The attempt at a solution

    ## f_x (x,y) = 6x^2y^3##
    ## f_y (x,y) = 6x^3y^2##

    ## f_x (3,4) = 3456 ##
    ## f_y (3,4) = 2592 ##

    ## D_u f(x_0, y_0) = 3456u_1 +2592 u_2 ##

    ##u = \frac {a} {||a||} = \frac {\langle 3,4 \rangle} {5} = \langle \frac {3} {5}, \frac {4} {5} \rangle##

    ##D_u f(x_0, y_0) = 3456(\frac {3} {5}) + 2592(\frac {4} {5}) ##

    ##D_u f(x_0, y_0) = \frac {20736} {5}##

    Now, my program wants this an exact number, no tolerance. It won't accept division either, so I don't know how to put in 20736/5. Just wondering if I made a mishap somewhere within the solution.
     
  2. jcsd
  3. Feb 21, 2016 #2

    SteamKing

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    You can't enter in a decimal number?
     
  4. Feb 21, 2016 #3
    Wouldn't that not be exact, but approximate form though? Or if I decimal isn't repeating is it considered exact?
     
  5. Feb 21, 2016 #4

    SteamKing

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    So, you're saying that (1/2) = 0.5 is only an approximation and not an exact representation? Interesting.
     
  6. Feb 21, 2016 #5
    So, I'm guessing it's not an approximation? Makes sense, good to learn something new. I will attempt to insert my answer.
     
  7. Feb 21, 2016 #6
    I attempted the answer of 4147.2, and it was incorrect. Therefore, my work must be incorrect somewhere.
     
  8. Feb 21, 2016 #7

    SteamKing

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    What if the wrong answer has been programmed into the software you're using?
     
  9. Feb 21, 2016 #8
    I've spoken to other students who have the same problem, but with different numbers, so I'm pretty positive that the programs solution is correct, but now that I am at home I don't have access to see any of their solutions.
     
  10. Feb 21, 2016 #9

    Ray Vickson

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    ##u_y \neq 4/5##; go back and check your work.
     
  11. Feb 21, 2016 #10
    Ah, thank you I missed that. I have remodeled my work, and the solution turns out to be 0, and is correct. Thank you both for your time.
     
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