Results 1 to 2 of 2

Math Help - Estimating partial derivatives using a table

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    80

    Estimating partial derivatives using a table

    So, here are the values given on the table (I tried to upload an image, but I couldn't get it any bigger):

    K = 25, L = 700 : 2663.4; K = 30, L = 700 : 2917.9; K = 35, L = 700 : 3151.7
    K = 25, L = 720 : 2688.8; K = 30, L = 720 : 2945.5; K = 35, L = 720 : 3181.5
    K = 25, L = 740 : 2713.5; K = 30, L = 740 : 2972.5; K = 35, L = 740 : 3210.7
    K = 25, L = 760 : 2737.5; K = 30, L = 760 : 2999.0; K = 35, L = 760 : 3239.3

    (On the table, K is on the left side of the table and L is along the top; i.e., K is y and L is x).
    The problem is to define the partial derivatives dP/dK and dP/dL at the point K = 30, L = 740.

    My work:

    For dP/dK, I did (2972.5-2713.5)/5 + (3210.7-2972.5)/5 = 99.44
    And for dP/dL, I did (2972.5-2945.4)/20 + (2999.0-2972.5)/20 = -0.325

    Did I do this right? Any input would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Rumor View Post
    So, here are the values given on the table (I tried to upload an image, but I couldn't get it any bigger):

    K = 25, L = 700 : 2663.4; K = 30, L = 700 : 2917.9; K = 35, L = 700 : 3151.7
    K = 25, L = 720 : 2688.8; K = 30, L = 720 : 2945.5; K = 35, L = 720 : 3181.5
    K = 25, L = 740 : 2713.5; K = 30, L = 740 : 2972.5; K = 35, L = 740 : 3210.7
    K = 25, L = 760 : 2737.5; K = 30, L = 760 : 2999.0; K = 35, L = 760 : 3239.3

    (On the table, K is on the left side of the table and L is along the top; i.e., K is y and L is x).
    The problem is to define the partial derivatives dP/dK and dP/dL at the point K = 30, L = 740.

    My work:

    For dP/dK, I did (2972.5-2713.5)/5 + (3210.7-2972.5)/5 = 99.44
    And for dP/dL, I did (2972.5-2945.4)/20 + (2999.0-2972.5)/20 = -0.325

    Did I do this right? Any input would be appreciated.
    That looks an acceptable method, though rather more arithmetic than you need since the central term cancells in both..

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 5th 2010, 09:09 AM
  2. Replies: 3
    Last Post: February 18th 2010, 06:09 AM
  3. Help with partial derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 12:57 AM
  4. Help - Partial Derivatives of f(x,y)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 26th 2009, 08:52 PM
  5. Partial Derivatives.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 24th 2008, 07:31 AM

Search Tags


/mathhelpforum @mathhelpforum