1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding specified partial derivatives

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    I follow the steps until I get to 2(x+y+z)(1 - sin(r+s) + cos(r+s))

    The actual derivation process isn't the problem, I get lost in trying to figure out when to plug values for r and s.

    Attached Files:

    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    Images don't load.
  4. Oct 24, 2007 #3
    images loaded fine for me...
    anyhow, you have done everything correctly thus far. just plug the given values for s and r into the final expression 2(x+y+z)(1 - sin(r+s) + cos(r+s))

    you will never go awry if you always plug in given values last
  5. Oct 24, 2007 #4
    I'm guessing this is after I replace x,y,z with their respective functions, correct?

    edit: Negative. I did that and got 0 as an answer. According to the answer guide (back of the book), the correct answer is 13.
    Last edited: Oct 24, 2007
  6. Oct 24, 2007 #5
    yes, i'm sorry, i should have specified.... plug in x,y,z and then s, r
  7. Oct 24, 2007 #6
    That's not correct though, I get 6(0) = 0, correct answer should be 13.
  8. Oct 24, 2007 #7
    i'm currently unable to view the images you attached... what were x,y, and z again?

    EDIT: i recall r and s being 1 and -1, right? if that's the case (1 - sin(r+s) + cos(r+s) should be equal to (1 - sin0 + cos0) = (1 - 0 +1) = 2, not zero..
    Last edited: Oct 24, 2007
  9. Oct 24, 2007 #8
    x = r - s
    y = cos(r+s)
    z = sin(r+s)
  10. Oct 24, 2007 #9
    so if r was 1 and s was -1, then you should have 2(2+1)(2) = 12... but you say the answer was 13?
  11. Oct 25, 2007 #10
    Yeah, that's what I got too. But then again, the answer guide has been known to be wrong on more than a couple of occasions. Disregard further posts, and thanks for your help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Finding specified partial derivatives