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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


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    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.
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