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Directional derivatives

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data

    find the direction of P sub 0 in the direction of A

    see second post for attachment, I forgot to place it on this one.


    3. The attempt at a solution

    I'm only worried about the part that says gy(x,y,z) = -3zexsin yz. I also don't understand the conversion of gx and gz.

    1. why does cos change to sin? obviously its the derivative but why isn't the derivative being used in gx and why is it also used in gz

    2. if they're taking the derivative then it doesn't look like the product rule is being used. why not?

    3. I would think they would be using u substitution with the x in ex but I'm better at using u substitution with integrals than with derivatives. I'm pretty sure this is the explanation for why there is a -3z and a -3y in gy and gz. I understand where the negative sign comes from, it's because the derivative of cos in -sin. But I don't understand where the z and y comes from.
     
    Last edited: Mar 2, 2012
  2. jcsd
  3. Mar 2, 2012 #2
    attachment is here.
     

    Attached Files:

  4. Mar 2, 2012 #3

    SammyS

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    attachment.php?attachmentid=44652&d=1330733329.png

    What is the problem that this is the solution for ?

    What id the function, g(x,y,z) ?
     
  5. Mar 2, 2012 #4
    see attachment for question.

    the function is 3excos xy
     
  6. Mar 2, 2012 #5

    Dick

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    It's actually 3*e^x*cos(yz). If you want the partial derivative of that with respect to y then you don't have to use the product rule because e^x is a constant. You do have to use the chain rule on the cos(yz) part.
     
  7. Mar 2, 2012 #6
    well if I use the chain rule on 3excos yz, then why is the answer for gx the same as the question?
     
  8. Mar 2, 2012 #7

    Dick

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    I really don't understand that. Just use the chain rule and tell me how you can get -3*e^x*z*sin(yz).
     
  9. Mar 2, 2012 #8
    well if use the chain rule for 3excos yz

    I'm pretty sure the derivative of ex is xex but I'm not 100% certain, I'll have to look it up. that would make the derivative with respect to x

    -3ex sin yz

    not 3excos yz like the book says.

    I'm not sure what to do about the yz part, sometimes it plays a factor in the answer sometimes it does not. Clearly in the derivative with respect to y and z, it does play a role, but I'm not sure what operation to use because when you take the derivative with respect to y, the y disappears and when taking it with respect to z, the z disappears.
     
  10. Mar 2, 2012 #9

    Dick

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    The derivative of e^x is not x*e^x. The rest of what you are saying makes even less sense.
     
  11. Mar 2, 2012 #10
    If i use chain rule on

    3excos yz

    why does cos not turn into -sin?

    What is the status of the yz? Why does it not factor into the answer for the derivative of

    gx but does play a role in y and gz ?

    i've already said all that but i'll say it again.
     
  12. Mar 3, 2012 #11

    SammyS

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    When you take
    [itex]\displaystyle \frac{\partial}{\partial x}\left(3e^x \cos(yz)\right)[/itex]​
    you treat y and z as if they're constants, so cos(yz) is treated as a constant times the constant 3 times ex .

    Therefore,
    [itex]\displaystyle \frac{\partial}{\partial x}\left(3e^x \cos(yz)\right)=3\cos(yz)\frac{d}{dx}e^x\,.[/itex]​
     
  13. Mar 3, 2012 #12
    ok, seems a bit weird, but that will work as an answer for now. thanks for your help.
     
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