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Homework Help: Finding the second derivative of multvariable function

  1. Nov 1, 2007 #1
    Hello I am trying to figure out the second derivative of

    [tex] \frac{\partial^2 z}{\partial x^2}[/tex] and [tex] \frac{\partial^2 z}{\partial t^2}[/tex]

    z(u) and u=x-vt

    i found the first derivate of [tex] \frac{\partial z}{\partial x}[/tex] to be
    [tex] \frac{\partial z}{\partial x}[/tex]= [tex] \frac{\partial z}{\partial u}[/tex] * [tex] \frac{\partial u}{\partial x}[/tex] = [tex] \frac{\partial z}{\partial u}[/tex] *1


    [tex] \frac{\partial z}{\partial t}[/tex] = [tex] \frac{\partial z}{\partial u}[/tex] * [tex] \frac{\partial u}{\partial t}[/tex]= [tex] \frac{\partial z}{\partial u}[/tex]*v

    after this i am clueless on how to compute
    ([tex] \frac{\partial }{\partial t}[/tex])[tex] \frac{\partial z}{\partial u}[/tex]*v
    ([tex] \frac{\partial }{\partial x}[/tex])[tex] \frac{\partial z}{\partial u}[/tex]*1

    i am new here am hoping to receive some help i will greatly appreciate it!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 1, 2007 #2
    Are you trying to say that z(u)=x-vt?
  4. Nov 1, 2007 #3
    i guess the answer should be for [tex] \frac{\partial^2 z}{\partial t^2}[/tex] = v^2 * [tex] \frac{\partial^2t }{\partial u^2}[/tex]

    im not sure at all how to get this tho any help?
  5. Nov 1, 2007 #4
    i guess im trying to say that z is a function of u and u is a function of x and t
  6. Nov 2, 2007 #5


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    Science Advisor

    No, since u= x- vt, [itex]\partial u/\partial t= -v[/itex], not v.

    The "v" and "1" are constants. So you are just differentiating a function of x and t again:
    [tex]\frac{\partial}{\partial x}\frac{\partial z}{\partial u}= \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial x}[/tex]
    and, of course, [itex]\partial u/\partial x[/itex] is 1.

    [tex]\frac{\partial}{\partial t}\frac{\partial z}{\partial u}= \frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial t}[/tex]
    and [itex]\partial u/\partial t[/itex] is -v.
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