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A second opinion for a tricky problem?

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]z=e^{i(\kappa x-\omega t)[/tex]

    [tex]\theta=x^2+t[/tex] and [tex]\phi=x-t[/tex]

    find [tex]\frac{\partial z}{\partial \theta}[/tex] and [tex]\frac{\partial z}{\partial \phi}[/tex] in terms of x and t only.

    2. Relevant equations

    3. The attempt at a solution
    for the first partial I thought to solve the [tex]\theta=x^2+t[/tex] for t and to substitute to the z and then to use the chain rule. So to do this

    [tex]t=\theta-x^2[/tex] and so [tex]z=e^{i(\kappa x-\omega (\theta-x^2))}[/tex]

    then in order to find [tex] \frac{\partial z}{\partial \theta}[/tex] I will do this [tex]\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta}[/tex]

    Now [tex]\frac{\partial z}{\partial x}=e^{i(\kappa x-\omega (\theta-x^2))}(i(\kappa x-\omega (\theta-x^2)))'=e^{i(\kappa x-\omega (\theta-x^2))}(i\kappa - 2\omega x)[/tex]

    the [tex]\frac{\partial x}{\partial \theta}=\frac{1}{2x}[/tex]

    so [tex]\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta}=\frac{e^{i(\kappa x-\omega (\theta-x^2))}(i\kappa - 2\omega x)}{2x}[/tex]

    Now we substitute again the t and we get

    [tex]\frac{\partial z}{\partial \theta}=\frac{e^{i(\kappa x-\omega t)}(i\kappa - 2\omega x)}{2x}[/tex]

    For the second partial I follow the same way

    [tex]x=\phi + t[/tex] I substitute to z and I get [tex]z=e^{i(\kappa (\phi + t) -\omega t)}[/tex]

    and then [tex]\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}[/tex]

    We find that [tex]\frac{\partial t}{\partial \phi}=-1[/tex] and

    [tex]\frac{\partial z}{\partial t}=e^{i(\kappa (\phi + t) -\omega t)}(i(\kappa (\phi + t) -\omega t))'=e^{i(\kappa (\phi + t) -\omega t)}(i \kappa - \omega)[/tex]

    so [tex]\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}=-e^{i(\kappa (\phi + t) -\omega t)}(i \kappa - \omega)[/tex]

    I am not sure if this is the solution, i need a second opinion. I am a mathematician and I need someone else to tell me if im wrong and where. Thank you for your time. Please read carefully before to answer. Thank you
     
  2. jcsd
  3. Oct 11, 2007 #2

    HallsofIvy

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    [tex]\frac{\partial z}{\partial \phi}\ne\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}[/tex]
    [tex]\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}+ \frac{\partial z}{\partial x}\frac{\partial x}{\partial \phi}[/tex]
    and the same for [itex]\theta[/itex].
     
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