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Partial derivatives

  1. Apr 13, 2009 #1

    Lyn

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    I am stuck on the question,

    'If f is a twice differentiable function of a single variable, find f = z(sqrt(x^2+y^2)) that satisfies d^2z/dx^2 +d^2z/dy^2 = x^2 +y^2

    (ALL d's ARE MEANT TO BE PARTIAL DERIVATIVES)

    i know dz/dx=(dz/du).(du/dx)
    i can find du/dx but i don't know how to find dz/du
     
  2. jcsd
  3. Apr 13, 2009 #2

    Hootenanny

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    Welcome to Physics Forums Lyn.

    HINT: Let u = (x2 + y2)1/2

    HINT (2): Using you definition of dz/dx above, can you write down an expression for d2z/dx2?
     
  4. Apr 13, 2009 #3
    Isn't this suppose to be [itex]z=f(\sqrt{x^2+y^2})[/itex] In which f is a twice differentiable function?
     
  5. Apr 13, 2009 #4

    Lyn

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    Thank you for replying.

    Yes I have set u=sqrt(x^2+y^2)

    Would d^2z/dx^2=d/dx.(dz/dx)?

    So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation
     
  6. Apr 13, 2009 #5

    Hootenanny

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    You don't need to find dz/du. Let's just start working though it. We have

    [tex]\frac{dz}{dx} = \frac{dz}{du}\frac{du}{dx}[/tex]

    Hence,

    [tex]\begin{aligned}
    \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
    & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right)
    \end{aligned}[/tex]

    Can you take the next couple of steps?
     
  7. Apr 13, 2009 #6

    Lyn

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    i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?
     
  8. Apr 13, 2009 #7

    Hootenanny

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    It doesn't matter, if you work the problem through you'll find that you don't actually need to know the explicit form of z=z(x,y).

    Simply expand the expression in the final line of my previous post.
     
  9. Apr 13, 2009 #8

    Lyn

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    ok i tried to expand your expression, not really sure but i get

    d^2z/dx^2=d/dx(dz/du.du/dx)
    =d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx)

    then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles
     
  10. Apr 13, 2009 #9

    Hootenanny

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    Not entirely sure what you're doing there. Let me do it for dz/dx,

    [tex]\begin{aligned}
    \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\
    & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \\
    & = \frac{du}{dx}\frac{d}{dx}\frac{dz}{du} + \frac{dz}{du}\frac{d}{dx}\frac{du}{dx} \\
    & = \frac{du}{dx}\left(\frac{d}{du}\frac{dz}{du}\right)\frac{du}{dx} + \frac{dz}{du}\frac{d^2u}{dx^2} \\
    & = \left(\frac{du}{dx}\right)^2\frac{d^2 z}{du^2} + \frac{dz}{du}\frac{d^2u}{dx^2}
    \end{aligned}[/tex]

    Do you follow?

    Can you now do the same for dz/dy?
     
  11. Apr 13, 2009 #10

    Lyn

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    I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest.

    I have done the same for dz/dy

    d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy
    =d^2z/du^2(du/dy)^2+dz/du.d^2u/dy^2

    but there is still the problem that i can't work out dz/du. i have a feeling i'm misunderstanding or missing a really important point but i don't know what?
     
  12. Apr 13, 2009 #11

    Hootenanny

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    In going from the second line to the third, I have simply used the product rule since both dz/du and du/dx are functions of x. Does that make sense?
    Good. Don't worry about dz/du, we'll come to that later. So now we have,

    [tex]\frac{d^2 z}{dx^2} + \frac{d^2 z}{dx^2} = \left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du}[/tex]

    Do you agree?

    Note that I have change the ordinary differentials du/dx and du/dx to partial differentials since u is a function of two variables, u=u(x,y). Technically I should have used partial differentials from the start.
     
  13. Apr 13, 2009 #12

    Lyn

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    Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2
     
  14. Apr 13, 2009 #13

    Hootenanny

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    You are told in the question that this quantity is equal to u2. So we now have,

    [tex]\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du} = u^2[/tex]

    Now it's time for you to do a little work. All that's left for you to do is evaluate the partial derivatives and solve the resulting ODE.
     
  15. Apr 13, 2009 #14

    Lyn

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    ok i think i'm getting there. thenk you so much for all your help, i'll try finish it off now
     
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