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Partial derivative proof

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose f: R^2 --> R is differentiable and (df/dt) = c(df/dx) for some nonzero constant c.
    Prove that f(x, t) = h(x + ct) for some function h.


    2. Relevant equations

    hint: use (u, v) = (x, x+ct)

    3. The attempt at a solution

    df/dt = limk-->0 (f(x, x+ct+k) - f(x, x+ct))/k
    multiplying this by c gives:
    limk-->0 (f(x, x+ct+ck) - f(x, x+ct))/k

    I'm not sure where to go from here
     
  2. jcsd
  3. Nov 12, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    Are you required to use the basic definition of the derivative? You were given a hint which works nicely with the chain rule but you have ignored it. If u= x, v= x+ ct, then [itex]\partial f/\partial x= (\partial f/\partial u)(\partial u/\partial x)+ (\partial f/\partial v)(\partial v/\partial x)[/itex][itex]= \partial f/\partial u+ \partial f/\partial v[/itex]. Similarly, [itex]\partial f/\partial t= (\partial f/\partial u)(\partial u/\partial t)+ (\partial f/\partial v)(\partial v/\partial t)[/itex][itex]= c\partial f/\partial v[/itex].

    Now, the original differential equation, in terms of u and v, becomes
    [itex]c\partial f/\partial v= c(\partial f/\partial u+ \partial f/\partial v)[/itex].

    We can now subtract [itex]c\partial f/\partial v[/itex] from both sides leaving [itex]c\partial f/\partial u= 0[/itex] which says that f does not depend upon u at all, it is simply a function of v.
     
  4. Nov 12, 2012 #3

    I like Serena

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    Homework Helper

    Hi The1TL! :smile:

    Can you calculate ##\frac d {du}f(x(u,v), t(u,v))## with partial derivatives?
     
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