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Partial differentiation

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

    y(x,t) = f(x-ct)

    verify this solution satisfies equation
    ∂y2/∂x2 = 1/c2*∂y2/∂t2

    2. Relevant equations



    3. The attempt at a solution

    ∂y/∂x = ∂f/∂x = 1
    ∂y2/∂x2 = 0


    ∂y/∂t = ∂f/∂t = -c
    ∂y2/∂t2 = 0

    Is this the way to do it???
     
  2. jcsd
  3. Jun 9, 2013 #2

    Ray Vickson

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    Start over: everything you did was wrong.
     
  4. Jun 9, 2013 #3
    Can please give me some directions as to how to start on it?
    I am really confused...
     
  5. Jun 9, 2013 #4

    Ray Vickson

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    OK: to see how to get ##\partial y / \partial x##, use the definition:
    [tex] \frac{\partial y(x,t)}{\partial x} = \lim_{h \to 0} \frac{y(x+h,t)-y(x,t)}{h},[/tex]
    substitute in the given form of y(x,t), carry out the steps, and see what you get.
     
  6. Jun 9, 2013 #5

    CAF123

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    Ray's way is nice, but alternatively you may use the chain rule.
     
  7. Jun 9, 2013 #6

    Ray Vickson

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    I agree, but did not want to suggest that. The OP seems to have a fundamental conceptualization problem, and just having him/her apply some "rules" without thinking seemed to me to be counterproductive. I would rather have the OP grind through things from first principles.
     
  8. Jun 9, 2013 #7

    CAF123

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    Ok, that makes sense
     
  9. Jun 9, 2013 #8
    I'm getting similar answer...

    ∂y/∂x = lim (x+h-ct-x+ct)/h
    = lim h/h
    = lim 1
    =1
    ∂y/∂t = lim (x-ct+h-x+ct)/h
    = lim 1
    = 1

    am i on the right track?
     
  10. Jun 9, 2013 #9

    Dick

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    No. ∂y/∂x = lim (f((x+h)-ct)-f(x-ct))/h=lim (f((x-ct)+h)-f(x-ct)). It's not 1. You left out f altogether. And I would suggest if you know the chain rule (and you probably should) then use it. Then look back at how that difference quotient will let you prove it.
     
  11. Jun 10, 2013 #10
    Ok..if I try doing chain rule

    ∂y/∂x = ∂f/∂x * ∂y/∂f

    right?

    I have only dealt with questions where the function is actually given.. like the terms are defined

    something like y(x,t) = ax + xt + xt^2 blah blah

    then I would keep one of the x and t as a constant and differentiate.

    but I am really stuck as to how to this question as it only says function of x-ct.

    :(
     
  12. Jun 10, 2013 #11

    Dick

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    y=f(x-ct). Define g=(x-ct). Then y=f(g). ∂y/∂x=f'(g)*∂g/∂x. f is a function of a single variable. I'm not sure you are getting this chain rule thing very well.
     
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