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Math Help - Partial derivatives and the chain rule

  1. #1
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    Partial derivatives and the chain rule

    Im revising for an exam i have next week, and cannot for my life understand the solutions to this previous exam question:

    http://i.imgur.com/hLGMwIp.jpg

    Is this essentially the same question as:
    z = f(x,y) , g = g(s,t), y = h(s,t)

    and then find dz/ds and dz/dt ?
    the solutions look like the employ implicit differentiation, but i am not sure. typically I am confused from the second line,

    HOw is du/dr * dr/dt = -c * df/dr ?
    My train of thought is dr/dt = -c
    du/dr is another way of saying df/dr, and using that same deduction we get the second half of that equation

    But then i am COMPLETELY lost on the third line

    Any help would be appreciated.
    Please explain to me, like I am an idiot though..
    thank MHF, you're always a huge help
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Partial derivatives and the chain rule

    Quote Originally Posted by 99.95 View Post
    du/dr is another way of saying df/dr
    Point of clarification: how is this line true?

    -Dan
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    Re: Partial derivatives and the chain rule

    I was just taking a guess, assuming that the change in notation allowed them to get the solution.
    Not sure what they are doing.
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  4. #4
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    Re: Partial derivatives and the chain rule

    Quote Originally Posted by 99.95 View Post
    Im revising for an exam i have next week, and cannot for my life understand the solutions to this previous exam question:

    http://i.imgur.com/hLGMwIp.jpg

    Is this essentially the same question as:
    z = f(x,y) , g = g(s,t), y = h(s,t)
    Typo: you mean x= g(s, t), right?

    and then find dz/ds and dz/dt ?
    the solutions look like the employ implicit differentiation, but i am not sure. typically I am confused from the second line,

    HOw is du/dr * dr/dt = -c * df/dr ?
    Yes, that's the chain rule. You are given that u(x, t)= f(x- ct)+ g(x+ ct) and they are setting r= x- ct, s(x+ ct) so that u(x, t)= f(r)+ g(s).

    My train of thought is dr/dt = -c
    Yes, that is true.

    du/dr is another way of saying df/dr, and using that same deduction we get the second half of that equation
    That's rather roughly what is happening. More specifically, since u= f(r)+ g(s) and r and s are functions or x and t,
    \partial u/\partial x= \partial f/\partial x+ \partial g/\partial x= (\partial f/\partial r)(\partial r/\partial x)+ (\partial g/\partial s)(\partial x)
    (The fact that f depends only on r and g only on s simplifies this. Otherwise we would have (\partial f/\partial s)(\partial s/\partial x) and (\partial g/\partial r)(\partial r/\partial x) as well.)

    = -c \partial f/\partial r)+ c\partial g/\partial s

    But then i am COMPLETELY lost on the third line
    Do it again.
    \partial^2 u/\partial x^2= \frac{\partial }{\partial x}(\partial u}{\partial x})= \frac{\partial}{\partial x}(-c\partial f/\partial r+ c\partial g/\partial s)

    = -c\frac{\partial}{\partial x}(\partial f/\partial r)+ c\partial}{\partial x}(\partial f/\partial s)
    And now use the fact that \frac{\partial F}{\partial x}= -c\partial F/\partial r+ c\partial F/\partial s
    where F is any differentialble function of x and t. Here, F= -c\partial f/\partial r+ c\partial f/\partial s

    Taking it one part at a time, -c\partial (-c \partial f/\partial r+ c\partial f/\partial s)/\partial r= c^2\partial^2f/\partial r^2- c^2\partial f/\partial r\partial s and c\partial(-c\partial f/\partial r+ c\partial f/\partial s)/\partial s= -c^2\partial^2 r/\partial r\partial s+ c^2\partial^2 f\partial s^2./
    Any help would be appreciated.
    Please explain to me, like I am an idiot though..
    thank MHF, you're always a huge help[/QUOTE]
    Thanks from topsquark and 99.95
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Partial derivatives and the chain rule

    Not in time!

    -Dan
    Last edited by topsquark; June 19th 2013 at 05:51 AM. Reason: Hallsl beat me to it
    Thanks from 99.95
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  6. #6
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    Re: Partial derivatives and the chain rule

    Thank you both a million times over!

    Makes sense now
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