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Thread: PDE's - Implicit Differentiation Query

  1. #1
    Senior Member bugatti79's Avatar
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    PDE's - Implicit Differentiation Query

    hi,
    I am looking at a question in 'advanced engineering mathematics'.

    if z=x^2-y^2 and x=r cos \theta, y=r sin \theta

    1) Do we say that z=u(x,y) or z=u(x(r,\theta),y(r,\theta)) or z=u(r,\theta)?

    2)I know we can get \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}and so on for \frac{\partial z}{\partial \theta}.
    Now lets say we have z=u(x(r,\theta),y(r,\theta))

    How do I get \frac{\partial z}{\partial r} and \frac{\partial z}{\partial \theta} in terms of \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y}?
    Do I differentiate both sides of the equation?

    Thanks
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  2. #2
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    Quote Originally Posted by bugatti79 View Post
    hi,
    I am looking at a question in 'advanced engineering mathematics'.

    if z=x^2-y^2 and x=r cos \theta, y=r sin \theta

    1) Do we say that z=u(x,y) or z=u(x(r,\theta),y(r,\theta)) or z=u(r,\theta)?

    2)I know we can get \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}and so on for \frac{\partial z}{\partial \theta}.
    Now lets say we have z=u(x(r,\theta),y(r,\theta))

    How do I get \frac{\partial z}{\partial r} and \frac{\partial z}{\partial \theta} in terms of \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y}?
    Do I differentiate both sides of the equation?

    Thanks
    Does the question say use a change of variables or in a section about a change of variables?
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dwsmith View Post
    Does the question say use a change of variables or in a section about a change of variables?
    The question in in the section on 'change of variables' and specifically relates to query 1 in my post.

    Query 2 is my own question. Im interested in getting del z del r and del z del theta in terms of del u del x and del u del y.

    Thanks
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  4. #4
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    Quote Originally Posted by bugatti79 View Post
    The question in in the section on 'change of variables' and specifically relates to query 1 in my post.

    Query 2 is my own question. Im interested in getting del z del r and del z del theta in terms of del u del x and del u del y.

    Thanks
    I know how to do a change of variables like this

    u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha})  =\omega(\xi,\eta)
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dwsmith View Post
    I know how to do a change of variables like this

    u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha})  =\omega(\xi,\eta)
    No, i beileve that is not the change of variable the book is referring to. Anyway, diferentiate the LHS and RHS of the first equation on post #1 I calculate
    \frac{\partial z}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} and

    \frac{\partial z}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}

    I wonder is this correct?
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  6. #6
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    Yes, that's correct.
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  7. #7
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    [QUOTE=bugatti79;605264]hi,
    I am looking at a question in 'advanced engineering mathematics'.

    if z=x^2-y^2 and x=r cos \theta, y=r sin \theta

    1) Do we say that z=u(x,y) or z=u(x(r,\theta),y(r,\theta)) or z=u(r,\theta)?
    Any one of those three is correct- with the understanding that x and are functions of u and v (and vice-versa) they all mean the same thing.

    2)I know we can get \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}and so on for \frac{\partial z}{\partial \theta}.
    Now lets say we have z=u(x(r,\theta),y(r,\theta))

    How do I get \frac{\partial z}{\partial r} and \frac{\partial z}{\partial \theta} in terms of \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y}?
    Do I differentiate both sides of the equation?
    If you know x and y as functions, of r and \theta, do the differentiations and put them into that equation. For example, if x= r cos(\theta) and y= r sin(\theta), then
    \frac{\partial z}{\partial r}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}
    = cos(\theta)\frac{\partial z}{\partial x}+ sin(\theta)\frac{\partial z}{\partial y}.


    Thanks
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