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Math Help - Second partial derivative?

  1. #1
    Member billym's Avatar
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    Second partial derivative?

    I have a bunch of questions like this but am a little mixed up with what the method is. Can somebody get me started on this?

    <br />
f(x,y)  , x = e^rcos(\theta), y=e^rsin(\theta)

    f^2_r+f^2_\theta=<br />
    Last edited by billym; October 2nd 2009 at 09:40 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by billym View Post
    I have a bunch of questions like this but am a little mixed up with what the method is. Can somebody get me started on this?

    <br />
f(x,y)  , x = e^rcos(\theta), y=e^rsin(\theta)

    f^2r+f^2\theta=<br />
    (not sure about the latex for that - its second derivative of f with respect to r and theta - I think)
    use \partial ok nvm


    \frac{\partial ^2 f}{\partial r^2} = \frac{\partial }{\partial r}(\frac{\partial f}{\partial r})

    \frac{\partial f}{\partial r} = \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}+\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r}

    but you did not give f(x,y) = ?? what is equal
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  3. #3
    Member billym's Avatar
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    if you look at the top its

    x=e^rcos(\theta), y=e^rsin(\theta)

    What is the difference between f^2_r and  frr ?
    Last edited by billym; October 2nd 2009 at 10:11 AM.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by billym View Post
    if you look at the top its

    x=e^rcos(\theta), y=e^rsin(\theta)

    What is the difference between f^2_r and  frr ?

    I do not know but I think the question should give you f(x,y) equal what for example

    f(x,y)=x+y or f(x,y) = xy or f(x,y) = \frac{x}{2} + \frac{y}{x} etc

    since you want to find the partial derivatives of f so you need the rule of f and the rule of x,y you give me x,y rules but I need f(x,y) = ?? to solve the question
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  5. #5
    Member billym's Avatar
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    Well that's what is causing me problems.

    Here is the question in all it's glory:


    If f=f(x,y),  x=e^rcos(\theta) and y=e^rsin(\theta), prove the following identidy:

    f^2_r+f^2_\theta=e^{2r}(f^2_x+f^2_y)



    Wouldn't this just work out to 0 = 0 ?
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by billym View Post
    Well that's what is causing me problems.

    Here is the question in all it's glory:


    If f=f(x,y),  x=e^rcos(\theta) and y=e^rsin(\theta), prove the following identidy:

    f^2_r+f^2_\theta=e^{2r}(f^2_x+f^2_y)



    Wouldn't this just work out to 0 = 0 ?
    ok I get it

    f_r ^2 + f_\theta ^2 = e^{2r}(f_x ^2 + f_y ^2 ) ok take the left side


    I will write d instead of partial to make is easier for me ok
    f_r = \frac{df}{dx} \cdot \frac{dx}{dr} + \frac{df}{dy} \cdot \frac{dy}{dr}

    f_r = f_x (x) + f_y (y) since dx/dr =x and dy/dr =y

    f_r ^2 = f_x ^2 x^2 + f_x f_y yx + f_y ^2 y^2

    f_r ^2 = f_x ^2 (e^{2r} \cos ^2 \theta ) + f_x f_y (e^{2r}\sin \theta \cos \theta ) + f_y ^2 (e^{2r} \sin ^2 \theta )

    this is the first thing now

    \frac{df}{d\theta} = f_x dx/d\theta + f_y dy/d\theta

    x_\theta = -y , y_\theta = x

    \frac{df}{d\theta} = f_x (-y) + f_y (x)

    simplify it and find the sum you will get the solution

    \sin ^2 x + \cos ^2 x =1 use this
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  7. #7
    MHF Contributor Amer's Avatar
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    x= e^r \cos \theta

    y=e^r \sin \theta

    x_\theta = \frac{\partial x}{\partial \theta } = e^r (-\sin \theta) = -e^r\sin \theta = -y

    y_\theta = \frac{\partial y}{\partial \theta } = e^r \cos \theta = x
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  8. #8
    Member billym's Avatar
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    er... could you start me off on this one?

    f_{rr}+f_{\theta\theta}=e^{2r}(f_{xx}+f_{yy})
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  9. #9
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by billym View Post
    er... could you start me off on this one?

    f_{rr}+f_{\theta\theta}=e^{2r}(f_{xx}+f_{yy})

    f_r = f_x \cdot x_r + f_y \cdot y_r

    f_{rr} = (f_{xx}\cdot x_r)x_r + f_x\cdot x_{rr} + (f_{yy} \cdot y_r)y_r + f_y \cdot y_{rr} now

    x_{rr} = x_r=x , y_{rr} = y_r=y

    f_{rr} = f_{xx}\cdot x^2 + f_x \cdot x + f_{yy}\cdot y^2 + f_y \cdot y .................(1)

    other one I will symbol theta with t ok

    f_{t} = f_x \cdot x_t + f_y \cdot y_t

    f_{tt}= (f_{xx} x_t)x_t + f_x \cdot x_{tt} + (f_{yy} \cdot y_t)y_t + f_y \cdot y_{tt}

    x_t = -y , y_t = x and x_{tt} = -x , y_{tt} = -y

    f_{tt} = f_{xx} \cdot y^2 + f_x \cdot -x + f_{yy} \cdot x^2 + f_y \cdot -y .........(2)



    sub x value and y value in 1,2 and fin sum and simplify

    f_{rr}+f_{tt} = f_{xx}\cdot x^2 + f_x \cdot x + f_{yy}\cdot y^2 + f_y \cdot y + f_{xx} \cdot y^2 + f_x \cdot -x + f_{yy} \cdot x^2 + f_y \cdot -y

    you can continue from here just do what I say best wishes
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  10. #10
    MHF Contributor Amer's Avatar
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    I made a mistake

    f_r = f_x (x_r) + f_y(y_r)

    f_{rr} = \frac{\partial f_x}{\partial r} (x_r ) + f_x (x_{rr}) +  \frac{\partial f_y}{\partial r}(y_r) + f_y(y_{rr})

    \frac{\partial f_x}{\partial r} = f_{xx} (x_r) + f_{xy} (y_r) \ne f_{xx} (x_r)

    same thing for all second partial for f with respect to r and theta
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