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Math Help - Partial derivatives

  1. #1
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    Partial derivatives

    What are the Partial derivatives for this function?

    I came up with these expressions:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3

    It's not logical, right?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    What are the Partial derivatives for this function?

    I came up with these expressions:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3

    It's not logical, right?
    u=e^{-x^2-y^2}(2x^2+3y^2)=2x^2e^{-x^2-y^2}+3y^2e^{-x^2-y^2}

    So \frac{\partial{u}}{\partial{x}}=4xe^{-x^2-y^2}-4x^3e^{-x^2-y^2}-6xy^2e^{-x^2-y^2}

    and

    \frac{\partial{u}}{\partial{y}}=-4yx^2e^{-x^2-y^2}+6ye^{-x^2-y^2}-6y^3e^{-x^2-y^2}
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    Quote Originally Posted by Mathstud28 View Post
    u=e^{-x^2-y^2}(2x^2+3y^2)=2x^2e^{-x^2-y^2}+3y^2e^{-x^2-y^2}

    So \frac{\partial{u}}{\partial{x}}=4xe^{-x^2-y^2}-4x^3e^{-x^2-y^2}-6xy^2e^{-x^2-y^2}

    and

    \frac{\partial{u}}{\partial{y}}=-4yx^2e^{-x^2-y^2}+6ye^{-x^2-y^2}-6y^3e^{-x^2-y^2}

    Ok, now I need to find Maxima and minima, so I compare those to 0, right?

    And then I come up with what I showed you...
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    Ok, now I need to find Maxima and minima, so I compare those to 0, right?

    And then I come up with what I showed you...
    Ok, so factoring a little we see that

    u_x=e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)
    and u_y=e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)

    So we need to find all (x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0

    I get (-1,0),(0,-1),(0,0),(0,1),(1,0)

    Now test them all in the discriminant.

    D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2

    And then draw the correct conclusions.
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Ok, so factoring a little we see that

    u_x=e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)
    and u_y=e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)

    So we need to find all (x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0

    I get (-1,0),(0,-1),(0,0),(0,1),(1,0)

    Now test them all in the discriminant.

    D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left\{f_{xy}(a,b)\right\}^2

    And then draw the correct conclusions.

    One second , you did (x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0
    and came up with this equations:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3


    right?

    How did you solved that? how did you come up with (-1,0),(0,-1),(0,0),(0,1),(1,0)?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    One second , you did (x,y)\backepsilon~u_x(x,y)=u_y(x,y)=0
    and came up with this equations:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3


    right?

    How did you solved that? how did you come up with (-1,0),(0,-1),(0,0),(0,1),(1,0)?
    u_x=0\Rightarrow{e^{-x^2-y^2}\left(4x-4x^3-6xy^2\right)}=0

    and

    u_y=0\Rightarrow{e^{-x^2-y^2}\left(-4yx^2+6y-6y^3\right)}=0

    So we need to simultaneously solve these. But we know the exponential function is never zero. so we just need to solve

    4x-4x^3-6xy^2=0

    and

    -4xy^2+6y-6y^3=0




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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    If this is a maxima/minima problem, have you tried using Lagrange Multipliers? I think that technique would make things a "little" easier. When I find the time later, I will solve the problem using this alternative way...

    --Chris
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  8. #8
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    isn't Lagrange Multipliers is over do for this question? I found the Partial derivatives, compared them to 0 and came up with 2 equation that I can't find solution to:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3


    This is the problem, what do I do from here?

    10x again.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by asi123 View Post
    isn't Lagrange Multipliers is over do for this question? I found the Partial derivatives, compared them to 0 and came up with 2 equation that I can't find solution to:

    2x^2 + 3y^2 = 2

    2x^2 + 3y^2 = 3


    This is the problem, what do I do from here?

    10x again.
    Where are you getting these equations?!

    {2x^2+3y^2=2}\brace{2x^2+3y^2=3} does not have any solutions!

    We can see this since if we see that the left hand sides of both are equivalent we can set them equatl to each other getting 2=3
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  10. #10
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    This is what I did (pic)
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  11. #11
    MHF Contributor kalagota's Avatar
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    ok.. so your system of equations is:

    \frac{\partial u}{\partial x} = -2xe^{-x^2-y^2}(2x^2+3y^2-2)=0--------(1)

    \frac{\partial u}{\partial y} = -2ye^{-x^2-y^2}(2x^2+3y^2-3)=0--------(2)

    if you set x=0 in (1), you should substitute that value in (2) to get values of y corresponding to x=0. from here, you will get three values of y corresponding to x=0.

    if in (1), x\not=0, then it must be
    2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2..
    if you substitute this in (2), you will get
    -2ye^{-x^2-y^2}(\underbrace{2x^2+3y^2}_{=2}-3)=-2ye^{-x^2-y^2}(2-3)=0
    you will get from here, y=0.
    substitute this back to
    2x^2+3y^2-2=0 \Longleftrightarrow 2x^2+3y^2=2
    to get the specific value of x which is not equal to 0 (corresponding to y=0).

    you should get the values what Mathstud28 got..
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