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Math Help - Higher order derivative

  1. #1
    Super Member
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    Higher order derivative

    Hi
    I need help on the following:
    1) Find \frac{d^{2}y}{dx^2} for x^2 + 4xy +3y^2 = 5 at points (-1,2) <br /> <br />
2x + 4x \frac{dy}{dx} +4y + 6y \frac{dy}{dx} = 0

    \frac{dy}{dx} = \frac{-4x -2x}{6y +4x}

    \frac{(-6y+4x)(-2 - 4\frac{d^{2}y}{dx^2}) - (-2x-4y)(6<br />
\frac{d^{2}y}{dx^2}+4)}{(6y+4x)^2} = 0

    \frac{d^{2}y}{dx^2}-24y+24y-16x+12x = 12y-16y+8x-8x

    \frac{d^{2}y}{dx^2} = \frac{-4y}{-4x}

    \frac{d^{2}y}{dx^2} = \frac{y}{x} * \frac{1}{(6y+4x)^2}

    when x=-1 and y=2 then:
    \frac{d^{2}y}{dx^2} =  \frac{2}{2(128)} = \frac{1}{64}

    book's answers says: \frac{5}{64}


    2) Find \frac{d^{2}y}{dx^2} for x= 1 - cos(t) y= t - sin(t)

    \frac{dx}{dt} =  sin(t)

    \frac{dy}{dt} = 1 - cos(t)

    \frac{dy}{dx} = \frac{1 - cos(t)}{sin(t)}

    \frac{d^{2}y}{dx^2} = \frac{(sin(t))^2 - cos(t) - (cos(t))^2}{(sin(t))^2}

    \frac{d^{2}y}{dx^2} = \frac{1 - cos(t)}{(sin(t))^2}

    answer says it is: \frac{1 - cos(t)}{(sin(t))^3}

    P.S
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  2. #2
    Junior Member piglet's Avatar
    Joined
    Feb 2010
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following:
    1) Find \frac{d^{2}y}{dx^2} for x^2 + 4xy +3y^2 = 5 at points (-1,2) <br /> <br />
2x + 4x \frac{dy}{dx} +4y + 6y \frac{dy}{dx} = 0

    \frac{dy}{dx} = \frac{-4x -2x}{6y +4x}

    Should be
     \frac{dy}{dx} = \frac{-4y-2x}{6y+4x}

    \frac{(-6y+4x)(-2 - 4\frac{d^{2}y}{dx^2}) - (-2x-4y)(6<br />
\frac{d^{2}y}{dx^2}+4)}{(6y+4x)^2} = 0

    \frac{d^{2}y}{dx^2}-24y+24y-16x+12x = 12y-16y+8x-8x

    \frac{d^{2}y}{dx^2} = \frac{-4y}{-4x}

    \frac{d^{2}y}{dx^2} = \frac{y}{x} * \frac{1}{(6y+4x)^2}

    when x=-1 and y=2 then:
    \frac{d^{2}y}{dx^2} =  \frac{2}{2(128)} = \frac{1}{64}

    book's answers says: \frac{5}{64}


    2) Find \frac{d^{2}y}{dx^2} for x= 1 - cos(t) y= t - sin(t)

    \frac{dx}{dt} =  sin(t)

    \frac{dy}{dt} = 1 - cos(t)

    \frac{dy}{dx} = \frac{1 - cos(t)}{sin(t)}

    \frac{d^{2}y}{dx^2} = \frac{(sin(t))^2 - cos(t) - (cos(t))^2}{(sin(t))^2}

    \frac{d^{2}y}{dx^2} = \frac{1 - cos(t)}{(sin(t))^2}

    answer says it is: \frac{1 - cos(t)}{(sin(t))^3}

    P.S
    You've made another simple error on question 2). I reckon if you attempt it one more time you'll get it. I'll tell you if you really can't spot it...

    Piglet
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  3. #3
    Super Member
    Joined
    Dec 2008
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    So for question 1 what is my problem?

    you wrote it should be \frac{dy}{dx} = \frac{-4y-2x}{6y+4x}

    however that is what i got as well.
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