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Math Help - second drivitive of a parametric

  1. #1
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    Lightbulb second drivitive of a parametric

    I have a parametrically defined curve where i am trying to find the second derivative.

    here is the problem:

    x(t) = \frac {1}{t}
    and
    y(t) = -2 + lnt

    I need to find the equation for the tangent line to the curve at the point defined by t = 1

    finding \frac{dy}{dx}

    the dirivitive of x =  -\frac{1}{t^2}

    the dirivitive of y =  \frac{1}{t}

    then, i evaluate t = 1 by plugging it into

    Sorry having some formatting issues below i hope you can read it..

     \frac{dy}{dt}          \frac{1}{t}
    _______
     \frac{dx}{dt}       \frac{-1}{t^2}

    after plugging in 1 for t

    I get a slope of -1

    so, then i need to take a second derivative
    \frac{d^2y}{dx^2}

    for the numerator its just 1
    for the denominator its the derivative of x = -\frac{1}{t^2}

    if so how do i evaluate that?
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  2. #2
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    Re: second drivitive of a parametric

    Quote Originally Posted by icelated View Post
    I have a parametrically defined curve where i am trying to find the second derivative.

    here is the problem:

    x(t) = \frac {1}{t}
    and
    y(t) = -2 + lnt

    I need to find the equation for the tangent line to the curve at the point defined by t = 1

    finding \frac{dy}{dx}

    the dirivitive of x =  -\frac{1}{t^2}

    the dirivitive of y =  \frac{1}{t}

    then, i evaluate t = 1 by plugging it into

    Sorry having some formatting issues below i hope you can read it..

     \frac{dy}{dt}          \frac{1}{t}
    _______
     \frac{dx}{dt}       \frac{-1}{t^2}

    after plugging in 1 for t

    I get a slope of -1

    so, then i need to take a second derivative
    \frac{d^2y}{dx^2}

    for the numerator its just 1
    for the denominator its the derivative of x = -\frac{1}{t^2}

    if so how do i evaluate that?
    You have a first derivative of \dfrac{dy(t)}{dx(t)} = -t

    To find the second derivative you need to differentiate the first derivative which should be quite simple
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    Re: second drivitive of a parametric

    Quote Originally Posted by e^(i*pi) View Post

    To find the second derivative you need to differentiate the first derivative which should be quite simple
    Thats what i thought - so there isnt a second derivative! or rather zero
    So im left with just

    \frac{-1}{t^2}

    for the second derivitive?
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    Re: second drivitive of a parametric

    Quote Originally Posted by icelated View Post
    Thats what i thought - so there isnt a second derivative! or rather zero
    So im left with just

    \frac{-1}{t^2}

    for the second derivitive?
    no ... you need to take the derivative w/ respect to x
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    Re: second drivitive of a parametric

    Quote Originally Posted by icelated View Post
    Thats what i thought - so there isnt a second derivative! or rather zero So im left with just
    \frac{-1}{t^2} for the second derivitive?
    You have the wrong idea about the second derivative.
    \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}.

    This may be simpler notation, is not strictly correct.

    Let y'=\frac{dy}{dx} then \frac{d^2y}{dx^2}=\frac{dy'/dt}{dx/dt}
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    Re: second drivitive of a parametric

    Hi plato - oh, lets see if i got this right..

    so i take the derivative of y prime
    over the derivative of x

    Y`` =  \frac{1}{t}

    because i dont need to take the 2nd derivative of denominator.

    so i end up with the numerator as the same as the denominator?
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    Re: second drivitive of a parametric

    Quote Originally Posted by icelated View Post
    Hi plato - oh, lets see if i got this right..
    so i take the derivative of y prime
    over the derivative of x
    Y`` =  \frac{1}{t}
    because i dont need to take the 2nd derivative of denominator.
    so i end up with the numerator as the same as the denominator?
    The first derivative is:
    y'=\frac{dy/dt}{dx/dt}=-t.

    So \frac{d^2y}{dx^2}=~?
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  8. #8
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    Re: second drivitive of a parametric

    Well if \frac{dy}{dx}

    =  \frac{1}{t}

    then, i take the second derivitive of that in respect to x but i cant do a
    fraction over a fraction in latex.

    taking the derivitive of  \frac{1}{t} in respect to x

    I get  \frac{-1}{t^2}

    Which is exactly as the same as \frac{dx}{dt}

    So they cancel out for

    \frac{d^2y}{dx^2} = 0

    Right?
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  9. #9
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    Re: second drivitive of a parametric

    \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-t) = -1 \cdot \frac{dt}{dx} = -1 \cdot (-t^2) = t^2
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