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Math Help - help with parametric differentiation

  1. #1
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    Unhappy help with parametric differentiation

    Seeking help with finishing this problem, from line 6 from bottom (with ???).

    Find dy/dx and d^2y/dx^2 for,

    y = ln sin^4 t and x = ln cos^8 t

    y = ln sin^4 t
    dy/dt = 1/sin^4 t X (cos^4 t)
    = cos^4 t/sin^4 t
    = cot^4 t

    x = ln cos^8 t
    dx/dt = 1/cos^8 t X (-sin^8 t)
    = -sin^8 t/cos^8 t
    = -tan^8 t

    dy/dx = dy/dt X dt/dx
    = cot^4 t X -1/tan^8 t
    = cot^4 t X (-cot^8 t)
    = -(cot^4 t X cot^8 t)


    d^2y/dx^2 = d(dy/dx)/dx
    = d(-cot^4 t X cot^8 t)/dx
    dy/dt = -[-cot^4 t.(cosec^2)^8 t] + [cot^8 t.cosec^2)^4 t]
    = [cot^4 t.cosec^16 t] + [cot^8 t.cosec^8 t]

    = cot^4 t.cosec^8 t(cosec^2 t + cot^2 t) ???

    d^2y/dx^2 = dy/dt X dt/dx

    = [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -1/tan^8 t
    = [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -cot^8 t

    Thanks for your help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    Seeking help with finishing this problem, from line 6 from bottom (with ???).

    Find dy/dx and d^2y/dx^2 for,

    y = ln sin^4 t and x = ln cos^8 t

    y = ln sin^4 t
    dy/dt = 1/sin^4 t X (cos^4 t)
    = cos^4 t/sin^4 t
    = cot^4 t

    x = ln cos^8 t
    dx/dt = 1/cos^8 t X (-sin^8 t)
    = -sin^8 t/cos^8 t
    = -tan^8 t

    dy/dx = dy/dt X dt/dx
    = cot^4 t X -1/tan^8 t
    = cot^4 t X (-cot^8 t)
    = -(cot^4 t X cot^8 t)


    d^2y/dx^2 = d(dy/dx)/dx
    = d(-cot^4 t X cot^8 t)/dx
    dy/dt = -[-cot^4 t.(cosec^2)^8 t] + [cot^8 t.cosec^2)^4 t]
    = [cot^4 t.cosec^16 t] + [cot^8 t.cosec^8 t]

    = cot^4 t.cosec^8 t(cosec^2 t + cot^2 t) ???

    d^2y/dx^2 = dy/dt X dt/dx

    = [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -1/tan^8 t
    = [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -cot^8 t

    Thanks for your help.
    First off, your first derivatives are incorrect. You didn't use the chain rule properly.

    x(t) = ln \left ( cos^8(t) \right )
    y(t) = ln \left ( sin^4(t) \right )

    (I will do it your way, but note that: ln \left ( cos^8(t) \right ) = 8 ln(cos(t)), which makes the derivative easier to think about.)

    \frac{dx}{dt} = \frac{1}{cos^8(t)} \cdot (8cos^7(t)) \cdot (-sin(t)) = \frac{-8sin(t)}{cos(t)} = -8tan(t)

    \frac{dy}{dt} = \frac{1}{sin^4(t)} \cdot (4sin^3(t)) \cdot (cos(t)) = \frac{4cos(t)}{sin(t)} = \frac{4}{tan(t)}

    \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{\frac{4}{tan(t)}}{-8 tan(t)} = -\frac{1}{2tan^2(t)}

    There are two ways to proceed from here. I would suspect that the standard thing to do would be two write dy/dx as a function of x, rather than leaving it a function of t. But it depends on what you are shooting for. I would do it that way simply because its a bit shorter to take the next derivative. So we have:
    x(t) = ln \left ( cos^8(t) \right )

    e^x = cos^8(t)

    \pm \left ( e^x \right ) ^{1/8} = cos(t)

    t = acs \left ( \pm \left ( e^{x/8} \right ) \right )

    So
    \frac{dy}{dx} = -\frac{1}{2tan^2 \left ( acs \left ( \pm \left ( e^{x/8} \right ) \right ) \right )}
    and go from there.

    Otherwise I believe your method is correct, simply that your derivatives weren't correct.

    -Dan
    Last edited by topsquark; November 7th 2006 at 06:04 AM.
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  3. #3
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    Actually I think the derivative should be,
    y=\ln |\cos^8 x|
    Then you can write,
    y=8\ln |\cos x|
    However, when you have parantheses instead of absolute value then,
    y=\ln (\cos^8 x)=8\ln| \cos x|

    Sorry, topsquark. I hope all my formal corrections do not bother ye.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Actually I think the derivative should be,
    y=\ln |\cos^8 x|
    Then you can write,
    y=8\ln |\cos x|
    However, when you have parantheses instead of absolute value then,
    y=\ln (\cos^8 x)=8\ln| \cos x|

    Sorry, topsquark. I hope all my formal corrections do not bother ye.
    Oops. I missed that point. Ah well.

    I'm never bothered by corrections. As long as the corrections are correct.

    -Dan
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  5. #5
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    Here is a direct way to the second derivative giving the answer as a function of t.
    \frac{{d^2 y}}{{dx^2 }} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}
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  6. #6
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    Wink

    Thanks for the correction of function of a function.
    I'm getting dy/dx = -1/2tan^2(t) and not -2tan^2(t) from your working. Could you review this.

    dx/dt = -8tan(t)
    dy/dt = 4/tan(t)

    dy/dx = dy/dt X dt/dx
    = 4/tan(t) X -1/8tan(t)
    = -1/2tan^2(t)
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ashura View Post
    Thanks for the correction of function of a function.
    I'm getting dy/dx = -1/2tan^2(t) and not -2tan^2(t) from your working. Could you review this.

    dx/dt = -8tan(t)
    dy/dt = 4/tan(t)

    dy/dx = dy/dt X dt/dx
    = 4/tan(t) X -1/8tan(t)
    = -1/2tan^2(t)
    I have corrected this in my original post. Thanks for the catch and sorry about that!

    -Dan
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