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Math Help - More Difficult Chain Rule PRoblems!!!!!

  1. #1
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    More Difficult Chain Rule PRoblems!!!!!

    1. Let
    ?????

    2. Let .
    Find f ''(0)= ????

    3. Let .
    Then is = ?????

    is =??????

    and is =?????


    4. Suppose that the equation of motion for a particle (where is in meters and in seconds) is
    .
    (a) Find the time (other than at ) when the acceleration is 0.

    Acceleration is 0 at time = ??????
    (b) Find the position and velocity at this time.
    Position = ??????Velocity = ?????

    5. Suppose that
    Find = ?????
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  2. #2
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    Just in case a picture helps...

    The chain rule in general...



    ... in which straight continuous lines differentiate downwards with respect to x, and the dashed continuous line with respect to the dashed balloon expression, so that the triangular network on the right satisfies the chain rule.

    Then in particular we have, for 2...



    ... and then differentiating again with the product rule as well ...



    Don't integrate - balloontegrate!

    Balloon Calculus: worked examples from past papers
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  3. #3
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    1. f(x)=\frac{-8x}{1-x}

    f^{'}(x) = \frac{-8\cdot (1-x)-(-x) \cdot (-8x)}{(1-x)^2} = -\frac{8}{(1-x)^2}

    f^{''}(x) = -\frac{(0)\cdot (1-x)^2 -(2(1-x) \cdot (-1)\cdot (8)}{((1-x)^2)^2} = -\frac{16}{(1-x)^3}

    f^{'''}(x) = -\frac{(0) \cdot (1-x)^3 - (3(1-x)^2 \cdot (-1) \cdot (16)}{((1-x)^3)^2} = -\frac{48}{(1-x)^4}

    carrying on with the pattern we have -\frac{48\cdot 4}{(1-x)^5} = <br />
-\frac{192}{(1-x)^5}


    3. f(x) = \frac{1-5x}{1+5x}

    f^{'}(x) = \frac{(-5) \cdot (1+5x) -(5)\cdot (1-5x)}{(1+5x)^2} = -\frac{10}{(1+5x)^2}

    f^{''}(x) = -\frac{(0) \cdot (1+5x)^2) - (2(1+5x) \cdot (10)\cdot 5)}{((1+5x)^2)^2} = \frac{100}{(1+5x)^3}

    f^{'''}(x) = -\frac{1500}{(1+5x)^4} given the pattern that has emerged.


    5. y = 7(x^2-3)^3

    \frac{dy}{dx} = 7\cdot(3) \cdot (x^2-3)^2 \cdot (2x) = 42x(x^2-3)^2
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  4. #4
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    Thanks so much guys...i really appreciated your help. Thank you

    P.S : I do need help on four please...# 4

    4. Suppose that the equation of motion for a particle (where is in meters and in seconds) is

    .
    (a) Find the time (other than at ) when the acceleration is 0.

    Acceleration is 0 at time = ??????
    (b) Find the position and velocity at this time.
    Position = ??????
    Velocity = ?????
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  5. #5
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    Quote Originally Posted by Kayla_N View Post
    Thanks so much guys...i really appreciated your help. Thank you

    P.S : I do need help on four please...# 4

    4. Suppose that the equation of motion for a particle (where is in meters and in seconds) is

    .
    (a) Find the time (other than at ) when the acceleration is 0.

    Acceleration is 0 at time = ??????
    (b) Find the position and velocity at this time.
    Position = ??????
    Velocity = ?????
    (a) Solve a = \frac{d^2 s}{dt^2} = 0 for t.

    (b) Substitute the value of t found in (a) into s and v = \frac{ds}{dt}.
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  6. #6
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    Thanks Mr. Fantastic.
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  7. #7
    MHF Contributor matheagle's Avatar
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    v=4t^3-9t^2
    so a=12t^2-18t=6t(2t-3)
    setting this equal to 0 gives t=0, but they want the other which is t=1.5.


    NOW get s(1.5) and v(1.5).
    Last edited by mr fantastic; March 2nd 2009 at 05:13 PM. Reason: Merged posts
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  8. #8
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    ok for number 4. I got t=1.5 and the velocity is = -6.75. How can i get the position or find it? please help me out. thanks


    NEVERMIND i got it now. sorry.
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