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Math Help - c curve,work done by force........

  1. #1
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    c curve,work done by force........

    Hello,
    plz try do these questions.
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  2. #2
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    1)The path is from (0,6) to (7,0).
    This is the equation,
    y=-\frac{7}{6}x+6
    Thus, if x=t
    And it goes from 0\leq t\leq 7.
    Thus,
    \b{C}=t\b{i}-(\frac{7}{6}t+6)\b{j}
    Thus, the line integral is,
    \int_0^7 (3t)(t)' dt+\int_0^7 2\left(-\frac{7}{6}t+6\right)\left(-\frac{7}{6}t+6\right)' dt
    Now you can compute it.
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  3. #3
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    2)First we find,
    x'=1
    y'=t^{1/2}
    Thus,
    \sqrt{(x')^2+(y')^2}=\sqrt{1+t}
    Thus, the line integral is,
    \int_0^3 \frac{1}{1+t} \cdot \sqrt{1+t} dt=\int_0^3 (1+t)^{-1/2} dt
    Use subsitution, u=1+t to get answer.
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  4. #4
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    3)
    Are you sure there is supposed to be a four? It seems it is asking for path independence, but the cross pratials test does not work if there is not four.

    M(x,y)=\frac{1}{x^2+y^2}
    N(x,y)=\frac{4}{x^2+y^2}
    To have path indepence (what it seems)
    It is necessary and suffienct for,
    \frac{\partial M}{ \partial y}=\frac{\partial N}{\partial x}
    But that is not work here.

    Thus, what is the path? A straight line?
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  5. #5
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    reply.........

    Hello,
    yes, i am sure it is four.path is not straight line it is like a curve shape.
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  6. #6
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    correct question:

    Hello,
    the correct question no 3 is attached.
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  7. #7
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    Quote Originally Posted by m777 View Post
    Hello,
    the correct question no 3 is attached.
    I am unable to load the document perhaps it is defective.
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