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Math Help - Vertor Line intergral

  1. #1
    Member Ranger SVO's Avatar
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    Vertor Line intergral

    Let C be the portion of the curve y = 2*sqrt(x) between (1,2) and (9,6) (explination is more important than the answer)

    Find
    Attached Thumbnails Attached Thumbnails Vertor Line intergral-intergral.bmp  
    Last edited by Ranger SVO; November 16th 2006 at 04:27 PM. Reason: More explination
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    Quote Originally Posted by Ranger SVO View Post
    Let C be the portion of the curve y = 2*sqrt(x) between (1,2) and (9,6) (explination is more important than the answer)

    Find
    When I see the ds I take that to mean the line intetgral with \sqrt{[x'(t)]^2+[y'(t)]^2}

    First we have to parametrize the curve,
    If you let,
    x=t and y=2\sqrt{t}
    1\leq t\leq 9
    Thus,
    \sqrt{[x'(t)]^2+[y'(t)]^2}=\sqrt{1+(3\sqrt{t})^2}=\sqrt{1+9t^2}
    Thus,
    \int_C 2yds=\int_1^9 t\sqrt{1+9t^2}dt
    But, this integral is easy to evaluate, use substitution,
    u=1+9t^2
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    When I see the ds I take that to mean the line intetgral with \sqrt{[x'(t)]^2+[y'(t)]^2}

    First we have to parametrize the curve,
    If you let,
    x=t and y=2\sqrt{t}
    1\leq t\leq 9
    Thus,
    \sqrt{[x'(t)]^2+[y'(t)]^2}=\sqrt{1+(3\sqrt{t})^2}=\sqrt{1+9t^2}
    Thus,
    \int_C 2yds=\int_1^9 t\sqrt{1+9t^2}dt
    But, this integral is easy to evaluate, use substitution,
    u=1+9t^2

    Good explanation, but a few mistakes.
    Let x = t and y = 2 \sqrt{t}.

    Then
    ds = \sqrt{x'^2 + y'^2}dt = \sqrt{1 + \frac{1}{t}}dt

    So
    \int_C3yds = \int_1^9 3 \cdot 2 \sqrt{t} \sqrt{1 + \frac{1}{t}}dt

    =  6 \int_1^9 \sqrt{t + 1}dt

    -Dan
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