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Thread: Vertor Line intergral

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    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; Nov 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 $\displaystyle ds$ I take that to mean the line intetgral with $\displaystyle \sqrt{[x'(t)]^2+[y'(t)]^2}$

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

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

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

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

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

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

    -Dan
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