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Thread: Find the Length

  1. #1
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    Find the Length

    $\displaystyle x=\frac{(2t+3)^\frac{3}{2}}{3}$, $\displaystyle y=t+\frac{t^2}{2}$, $\displaystyle 0\leq t\leq 3$ They want me to find the Length

    right now ive gotten to $\displaystyle L=\int^3_0 \sqrt{t^2+5t+4} dt$ I dont know if this is even right, and I dont know how to take this integral either :-(
    Last edited by jobebob; Oct 28th 2007 at 09:42 PM. Reason: 2t+3 not 3t+2 ><
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jobebob View Post
    $\displaystyle x=\frac{(2t+3)^\frac{3}{2}}{3}$, $\displaystyle y=t+\frac{t^2}{2}$, $\displaystyle 0\leq t\leq 3$ They want me to find the Length

    right now ive gotten to $\displaystyle L=\int^3_0 \sqrt{t^2+5t+4} dt$ I dont know if this is even right, and I dont know how to take this integral either :-(
    You need:

    $\displaystyle
    L=\int_0^3 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2}~dt
    $

    $\displaystyle
    \frac{dx}{dt}=\frac{1}{2}\sqrt{2t+3}
    $

    $\displaystyle
    \frac{dy}{dt}=1+t
    $

    $\displaystyle
    L=\frac{1}{2}~\int_0^3 \sqrt{4t^2+10t+7}~dt
    $

    RonL
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  3. #3
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    so $\displaystyle L=\frac{1}{2}(\frac{1}{\sqrt{4}}sinh^{-1} \frac{8t+10}{\sqrt{10}})$ from t=0 to t=3?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jobebob View Post
    so $\displaystyle L=\frac{1}{2}(\frac{1}{\sqrt{4}}sinh^{-1} \frac{8t+10}{\sqrt{10}})$ from t=0 to t=3?
    Not sure what you mean?

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Not sure what you mean?

    RonL
    Yea im not really sure either, but wouldn't that be the integral of $\displaystyle \sqrt{4t^2+10t+7}$
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