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Math Help - Calc III: Arc Length

  1. #1
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    Calc III: Arc Length

    Find the arclength of the curve:
    r(t) = <6t^2, 2sqrt(6)t, ln(t)> when
    t is less than or equal to 2, but greater than or equal to 1.

    I know that the arc length formula is the integral of the magnitude of the derivative of the function r, but I don't know how to integrate it to find the arc length.

    Thank you!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by saxyliz View Post
    Find the arclength of the curve:
    r(t) = <6t^2, 2sqrt(6)t, ln(t)> when
    t is less than or equal to 2, but greater than or equal to 1.

    I know that the arc length formula is the integral of the magnitude of the derivative of the function r, but I don't know how to integrate it to find the arc length.

    Thank you!
    r(t) = \left<6t^2, 2\sqrt 6 t, \ln(t)\right>

    therefore, x(t)=6t^2, y(t)=2\sqrt 6 t and z(t)=\ln t

    Thus, \frac{\,dx}{\,dt}=12t, \frac{\,dy}{\,dt}=2\sqrt 6 and \frac{\,dz}{\,dt}=\frac{1}{t}

    Thus, L=\int_1^2\sqrt{144t^2+24+\frac{1}{t^2}}\,dt

    Note that 144t^2+24+\frac{1}{t^2}=144t^2\cdot{\color{red}\fr  ac{t^2}{t^2}}+24\cdot{\color{red}\frac{t^2}{t^2}}+  \frac{1}{t^2}=\frac{144t^4+24t^2+1}{t^2}=\frac{(12  t^2+1)^2}{t^2}

    Thus, L=\int_1^2\sqrt{\frac{(12t^2+1)^2}{t^2}}\,dt=\int_  1^2\left(12t+\frac{1}{t}\right)\,dt

    Can you take it from here?

    --Chris
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  3. #3
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    Wow! Thank you so much! I got to that integral, but I had no idea how to solve it, I found the answer from there.

    Thank you again!
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