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Math Help - arc length parametric

  1. #1
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    arc length parametric

    Find arc length
    x=t^2, y=4t^3-1 -1<t<1

    dx/dt=2t dy/dt= 12t^2

    integral from -1 to 1 sq root (4t^2 +144t^4)dt

    then, integral from -1 to 1 2t*sq root (1+36t^2)dt

    Please help me find the answer. I know i have to use trig substitution does u=6tantheta?
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  2. #2
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    Quote Originally Posted by andrewsx View Post
    Find arc length
    x=t^2, y=4t^3-1 -1<t<1

    dx/dt=2t dy/dt= 12t^2

    integral from -1 to 1 sq root (4t^2 +144t^4)dt

    then, integral from -1 to 1 2t*sq root (1+36t^2)dt

    Please help me find the answer. I know i have to use trig substitution does u=6tantheta?
    Make the substitution u = 1 + 36t^2 \Rightarrow dt = \frac{du}{72t}.
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