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Find the arc length for the given interval (parametric curve)

  1. Aug 9, 2009 #1
    Find the arc length.

    x = sqrt(t)
    y = 6t - 2

    Interval from 0 to 5 inclusive.

    Whenever I do this, I get a long answer with big numbers in the numerator all divided by 48. Can someone walk me through the steps? THanks.
     
  2. jcsd
  3. Aug 9, 2009 #2

    mathman

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    Science Advisor
    Gold Member

    It would be helpful if you could show what you did to get the "long answer ". In any case the derivative of arc length with respect to t (I assume the interval from 0 to 5 is a t interval) is infinite at t=0, so it is not surprising you get a big number.
     
  4. Aug 10, 2009 #3
    Given parametric equation implies that [tex] y=6x^2 -2 [/tex] , then you should use arc length formula: [tex]L=\int\limits_a^b {\sqrt {1 + [f'(x)]^2 } dx}[/tex] where [tex]f(x)=y=6x^2-2[/tex], a=0, b=[tex]\sqrt5[/tex] , f'(x)=12x
    [tex]L=\int\limits_0^{\sqrt 5 } {\sqrt {1 + 144x^2 } dx} [/tex] ,after [tex]x = \frac{1}{{12}}\tan \theta [/tex] substution
    [tex]\frac{1}{{12}}\int {\sec ^3 \theta d\theta } = \frac{1}{{24}}(\ln \left| {\tan \theta + \sec \theta } \right| + \tan \theta \sec \theta )
    \][/tex]by using integration by parts
    [tex]\[
    \left. {\frac{1}{{24}}(\ln \left| {12x + \sqrt {1 + 144x^2 } } \right| + 12x\sqrt {1 + 144x^2 } )} \right|_0^{\sqrt 5 }
    \]
    [/tex]
    [tex]\[
    L = \frac{1}{{24}}(\ln \left| {12\sqrt 5 + \sqrt {721} } \right| + 12\sqrt 5 \sqrt {721} )
    \][/tex]
    approximately L=30.19 units
     
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