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

  • Thread starter wh00ps
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  • #1
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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.
 

Answers and Replies

  • #2
mathman
Science Advisor
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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.
 
  • #3
2
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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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