Arc length of

  • Thread starter Rozenwyn
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  • #1
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[tex]c(t)=t^2\cos{t}.i+t^2\sin{t}.j+2t.k[/tex], and [tex] 0 \leq{t} \leq1[/tex]

Firstly, [tex] L(c) = \int_{0}_{1} || c'(t) ||.dt[/tex]

Then, [tex]=\int_{0}^{1} || (-t^{2} \sin{t} + 2t \cos{t} ) i + (t^{2} \cos{t} + 2t \sin{t}) j + 2k || dt [/tex]

Skipping one step, this simplifies to;
[tex]\int_{0}^{1} \sqrt{t^4 + 4t^2 + 4} .dt[/tex]

[tex]\int_{0}^1 t^2+2.dt[/tex]

[tex] \left[ \frac{t^3}{3} + 2t \right]_{0}^{1} = 1/3 + 2 = 7/3 [/tex]

Well, yeah is this correct ?

Let's see, I hope I didn't mess up the latex code.
 
Last edited:

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
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Remember that |c'(t)| is the square root of <c(t),c(t)>, so your integrand should be the square root of what you have now.
 
  • #3
31
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You're absolutely right. How did I miss that ? :P... It should be correct now.
 

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