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TomLikesPhysics Group Title

I want to find the length of a curve which is given by the equation: r=(t-sint,1-cost, 2). To calculate the length I first calculate the first derivative of r which is v=(1-cost, sint, 0). Than I went on to integrate the absolulte value of v. This gave me the integral of the squareroot of (2-2cost). Than I ended up with -2*squareroot of (2+2cost). Which is the same as Wolfram Alpha got. However, if I want to integrate this curve from 0 to 2pi than I end up with 0, but it should be 8. If I plug in 0 and 2pi I get -4-(-4) which is 0, but the curve can not have a lenght of 0.

  • 2 years ago
  • 2 years ago

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  1. TomLikesPhysics Group Title
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    In a nutshell: I ended up with the right integral but if I plug in my boundaries (0 and 2pi) I get 0 as the lenght of the curve. So somehow the lenghts eat each other up? So, how do I "use" the integral in the right way to get 8 as a lenght?

    • 2 years ago
  2. dumbcow Group Title
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    hmm to make the math work maybe you should split it into 2 integrals with limits (0,pi) and (pi,2pi)

    • 2 years ago
  3. Fellowroot Group Title
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    This is very strange, wolfram does indeed give the answer as 8 yet whenever i try and evaluate the integral I get zero just think 0 and 2pi are the same value when evaluated at cos(t) no wonder its zero, yet wolfram says otherwise.

    • 2 years ago
  4. TomLikesPhysics Group Title
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    Changing the boundaries from before 0 to 2pi to 0 to pi and pi to 2pi did not help at all. I only started learning about calculating the length of a curve so I have no clue what this is supposed to tell me. The Curce must clearly have some length.

    • 2 years ago
  5. TomLikesPhysics Group Title
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    Here is my (wrong) calculation so far:

    • 2 years ago
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  6. TomLikesPhysics Group Title
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    The second integral goes from pi/2 to what?

    • 2 years ago
  7. sirm3d Group Title
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    correction: it's \[\int\limits_{0}^{\pi} \] and \[\int\limits_{\pi}^{2\pi}\]

    • 2 years ago
  8. sirm3d Group Title
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    here's the corrected equation: \[\sqrt{2-2\cos t}=2\cos \frac{ t }{ 2 }\] when \[0 \le t \le \pi\] and \[\sqrt{2-2\cos t}=-2\cos \frac{ t }{ 2 }\] in \[\pi \le t \le 2\pi\]

    • 2 years ago
  9. TomLikesPhysics Group Title
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    So I did the integral wrong?

    • 2 years ago
  10. sirm3d Group Title
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    so that \[\int\limits_{0}^{2\pi}\sqrt{2-2\cos t}dt = \int\limits_{0}^{\pi}2\cos \frac{ t }{ 2 }dt + \int\limits_{\pi}^{2 \pi} -2\cos \frac{ t }{ 2 }dt\]

    • 2 years ago
  11. TomLikesPhysics Group Title
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    Yeah, if I rewrite that Integral I get 8. However, I could nowhere find that Trig. Identity that you used.

    • 2 years ago
  12. TomLikesPhysics Group Title
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    Ah... now I see where you got that from. If you don´t mind could you check where my substitution went wrong?

    • 2 years ago
  13. TomLikesPhysics Group Title
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    Thank you so far for your help sirm3d.

    • 2 years ago
  14. sirm3d Group Title
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    \[\cos^2 A=\frac{ 1 }{ 2 }(1+\cos 2A)\] that's half-angle identity

    • 2 years ago
  15. sirm3d Group Title
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    im still checking your solution.

    • 2 years ago
  16. sirm3d Group Title
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    here's the mistake: \[\sqrt{2-2\cos t}\] \[=\sqrt{\frac{ (2-2\cos t)(2+2\cos t) }{ (2 + 2\cos t) }}\]\[=\frac{ \sqrt{4-4\cos^2 t} }{ \sqrt{2+2\cos t} }\] up to this point it is still correct.

    • 2 years ago
  17. TomLikesPhysics Group Title
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    I guess it might be the point where I set dt=du/2sint - sint could be zero so I am dividing by zero!?!?!?!?

    • 2 years ago
  18. sirm3d Group Title
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    \[\sqrt{4-4\cos^2 t} \neq 2\sin t\] due to the fact that \[2\sin \] is negative in \[\pi < t < 2\pi\]

    • 2 years ago
  19. sirm3d Group Title
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    \[\sqrt{4-4\cos^2t}=2\sin t\] in \[0<t<\pi\] and \[\sqrt{4-4\cos^2 t}=-2 \sin t\] in \[\pi < t < 2 \pi\]

    • 2 years ago
  20. sirm3d Group Title
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    a common mistake in definite integrals involving trig function is the choice of + over minus. be careful next time.

    • 2 years ago
  21. TomLikesPhysics Group Title
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    Thank you for your help, sirm3d. I try to be more carefull next time. :) Thx a lot.

    • 2 years ago
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