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Math Help - need explaining

  1. #1
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    need explaining

     L = \int^{2\pi}_{0}\sqrt{(-sin t)^2 + (cos t)^2 + (\frac{1}{\pi})^2} dt

    = \int^{2\pi}_{0}\sqrt{sin^2t+cos^2t+(\frac{1}{\pi^2  })} dt

    this is an example from the book it skipped a step, i dont know how they got the next step:

     = \int^{2\pi}_{0}\sqrt{1+\frac{1}{\pi^2}}dt

     = 2\pi\sqrt{1+\frac{1}{\pi^2}} is the arc length
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
     L = \int^{2\pi}_{0}\sqrt{(-sin t)^2 + (cos t)^2 + (\frac{1}{\pi})^2} dt

    = \int^{2\pi}_{0}\sqrt{sin^2t+cos^2t+(\frac{1}{\pi^2  })} dt

    this is an example from the book it skipped a step, i dont know how they got the next step:

     = \int^{2\pi}_{0}\sqrt{1+\frac{1}{\pi^2}}dt

     = 2\pi\sqrt{1+\frac{1}{\pi^2}} is the arc length
    i hope that by calc 2 you would know that \sin^2 x + \cos^2 x = 1

    also, \sqrt{1 + \frac 1{\pi^2}} is a constant, so the integral is just \left[ \sqrt{1 + \frac 1{\pi^2}} \right]t and the rest follows
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  3. #3
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    its been awhile since i took calc2, thanks for the reminder Jhevon
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    its been awhile since i took calc2, thanks for the reminder Jhevon
    really? what math are you doing now? (well, to be fair, you should have known that identity long before calc 2 even)
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    really? what math are you doing now?
    Probably a Civil Engineering course. Maybe calculating length of certain cables.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Probably a Civil Engineering course. Maybe calculating length of certain cables.
    lol nah im not smart enough to be taking that course, im taking calc3 online. was reading the book and got a lil lost on one of the example.
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