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Math Help - Arc Length and Curvature

  1. #1
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    Arc Length and Curvature

    This is my first post, so hello forums . Anyway, I am trying to relearn calculus, took a hiatus for a while, and I am currently taking beginners multi-variable calculus and I hit an example that I do not understand in my book (probably cause my diff/integral calculus could use some boning up ).

    This involves Arc Length and Curvature:

    r(t) = cos(t)i + sin(t)j + tk from point (1,0,0) to (1,0,2 \pi)

    they show that \mid r\prime (t)\mid = \sqrt{(-sin (t))^2 + cos^2 (t) + 1} = \sqrt{2}

    and then they say "The arc from (1,0,0) to (1,0,2 \pi) is described by the parameter interval 0 \leq t \leq 2\pi and so [from formula 3], we have...

    L = \int_0^{2\pi} \mid r\prime (t)\mid dt\ = \int_0^{2\pi} \sqrt{2}  dt = 2\sqrt{2\pi}

    I do not understand how they got 2\sqrt{2\pi}?
    shouldn't it be 2\pi\sqrt{2}?







    Thank you for any and all input
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  2. #2
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    Quote Originally Posted by Warrenx View Post
    This is my first post, so hello forums . Anyway, I am trying to relearn calculus, took a hiatus for a while, and I am currently taking beginners multi-variable calculus and I hit an example that I do not understand in my book (probably cause my diff/integral calculus could use some boning up ).

    This involves Arc Length and Curvature:

    r(t) = cos(t)i + sin(t)j + tk from point (1,0,0) to (1,0,2 \pi)

    they show that \mid r\prime (t)\mid = \sqrt{(-sin (t))^2 + cos^2 (t) + 1} = \sqrt{2}

    and then they say "The arc from (1,0,0) to (1,0,2 \pi) is described by the parameter interval 0 \leq t \leq 2\pi and so [from formula 3], we have...

    L = \int_0^{2\pi} \mid r\prime (t)\mid dt\ = \int_0^{2\pi} \sqrt{2} dt = 2\sqrt{2\pi}

    I do not understand how they got 2\sqrt{2\pi}?
    shouldn't it be 2\pi\sqrt{2}?







    Thank you for any and all input
    Dear Warrenx,

    It should be 2\pi\sqrt{2}. I think in the book they mean, 2\sqrt{2}\pi but accidently made the squareroot sign a little longer making it 2\sqrt{2\pi}

    Hope this helps you.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear Warrenx,

    It should be 2\pi\sqrt{2}. I think in the book they mean, 2\sqrt{2}\pi but accidently made the squareroot sign a little longer making it 2\sqrt{2\pi}

    Hope this helps you.
    Thank you very much Mr. Sudharaka, was bugging me so much I stopped reading the section.
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  4. #4
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    That's the reason it is standard (except in this textbook, apparently!) to put numbers or variable before the square root symbol- that is while we typically write "2x", we write " x\sqrt{2}".
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