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  1. #71
    Member Lazarus's Avatar
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    Quote Originally Posted by Blizzard View Post
    Yup, they both come from that.
    Oh, thanks. We just did hyperbolic functions in school, so I'm glad I could actually use some of it xD
    I hear integrating hyperbolic functions is probably the hardest topic we learn at my level (upper sixth - thats the British equivalent of senior at high school. Going to uni next year!). Just looking at all of those standard forms....

  2. #72
    Junior Member Super Neko's Avatar
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    assassin engineer


    So I'm doing an arc length problem (parametric) that is
    X=e^(t)cost
    Y=e^(t)sint
    0<t<pi (in radians)
    My teacher gave the solution of 2^(1/2)(e^(pi)-1)
    BUT he's old and his answers are wrong sometimes, and i got [2^(1/2)/2][e^(2pi)-1]
    So basically it comes down to if i did my algebra right, so my question is if i squared the following terms correctly (the derivatives of the x & y)
    [(e^(t))(cost-sint)]^2 = e^(2t)(cost^(2)-2sintcost+sint^(2)
    And the second is very similar but i don't feel like typing it.
    Basically i forget algebra basics sometimes, but that e gets raised to the 2t, when squaring, right? The 2 in the denominator on my solution's outside term comes from simplifying the two squared derivatives and adding them, then factoring.
    Sorry if my notation is confusing, in hindsight a picture would have been better

  3. #73
    Common Sense Tingling Shaki's Avatar
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    Noooo fucking clue, forgot that shit. @Stein
    Yeah. That is a gun in my pants. But that doesn't mean I'm not happy to see you...

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  4. #74
    Stein's Avatar
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    @Super Neko yes that basic algebra is of course correct. Perhaps your mistake is elsewhere.

    Usually these are solved with euler's and some calculations so maybe you have made a careless miscalculation somewhere.
    Last edited by Stein; Dec 9th '16 at 09:05 AM.
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  5. #75
    ShichiDrewKai's Avatar
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    Quote Originally Posted by Super Neko View Post
    So I'm doing an arc length problem (parametric) that is
    X=e^(t)cost
    Y=e^(t)sint
    0<t<pi (in radians)
    My teacher gave the solution of 2^(1/2)(e^(pi)-1)
    BUT he's old and his answers are wrong sometimes, and i got [2^(1/2)/2][e^(2pi)-1]
    So basically it comes down to if i did my algebra right, so my question is if i squared the following terms correctly (the derivatives of the x & y)
    [(e^(t))(cost-sint)]^2 = e^(2t)(cost^(2)-2sintcost+sint^(2)
    And the second is very similar but i don't feel like typing it.
    Basically i forget algebra basics sometimes, but that e gets raised to the 2t, when squaring, right? The 2 in the denominator on my solution's outside term comes from simplifying the two squared derivatives and adding them, then factoring.
    Sorry if my notation is confusing, in hindsight a picture would have been better
    Ya, a picture would have been way clearer. The difference betwee your answer and his seems to be whether or not you're multiplying by root 2(his answer) or dividing by root 2 (your answer).

    I know that you can take this: e^(2t)(cost^(2)-2sintcost+sint^(2) (what you had above)

    and make this: e^(2t)(1+sin(2t)). You can then distribute through the parentheses

    cos^2 + sin^2 = 1, and 2sintcost = sin2t. Pretty sure those are just basic trig identities. Does that help you figure out if you're right or the teacher is right?
    Quote Originally Posted by Stein View Post
    @Super Neko yes that basic algebra is of course correct. Perhaps your mistake is elsewhere.

    Usually these are solved with euler's and some calculations so maybe you have made a careless miscalculation somewhere.
    Love euler. His calculus stuff was boring to me, but I really liked his stuff on graph theory and topology. Ever heard of the 7 bridges of Konigsberg problem?

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  6. #76
    Stein's Avatar
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    Quote Originally Posted by ShichiDrewKai View Post
    Love euler. His calculus stuff was boring to me, but I really liked his stuff on graph theory and topology. Ever heard of the 7 bridges of Konigsberg problem?
    Yeap does ring a bell from graphs iirc.
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  7. #77
    AppleCider's Avatar
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    swordsman doctor


    Quote Originally Posted by Stein View Post
    Yeap does ring a bell from graphs iirc.
    you math?

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