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rosho

  • 2 years ago

CalcII how to solve the arc length: a) int{a,b}sqrt(1+16x^4) b) int{a,b}sqrt(1+36cos^2(2x)) ans a)f(x)=+/-4x^3/3+C b)+/-3sin2x+C

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  1. rosho
    • 2 years ago
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    @zepdrix @electrokid

  2. rosho
    • 2 years ago
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    do you want to me to draw the equation?

  3. rosho
    • 2 years ago
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    |dw:1363864864886:dw|

  4. rosho
    • 2 years ago
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    start with that problem, i'll draw the second when someone has explained the answer.

  5. zepdrix
    • 2 years ago
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    Are you sure you entered that correctly..? This is the solution Wolfram is giving, http://www.wolframalpha.com/input/?i=integral+sqrt%281%2B16x%5E4%29dx There must be a mistake somewhere in there.

  6. rosho
    • 2 years ago
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    we are trying to find the arc length. i think.

  7. zepdrix
    • 2 years ago
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    So find the arclength of this function, \(\large f(x)=\sqrt{1+16x^4}\) From \(\large a\) to \(\large b\) ?

  8. rosho
    • 2 years ago
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    i'll type the text verbatim from the book. what differentiable function have an arc length on the interval [a,b] given by the following integrals?

  9. zepdrix
    • 2 years ago
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    Oh oh oh ok I see what's going on.

  10. rosho
    • 2 years ago
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    okay, that's good because i have no idea.

  11. zepdrix
    • 2 years ago
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    The formula for arc length is,\[\large \int\limits ds \qquad = \qquad \int\limits \sqrt{1+\color{orangered}{\left(\frac{dy}{dx}\right)^2}}dx\] And we're given this,\[\large \int\limits \sqrt{1+\color{orangered}{16x^4}}dx\]

  12. rosho
    • 2 years ago
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    oh, i get it now... thanks a bunch

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