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 one year 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
 one year 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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rosho
 one year ago
Best ResponseYou've already chosen the best response.0do you want to me to draw the equation?

rosho
 one year ago
Best ResponseYou've already chosen the best response.0start with that problem, i'll draw the second when someone has explained the answer.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Are 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.

rosho
 one year ago
Best ResponseYou've already chosen the best response.0we are trying to find the arc length. i think.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So find the arclength of this function, \(\large f(x)=\sqrt{1+16x^4}\) From \(\large a\) to \(\large b\) ?

rosho
 one year ago
Best ResponseYou've already chosen the best response.0i'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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh oh oh ok I see what's going on.

rosho
 one year ago
Best ResponseYou've already chosen the best response.0okay, that's good because i have no idea.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The 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,\[\large \int\limits \sqrt{1+\color{orangered}{16x^4}}dx\]

rosho
 one year ago
Best ResponseYou've already chosen the best response.0oh, i get it now... thanks a bunch
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