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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0oh, i get it now... thanks a bunch
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