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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
 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
 one year ago

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

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

zepdrixBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

roshoBest ResponseYou've already chosen the best response.0
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?
 one year ago

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

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

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

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