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anonymous
 one year ago
Find the arc lenght
anonymous
 one year ago
Find the arc lenght

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442618898018:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442618963230:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442619001455:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is where i got stuck..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3its a good place to get stuck at for sure ... a trig sub might be a good shot, maybe.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im using this formula..dw:1442619138404:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we havnt gotten into using trigs yet...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my professor said something about making whats inside the sqrt to a perfect square to be able to cancel the sqrt, but i dont see how i can make it that way...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.34x^2 + kx + 1 (2x+1)^2 seems good 4x^2 + 1 + (4x4x) = (2x+1)^2 4x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3i dont believe that you can just 'make' something into a perfect square, you can rewrite it so that it 'contains' a perfect square

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3if u = 2x+1, du=2dx and x=(u1)/2 (2x+1)^2 4x u^2 4(u1)/2 u^2 2u +2 is still not a perfect square

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2B4x^2%29+dx notice that the wolfs integral isnt that pretty without knowing some hypertrigs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well thanks for the help ill email my professor.. so she can solve it lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3if y=t, then x^2 = t, and x=sqrt(t) y' = 1, x' = 1/(2sqrt(t)) y'^2 = 1, x'^2 = 1/(4t) but that doesnt make for a pretty integral either

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg i have tried to solve this problem over 100 times and guess what i read the question wrong. the questions sais to use a graphing utility to evaluate it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but again using in getting this https://www.wolframalpha.com/input/?i=arc+length+of+y%3Dsqrt%281%2B4x^2%29+from+x%3D0+to+3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when the the answer is 9.747

amistre64
 one year ago
Best ResponseYou've already chosen the best response.3https://www.wolframalpha.com/input/?i=arc+length+of+y%3Dx^2+from+x%3D0+to+3

wmj259
 one year ago
Best ResponseYou've already chosen the best response.0Are you not able to use the ArcLength formula of vectors to solve this?
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