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anonymous
 4 years ago
Find the perimeter of the region bounded by y=x and y=x^2
anonymous
 4 years ago
Find the perimeter of the region bounded by y=x and y=x^2

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, the downside of being level 80 lol do you know the arc length formula?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328027797447:dw

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0so the problem is integrating it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm no i need help setting up the equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm just a lil confused

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328027896437:dwhere's you pic, intersection (1,1) the length of the y=x can be found with the pythagorean theroem...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya that is what i got for the function y=x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the other one is a lil more scary

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int^{1}_{0} \sqrt{1+(2x)^2}dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}=2x\]\[L=\int\sqrt{1+(\frac{dy}{dx})^2}dx=\int_{0}^{1}\sqrt{1+4x^2}dx\]off the bat I'm thinking trig sub\[x=\frac12\tan\theta\to dx=\frac12\sec^2\theta d\theta\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yep, lengths go to trig stuff usually

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0so you wind up with\[\int_{0}^{1}\sec^3\theta d\theta\]which is a pretty wellknown integral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok but what abt the second function?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ive never meet that integral in me life :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the second function is just the distance from 0,0 to 1,1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yesterday someone told me you need to find the arc lengths of both functions

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if need be: \[\int_{0}^{1}\sqrt{1+(1)^2}dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0it's a little tricky the first time you see it. It can be done by integration by parts. You need to know how to do the integral of secant though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm i guess i can just solve it numerically if need be

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0talking about integral sec cubed here example 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0takes too long to type myself

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \int \sec x \;dx = \ln \sec x +\tan x +c \] I remember correctly?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1for a sec yes; tricky mathikers i tell ya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm so that wld be the answer? like i forget abt the other function?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0Well you have the straight part... I don't really have time to go through the whole undoing the trig sub thing, I gotta get to class

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but one last question though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like y is that the perimeter?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we only found the arc length of the line but not the curve

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328028665611:dwhere's the triangle for your trig sub I set up the curved part for you, remember?\[\frac12\tan\theta=\ x\]\[\frac12\sec^2\theta d\theta=dx\]winds up as \[\int\sec^3\theta d\theta\]Use example 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx and add that to the straight part

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0oh my triangle is wrong :/

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328028961912:dwthere we go

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Btw Turing Pippa is right, this integral gives the length of the parabola \(y=x^2\) right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We still need to sum up the straight line right?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I think amistre did that two different ways above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes that is correct LOL

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0pythagorean theorem, then he used arc length same answer for the straight bit both ways

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alrighty Thanks :D Dont be late for class

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 2 min late already!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand, :(

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0later fellas! @FFM figure it out :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Duh got it :D It's simple Pythagorean theorem no need of integrating ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hehehe I dont follow that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya but i get what to do
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