anonymous
  • anonymous
Find the perimeter of the region bounded by y=x and y=x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Hey u r pink :P
TuringTest
  • TuringTest
yeah, the downside of being level 80 lol do you know the arc length formula?
anonymous
  • anonymous
yes

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More answers

anonymous
  • anonymous
|dw:1328027797447:dw|
TuringTest
  • TuringTest
so the problem is integrating it?
anonymous
  • anonymous
umm no i need help setting up the equation
anonymous
  • anonymous
umm just a lil confused
TuringTest
  • TuringTest
|dw:1328027896437:dw|here's you pic, intersection (1,1) the length of the y=x can be found with the pythagorean theroem...
amistre64
  • amistre64
sqrt(2) :)
anonymous
  • anonymous
ya that is what i got for the function y=x
anonymous
  • anonymous
the other one is a lil more scary
amistre64
  • amistre64
\[\int^{1}_{0} \sqrt{1+(2x)^2}dx\]
TuringTest
  • TuringTest
\[\frac{dy}{dx}=2x\]\[L=\int\sqrt{1+(\frac{dy}{dx})^2}dx=\int_{0}^{1}\sqrt{1+4x^2}dx\]off the bat I'm thinking trig sub\[x=\frac12\tan\theta\to dx=\frac12\sec^2\theta d\theta\]
amistre64
  • amistre64
yep, lengths go to trig stuff usually
TuringTest
  • TuringTest
so you wind up with\[\int_{0}^{1}\sec^3\theta d\theta\]which is a pretty well-known integral
anonymous
  • anonymous
ok but what abt the second function?
amistre64
  • amistre64
ive never meet that integral in me life :)
amistre64
  • amistre64
the second function is just the distance from 0,0 to 1,1
anonymous
  • anonymous
lol
anonymous
  • anonymous
Yesterday someone told me you need to find the arc lengths of both functions
amistre64
  • amistre64
if need be: \[\int_{0}^{1}\sqrt{1+(1)^2}dx\]
TuringTest
  • TuringTest
it'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
  • anonymous
ummm i guess i can just solve it numerically if need be
TuringTest
  • TuringTest
talking about integral sec cubed here example 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx
TuringTest
  • TuringTest
takes too long to type myself
anonymous
  • anonymous
\[ \int \sec x \;dx = \ln |\sec x +\tan x| +c \] I remember correctly?
TuringTest
  • TuringTest
you do :D
amistre64
  • amistre64
for a sec yes; tricky mathikers i tell ya
anonymous
  • anonymous
Great :D
anonymous
  • anonymous
ummm so that wld be the answer? like i forget abt the other function?
TuringTest
  • TuringTest
Well 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
  • anonymous
LOL ok Thanks
anonymous
  • anonymous
but one last question though
anonymous
  • anonymous
like y is that the perimeter?
anonymous
  • anonymous
we only found the arc length of the line but not the curve
TuringTest
  • TuringTest
|dw:1328028665611:dw|here'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 10http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx and add that to the straight part
TuringTest
  • TuringTest
oh my triangle is wrong :/
TuringTest
  • TuringTest
|dw:1328028961912:dw|there we go
anonymous
  • anonymous
Btw Turing Pippa is right, this integral gives the length of the parabola \(y=x^2\) right?
TuringTest
  • TuringTest
...bounds 0 to 1
anonymous
  • anonymous
We still need to sum up the straight line right?
TuringTest
  • TuringTest
I think amistre did that two different ways above
anonymous
  • anonymous
ohhh ok i see
anonymous
  • anonymous
yes that is correct LOL
anonymous
  • anonymous
I forgot abt that :D
TuringTest
  • TuringTest
pythagorean theorem, then he used arc length same answer for the straight bit both ways
anonymous
  • anonymous
Alrighty Thanks :D Dont be late for class
TuringTest
  • TuringTest
yeah, 2 min late already!
anonymous
  • anonymous
I don't understand, :(
TuringTest
  • TuringTest
later fellas! @FFM figure it out :P
anonymous
  • anonymous
hehe thanks
anonymous
  • anonymous
Duh got it :D It's simple Pythagorean theorem no need of integrating ;)
anonymous
  • anonymous
hehehe I dont follow that
anonymous
  • anonymous
ya but i get what to do
anonymous
  • anonymous
peepppaaa :D
anonymous
  • anonymous
lol yes

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