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anonymous

  • 4 years ago

Find the perimeter of the region bounded by y=x and y=x^2

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  1. anonymous
    • 4 years ago
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    Hey u r pink :P

  2. TuringTest
    • 4 years ago
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    yeah, the downside of being level 80 lol do you know the arc length formula?

  3. anonymous
    • 4 years ago
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    yes

  4. anonymous
    • 4 years ago
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    |dw:1328027797447:dw|

  5. TuringTest
    • 4 years ago
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    so the problem is integrating it?

  6. anonymous
    • 4 years ago
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    umm no i need help setting up the equation

  7. anonymous
    • 4 years ago
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    umm just a lil confused

  8. TuringTest
    • 4 years ago
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    |dw:1328027896437:dw|here's you pic, intersection (1,1) the length of the y=x can be found with the pythagorean theroem...

  9. amistre64
    • 4 years ago
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    sqrt(2) :)

  10. anonymous
    • 4 years ago
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    ya that is what i got for the function y=x

  11. anonymous
    • 4 years ago
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    the other one is a lil more scary

  12. amistre64
    • 4 years ago
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    \[\int^{1}_{0} \sqrt{1+(2x)^2}dx\]

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

  14. amistre64
    • 4 years ago
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    yep, lengths go to trig stuff usually

  15. TuringTest
    • 4 years ago
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    so you wind up with\[\int_{0}^{1}\sec^3\theta d\theta\]which is a pretty well-known integral

  16. anonymous
    • 4 years ago
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    ok but what abt the second function?

  17. amistre64
    • 4 years ago
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    ive never meet that integral in me life :)

  18. amistre64
    • 4 years ago
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    the second function is just the distance from 0,0 to 1,1

  19. anonymous
    • 4 years ago
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    lol

  20. anonymous
    • 4 years ago
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    Yesterday someone told me you need to find the arc lengths of both functions

  21. amistre64
    • 4 years ago
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    if need be: \[\int_{0}^{1}\sqrt{1+(1)^2}dx\]

  22. TuringTest
    • 4 years ago
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    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

  23. anonymous
    • 4 years ago
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    ummm i guess i can just solve it numerically if need be

  24. TuringTest
    • 4 years ago
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    talking about integral sec cubed here example 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx

  25. TuringTest
    • 4 years ago
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    takes too long to type myself

  26. anonymous
    • 4 years ago
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    \[ \int \sec x \;dx = \ln |\sec x +\tan x| +c \] I remember correctly?

  27. TuringTest
    • 4 years ago
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    you do :D

  28. amistre64
    • 4 years ago
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    for a sec yes; tricky mathikers i tell ya

  29. anonymous
    • 4 years ago
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    Great :D

  30. anonymous
    • 4 years ago
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    ummm so that wld be the answer? like i forget abt the other function?

  31. TuringTest
    • 4 years ago
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    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

  32. anonymous
    • 4 years ago
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    LOL ok Thanks

  33. anonymous
    • 4 years ago
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    but one last question though

  34. anonymous
    • 4 years ago
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    like y is that the perimeter?

  35. anonymous
    • 4 years ago
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    we only found the arc length of the line but not the curve

  36. TuringTest
    • 4 years ago
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    |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 10 http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx and add that to the straight part

  37. TuringTest
    • 4 years ago
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    oh my triangle is wrong :/

  38. TuringTest
    • 4 years ago
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    |dw:1328028961912:dw|there we go

  39. anonymous
    • 4 years ago
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    Btw Turing Pippa is right, this integral gives the length of the parabola \(y=x^2\) right?

  40. TuringTest
    • 4 years ago
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    ...bounds 0 to 1

  41. anonymous
    • 4 years ago
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    We still need to sum up the straight line right?

  42. TuringTest
    • 4 years ago
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    I think amistre did that two different ways above

  43. anonymous
    • 4 years ago
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    ohhh ok i see

  44. anonymous
    • 4 years ago
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    yes that is correct LOL

  45. anonymous
    • 4 years ago
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    I forgot abt that :D

  46. TuringTest
    • 4 years ago
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    pythagorean theorem, then he used arc length same answer for the straight bit both ways

  47. anonymous
    • 4 years ago
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    Alrighty Thanks :D Dont be late for class

  48. TuringTest
    • 4 years ago
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    yeah, 2 min late already!

  49. anonymous
    • 4 years ago
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    I don't understand, :(

  50. TuringTest
    • 4 years ago
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    later fellas! @FFM figure it out :P

  51. anonymous
    • 4 years ago
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    hehe thanks

  52. anonymous
    • 4 years ago
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    Duh got it :D It's simple Pythagorean theorem no need of integrating ;)

  53. anonymous
    • 4 years ago
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    hehehe I dont follow that

  54. anonymous
    • 4 years ago
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    ya but i get what to do

  55. anonymous
    • 4 years ago
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    peepppaaa :D

  56. anonymous
    • 4 years ago
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    lol yes

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