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Polar Arc Length Question

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the length of the cardioid with equation r = 1 + cos (theta) located in the first quadrant


    2. Relevant equations
    f (theta) = 1 + cos (theta) f'(theta) = -sin (theta) s = antiderivative (0 to (pi/2)) sq rt (f(theta)^(2) + f'(theta)^(2)) d(theta)


    3. The attempt at a solution
    I have worked thru the problem and have arrived at sq rt (2) antiderivative (0 to (pi/2)) sq rt (1-sin (theta)) d(theta) but have hit a road block here. Is there a trig identity that will help solve this? I haven't had any luck finding anything and don't know how else to proceed. Thanks in advance!
     
  2. jcsd
  3. Jun 23, 2010 #2

    rock.freak667

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    [tex]1-sin x \times \frac{1+sinx}{1+sinx} = ?[/tex]

    then use cos2x+sin2x = 1

    then a trig substitution.
     
  4. Jun 23, 2010 #3
    It's been a long day, but I don't understand where the (1+sin x)/(1+sin x) comes from? From what I see that equals 1 and again leaves me with 1-sin x?
     
  5. Jun 23, 2010 #4

    Mark44

    Staff: Mentor

    Yes, 1 + sinx over itself equals 1, and you can always multiply by 1. If you carry out the multiplication suggested by rock.freak667, what do you get?
     
  6. Jun 23, 2010 #5

    rock.freak667

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    multiply it by 1-sinx and then what would you get? (you are multiplying by '1' so that you are not changing the integral)
     
  7. Jun 23, 2010 #6
    I'm getting ((1-sinx)*(1+sinx))/1+sinx = (1+sinx-sinx-sinx^(2))/(1+sinx) = (1-sinx^(2))/(1+sinx) = (cosx^(2))/(1+sinx). Is this correct, or did I mess it up?
     
  8. Jun 23, 2010 #7

    rock.freak667

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    Yes that is correct. So now instead of integrating √(1-sinx) wrt x, you can now integrate

    [tex]\int \sqrt{\frac{cos^2x}{1+sinx}} dx [/tex]

    now remember that √(a/b) = √a/√b

    So what does the new integral become?
     
  9. Jun 23, 2010 #8
    I'm getting the antiderivative of (cosx)/(1+(sq rt sinx))
     
  10. Jun 23, 2010 #9
    For your denominator,
    [tex]\sqrt{1 + \sin x} \neq 1 + \sqrt{\sin x}[/tex]
     
  11. Jun 23, 2010 #10
    How about this: (cosx)/(square root cosx)
     
  12. Jun 24, 2010 #11

    Mark44

    Staff: Mentor

    No, not even close. Following rock.freak667's suggestion you should have
    [tex]\int {\frac{cos~x}{\sqrt{1+sin~x}}} dx [/tex]

    The reason for his suggestion was to get to an integrand that is relatively easy to antidifferentiate.
     
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