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Homework Help: Find the arc length of the curve (Polar)

  1. Jul 24, 2012 #1
    1. The problem statement, all variables and given/known data
    I was wondering if I did this problem correctly as I don't have the solution, also wanted to make sure that my limits of integration were correct as they tend to be tricky in finding arc length in polar coordinates.


    2. Relevant equations
    S= integral from a-b of

    3. The attempt at a solution
    adding (dx/dt)^2+(dy/dt)^2
    I get 1/(1-t^2)^2
    Put all of this into the square root as said by the formula
    I simplified it to the integral from 0 to 1/2 of dt/(1-t^2)
    Factoring the bottom I get dt/((1-t)(1+t))
    by Partial Fractions I get
    2 separate integrals
    Finally integrating this I get
    Plugging in my limits of integration I get
    Using the log rule
    I get ln(3/4)^(1/2)

    Thank you so much to anyone who read through this long problem!
  2. jcsd
  3. Jul 24, 2012 #2


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    You missed a minus sign in the first integral.
    What are the limits a, b?

  4. Jul 24, 2012 #3
    Ah yes you're right. I always make those negative sign mistakes. As for the limits, I'm not 100% sure what they are but it says the curve is defined by the interval 0≤t≤1/2
    so I assumed they're 0 to 1/2.
    Another mistake I just saw, I never plugged in 0 into my log functions.
  5. Jul 24, 2012 #4
    Alright going back and fixing my mistakes.
    using the fundamental theorum of calculus, I have.

    On a side note, (1/2)^(-1)=2 right?
    If so I end up with
  6. Jul 24, 2012 #5


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    Why didn't you simplify -ln(1-t)+ln(1+t)?

    ln(1/2)^(-1) means the reciprocal of ln(1/2) and not the logarithm of the reciprocal of 1/2. Write ln[(1/2)^(-1)]. And yes, (1/2)^(-1)=2. And do not forget the closing parenthesis from (1/2)(ln(1/2)^(-1)+ln(3/2))

    If you meant [itex]\sqrt{\ln(3)}[/itex] then your solution is correct now.

  7. Jul 24, 2012 #6
    I didn't simplify because I was unsure of the effect of the - in front of ln, so I didn't continue with something I was unsure of to avoid mistakes early on.
    Thank you so much though for taking the time to help me!
  8. Jul 24, 2012 #7


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    Remember, ln(y)-ln(x)=ln(y/x). It appears quite often in case a definite integral.

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