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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.

    x(t)=arcsint
    y(t)=ln(sqrt(1-t^2))


    2. Relevant equations
    S= integral from a-b of
    sqrt((dx/dt)^2+(dy/dt)^2)dt


    3. The attempt at a solution
    (dx/dt)^2=1/(1-t^2)
    (dy/dt)^2=t^2/(1-t^2)^2
    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
    (1/2)∫dt/(1-t)+(1/2)∫dt/(1+t)
    Finally integrating this I get
    (1/2)(ln(1-t)+ln(1+t))
    Plugging in my limits of integration I get
    (1/2)(ln(1/2)+ln(3/2))
    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

    ehild

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    Homework Helper

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

    ehild
     
  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.
    (1/2)(-ln(1/2)+ln(3/2)+ln(1)-ln(1))
    =(1/2)(ln(1/2)^(-1)+ln(3/2)

    On a side note, (1/2)^(-1)=2 right?
    If so I end up with
    ln(3)^(1/2)
     
  6. Jul 24, 2012 #5

    ehild

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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.

    ehild
     
  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

    ehild

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    Homework Helper

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

    ehild
     
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