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Calculus/Arc length integral

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Expert: - 1/14/2011


Question

length
I think this question requires some simple transformation, to evaluate the integral for calculating then arc length; if not it can be very messy . Wonder how it be done. Please help!
Thanks

Answer
It is messy.  See below.
(t is phi or theta, or whatever.)

t = 1/2(r + 1/r),   r = 1 to r = 3

arc length  s =

{r=3
|  sqrt(r^2 + (dr/dt)^2) dt
}r=1


dr/dt = 1/(dt/dr)

dt/dr = 1/2(1 - 1/r^2)

dt/dr = (r^2 - 1)/2r^2

          2r^2         
dr/dt = -------------
       (r^2 - 1)

            4r^4
dr/dt^2 = ------------
         (r^2 - 1)^2


dt = (dt/dr) dr

dt = 2r^2/(r^2 - 1) r

{r=3
|sqrt(r^2 + 4r^4/(r^2 - 1)^2) (r^2 - 1)/2r^2 dr
}r=1


{r=3    r^2(r^2 - 1)^2 + 4r^4      r^2 - 1
|  sqrt(-----------------------) ------------- dr
}r=1          (r^2 - 1)^2            2r^2


{r=3    r^2(r^2 - 1)^2 + 4r^4       1
|  sqrt(-----------------------) ------ dr
}r=1            1                 2r^2

{r=3    r^2(r^4 - 2r^2 + 1) + 4r^4       1
|  sqrt(---------------------------) ------ dr
}r=1            1                      2r^2

{r=3    r^2(r^4 + 2r^2 + 1)       1
|  sqrt(--------------------) ------ dr
}r=1            1               2r^2


{r=3    r^2(r^2 + 1)^2           1
|  sqrt(---------------) ------ dr
}r=1            1               2r^2

{r=3  r(r^2 + 1)     
|  (---------------) dr
}r=1    2r^2

{r=3  (r^2 + 1)     
|  (---------------) dr
}r=1     2r

{r=3   1    
|     (-)(r + 1/r) dr
}r=1   2

1/2(r^2/2 + ln r)  ,  1 to 3

1/2(9/2 + ln 3) - 1/2(1/2 + ln 1)

1/2(8/2 + ln 3)

1/2(4 + ln 3)  

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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