Line integral - finding the arc length

  • Thread starter nb89
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  • #1
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a curve is given as 3 parameters of t:
x=a(3t - t^3), y=3a(t^2), z=a(3t + t^3)

i have to find the arc length measured from origin and curvature as functions of t.

would i be correct in using the integral at the bottom of page 2 here: http://homepages.ius.edu/wclang/m311/fall2005/notes17.2.pdf
 

Answers and Replies

  • #2
arildno
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Sure enough, with f=1.
Then you get:
[tex]\int3|a|\sqrt{(1-t^{2})^{2}+(2t)^{2}+(1+t^{2})^{2}}[/tex]
If you are clever, you'll find a perfect square buried within the radicand.
 
  • #3
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why is f=1?

would my limits be 0 to t?
 
  • #4
arildno
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why is f=1?
Because you are to find [tex]\int{ds}=\int{1}*ds[/tex]
would my limits be 0 to t?
Yes.
 
  • #5
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ive ended up with integral 3a√2 (t^2 +1) dt
so i take the 3a√2 out and integrate t^2 +1.

but im confused since the limits are 0 to t, how would i replace the t with the limit t once i have integrated it?
 
  • #6
HallsofIvy
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Don't worry about the limits of integration: you are finding the arclength as a function of t so find an anti-derivative and take the constant so that the integral is 0 when t= 0.
 
  • #7
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then i get 3a√2 (t^3/3 + t). thats wrong though?
 

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