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Calculate the arc length of the vector function

  • Thread starter jaydnul
  • Start date
  • #1
516
11

Homework Statement


Calculate the arc length of [itex]<2t,t^2,lnt>[/itex] from [itex]1=<t=<e[/itex]

Homework Equations


Arc length=[itex]∫√{(x')^2 + (y')^2 + (z')^2}[/itex]

The Attempt at a Solution


So I have gotten to this point:
[itex]∫√{4 + 4t^2 + \frac{1}{t^2}}[/itex]

Am I on the right track, and if so, how do I integrate that?
 

Answers and Replies

  • #2
33,510
5,194

Homework Statement


Calculate the arc length of [itex]<2t,t^2,lnt>[/itex] from [itex]1=<t=<e[/itex]

Homework Equations


Arc length=[itex]∫√{(x')^2 + (y')^2 + (z')^2}[/itex]
Missing dt in the integrand.

Also, if you use \sqrt instead of √, it will look better.
$$\int \sqrt{x'^2 + y'^2 + z'^2}dt $$

The Attempt at a Solution


So I have gotten to this point:
[itex]∫√{4 + 4t^2 + \frac{1}{t^2}}[/itex]

Am I on the right track, and if so, how do I integrate that?
Write the quantity in the radical as 4t2 + 4 + 1/t2, and then factor it.
 
  • #3
516
11
Thanks. Good advice so far. I think I might just be an idiot by how do you factor that? Haven't had to do that for a while.
 
  • #4
516
11
Nevermind, got it. multiply t^2
 
  • #5
516
11
Thanks for the help!
 
  • #6
33,510
5,194
Nevermind, got it. multiply t^2
You can't do that, but you can multiply by t2 over itself. OTOH, the expression can be factored directly, to (2t + 1/t)2.
 

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