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Length of parametic curves

  1. Mar 13, 2004 #1
    Find the length of the parametrized curve given by


    for t between 0 and 1.

    ok...thats the question. I have tried using the formula L = integral of (dx/dt)2 + (dy/dt)2 all square rooted but im not gettin the rite answer...i have a bunch of these questions havent been able to get ne of them...if sum1 cud jst walk me thru this 1 it wud b a lot of help...thx
  2. jcsd
  3. Mar 13, 2004 #2
    I've never seen a length formula like that.

    I don't know what you may be missing, so let's start from the beginning.

    [tex](dL)^2 = (dx)^2 + (dy)^2 [/tex]

    Now divide through by [tex](dt)^2[/tex]

    [tex](\frac{dL}{dt})^2 = (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2[/tex]

    Take the positive square root of both sides:

    [tex] dL/dt = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}[/tex]

    Here I assume you can now see how to get the correct length formula from that equation.

    Compute dx/dt and dy/dt, square both quantities and substitute the results in the correct length formula.

    Show what you get after these steps. I will be able to check my email tonight and Sunday after pacific coast time 2PM.
  4. Mar 13, 2004 #3
    That's actually exactly the formula sonya said--well, almost. The "all square rooted" part kind of implied that the integral was square rooted, but it's pretty clearly what's inside the integral.

    I just thought I'd mention that outandbeyond2004 began with the Pythagorean Theorem and went from there. It's all triangles!

  5. Mar 13, 2004 #4
    Cookiemonster, do some people write

    [tex]\int f(x)[/tex]

    instead of

    [tex]\int f(x) dx [/tex]


    If so, that makes me feel uncomfortable. The area element (area under the f(x) curve) is not evident in the first kind of notation above. Well, come to think of it maybe Sonya did mean dt in there, tho not explictly.
  6. Mar 13, 2004 #5
    My PDEs professor is notorious for

    [tex]\int f(x)[/tex]

    You can imagine the headache, I'm sure. I guess I just kinda got used to guessing the proper variable.

    And I think I'd call it a length element in this case. ;)

  7. Mar 14, 2004 #6
    formula is correct may be u are following a wrong procedure or making some wrong moves in between
    \frac{dx}{dt}= 18(t+2)

    \frac{dy}{dt}= -3(t^2+4t-5)

    so u have length of curve

    [tex] s = \int_0^1{\sqrt{t^4+42t^2+32t+8t^3+169}}dt
  8. Mar 14, 2004 #7
    yes...that was the formula i meant outandbeyond...

    and thx 4 the help...now i jst have 2 factor that rite..
  9. Mar 15, 2004 #8
    ok...im not gettin newhere with this problem...if ne1 has ne suggestions that would b awesome
  10. Mar 15, 2004 #9
    Are you sure they didnt ask you to integrate a function along that path. For instance finding the weight of a wire if given a function for its density.
  11. Mar 15, 2004 #10
    What's the answer that you're getting? Or are you having trouble evaluating the integral?

  12. Mar 15, 2004 #11
    ok i got it...i was having problems with integrating but i figured it out...thx 4 all the help...much appreciated!
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