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Pleasse Help Me to Find Parameter of CIrcle by integration.

  1. Nov 12, 2009 #1
    untitled.jpg

    plz help me to findout this problem
    i am putting this question second time bcoz i was unable to findout its catagory i am new here
    answer should be 2(pi)a
     
  2. jcsd
  3. Nov 12, 2009 #2

    Mark44

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    What you want is the perimeter of your circle, not the parameter.
    In your integral for arc length, you are apparently going to calculate the arc length of the first quadrant of the circle, and then multiply by 4.

    You have x^2 + y^2 = a^2. Either solve for y as a function of x, and then differentiate to get dy/dx, OR differentiate the given equation implicitly to get dy/dx. Put that expression in for dy/dx in your formula for arc length.

    Finally, you need to put in limits of integration, and then calculate the integral.
     
  4. Nov 12, 2009 #3
    yes i got your point, but i don't know what limit i have to apply. i was applying limit
    x=0 and x=2(pi)a/4 am i right here, i have solve it approximately at end but i am unable to solve last four steps
     
  5. Nov 12, 2009 #4

    Dick

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    It looks like you are trying to integrate the arclength over the first quadrant and multiply by 4. That's fine. That means your x limits are 0 to a. But you aren't really showing what you are doing or following Mark44's advice. What is y'?
     
  6. Nov 12, 2009 #5
    we will find value of y from eq 1 + y^2 = a^2 - x^2
    i will put it (dy/dx)^2
     
  7. Nov 12, 2009 #6

    Dick

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    Ok, I think. Do so.
     
  8. Nov 23, 2009 #7
    Please Help me to solve this Parameter of the circle

    1. i have try my best and i and reach at a point and unable to solve it complete please help me to find the parameter of the circle

    .2 also what limit i should apply ( i am thinking to apply 2(pi)a/4) am i right

    3.What i further do

    aaa.gif

    If you are unable to see http://i481.photobucket.com/albums/rr178/urduworld/aaa.gif" [Broken]

    Thanks in advance
     
    Last edited by a moderator: May 4, 2017
  9. Nov 23, 2009 #8

    tiny-tim

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    Hi urduworld! :smile:

    (have an integral: ∫ and a square-root: √ and a pi: π and a theta: θ :wink:)

    ok, you're trying to find the circumference of a circle of radius a by integrating arc-length round the circle,

    and you've arrived at 4∫ dx/√(a2 - x2)

    Now do a trig substitution. :smile:
    Since you've decided to find the length of just the first quadrant, your x is going from 0 to a, isn't it? So those are your limits: ∫0a … dx. :wink:

    (Incidentally, if you'd started with θ as your variable, and dθ as your arc-length, then you could have done the whole circle, with limits of 0 ≤ θ ≤ 2π, or one quadrant with limits of 0 ≤ θ ≤ π/2, the result multiplied by 4, which would give the same answer).
     
  10. Nov 23, 2009 #9

    Mark44

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    Re: Please Help me to solve this Parameter of the circle

    You posted the same problem in another thread: https://www.physicsforums.com/showthread.php?t=354071.

    You shouldn't start a new thread for the same problem.

    Also, the distance around a circle is its circumference, or perimeter. Parameter means something different.
     
  11. Nov 23, 2009 #10
    Re: Please Help me to solve this Parameter of the circle

    yes i have posted this second time because that time i cant got answer :(
     
  12. Nov 23, 2009 #11
    Re: Please Help me to solve this Parameter of the circle

    ohh i realy got point i have got that i have have to substitute a^2 - x^2 with (u or any alphabet) and then i have to apply limit am i right Thanks tiny :)
     
  13. Nov 23, 2009 #12

    Mark44

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    Re: Please Help me to solve this Parameter of the circle

    If you don't get an answer in a thread, don't post a new thread, "bump" the thread. That will bring it back up to the top of the displayed posts, and someone will usually respond to it.

    In any case, Dick responded to your other thread with what you needed to do. Did you follow his suggestion?
     
  14. Nov 23, 2009 #13

    berkeman

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    Re: Please Help me to solve this Parameter of the circle

    Threads merged. Do NOT do this again.
     
  15. Nov 23, 2009 #14

    Mark44

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    Re: Please Help me to solve this Parameter of the circle

    No, not an ordinary substitution - a trig substitution. Do you know how to do one of these substitutions?
     
  16. Nov 24, 2009 #15

    tiny-tim

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    No.

    That would give you ∫ 2√(a2 - u)du/√u , which is even worse. :frown:
     
  17. Nov 24, 2009 #16
    then what i have to do i cant understand trig substitution :(
     
  18. Nov 24, 2009 #17
    i dont know trig substitution. plz just hint me what i have to do
     
  19. Nov 24, 2009 #18

    berkeman

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    Trigonometric substitution means using trig identities to come up with a form of the equation that is easier to integrate. See this page for example, for some useful trig identities. I'm betting that your textbook also has a section like this....

    http://en.wikipedia.org/wiki/Trig_identities

    .
     
  20. Nov 25, 2009 #19

    tiny-tim

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    Or http://en.wikipedia.org/wiki/Trigonometric_substitution" [Broken] :wink:
     
    Last edited by a moderator: May 4, 2017
  21. Nov 25, 2009 #20
    Thanks Tiny-Tim, Mark 44 and Berkman :) You realy help me and i have solve this question, PF really helps me :) Again Thanks in Last
     
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