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Math Help - Polar Curves.. PLEASE HELP!

  1. #1
    zee
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    Unhappy Polar Curves.. PLEASE HELP!

    Question) The polar curve C is given by the polar equation r=1+cos(theta)
    Find the length of C.

    Answer) This is what I have so far:
    r = 1 + cos(theta)

    Since L = INT(sqrt[1 + (dr/d(theta))^2])d(theta)

    Therefore,
    L = int(sqrt[1+sin^2(theta)])d(theta) (with integral from 0 to 2)

    then Im stuck... please help to solve this!!
    zee
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  2. #2
    MHF Contributor
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    No.
    In polar coordinates,
    L = sqrt[r^2 +(dr/dT)^2]dT -------***

    It is not
    L = sqrt[1 +(dr/dT)^2]dT
    You are confused with
    L = sqrt[1 +(dy/dx)^2]dx
    in cartesian coordinates.

    ------
    r = 1 +cosT
    dr/dT = -sinT

    If the limits are from T=0 to T=2 radians,
    L = INT.(0->2)[sqrt{(1+cosT)^2 +(-sinT)^2}]dT
    L = INT.(0->2)[sqrt{(1 +2cosT +cos^2(T)) +sin^2(T)}]dT
    L = INT.(0->2)[sqrt{(1 +2cosT +{cos^2(T) +sin^2(T)}}]dT
    L = INT.(0->2)[sqrt{1 +2cosT +1}]dT
    L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)

    The "dT" outside of the bracket is not the derivative of the sqrt{2 +2cosT} inside the bracket.

    We need to express the sqrt{2 +2cosT} into a one-degree function.
    See if we can convert the (2 +2cosT) into a square of cos(nT) or sin(nT).

    cosT
    = cos(T/2 +T/2)
    = cos^2(T/2) -sin^2(T/2)
    = cos^2(T/2) -[1 -cos^2(T/2)]
    = 2cos^2(T/2) -1 ---***

    Substitute that into (1),
    L = INT.(0->2)[sqrt{2 +2cosT}]dT ---(1)
    L = INT.(0->2)[sqrt{2 +2(2cos^2(T/2) -1)}]dT
    L = INT.(0->2)[sqrt{2 +4cos^2(T/2) -2)}]dT
    L = INT.(0->2)[sqrt{4cos^2(T/2)}]dT
    L = INT.(0->2)[2cos(T/2)]dT

    Now we manipulate the "dT" outside.
    L = INT.(0->2)[2cos(T/2)](2dT/2)
    L = INT.(0->2)[2cos(T/2)](dT/2)(2)
    L = 2*2*INT.(0->2)[cos(T/2)](dT/2)
    L = 4*INT.(0->2)[cos(T/2)](dT/2)

    Now we can integrate,
    L = 4[sin(T/2)](0->2)
    L = 4[sin(2/2) -sin(0/2)]
    L = 4[sin(1) -sin(0)]
    L = 4[0.84147 -0]
    L = 3.366 ----answer.
    Last edited by ticbol; May 21st 2005 at 03:30 AM.
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  3. #3
    zee
    zee is offline
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    Thumbs up Ohh Thankyou!!!

    GOD BLESS YOU!!!

    IVE RACKED AND RACKED MY BRAINS AND
    YOUVE SOLVED IT LIKE A TRUE WIZ!!!

    THANKYOU TRULY SO MUCH!!!

    zee!
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