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Simplifying Trig Identities

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data

    The length of the curve r(t) = cos^3(t)j+sin^3(t)k, 0 =< t <= pi/2
    is

    2. Relevant equations

    AL in polar = ∫sqrt(r^2 + [dr/dθ]^2)

    3. The attempt at a solution

    I am having trouble simplifying the terms within the square root. What method should I use to deal with the pieces?

    r^2 = (cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2

    dr/dθ = 3(cos(t)^2)sin(t) + 3(sin(t)^2)cos(t)

    [dr/dθ]^2 = 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

    sqrt(cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2 + 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

    Simplified a bit:

    sqrt(1 + 4cos^3(t)sin^3(t) + 9 cos^4(t)sin^4(t) +9sin^4(t)cos^2(t))

    How would I further simplify from this?
     
  2. jcsd
  3. Jun 1, 2014 #2

    haruspex

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    There seem to be some inconsistencies in your notation.
    From the equation for r(t), I guess j and k are unit vectors, so r(t) is a vector. But your equation for arc length treats r as a scalar. This seems to have led to an error here:
    ##r^2 = (\cos^3(t))^2 + 2 \cos^3(t)\sin^3(t) + (\sin^3(t))^2##
     
  4. Jun 1, 2014 #3
    Oh, I see. Thanks.
     
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