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Math Help - length of astroid curve

  1. #1
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    length of astroid curve

    I'm stuck because the answer I keep arriving at is negative under a square root sign. In the original question the variable is theta but for ease of typing I'm going to make it x.

    If f(x) = 8cos^3(x) and g(x) = 8sin^3(x) , Find the total length of the astroid described by f(x) and g(x) (f(x), g(x)) as x ranges from x=0 to x=2pi

    So I found that f'(x) = -24x^2(sin(x^3))
    and g'(x) = 24x^2(cos(x^3))

    I multiplied through to get:

    so int_0^2pi sqrt[ 576x^4(sin(x^6)) + 576x^4(cos(x^6))]
    simplified:
    24 int_0^2pi x^2 sqrt[sin(x^6) + cos(x^6)]
    integrated:
    24 int_0^2pi 1/3x^3 sqrt[-cos(x^6) + sin(x^6)]

    when I plug in 2pi for x, I get -.0436499839 under the square root sign.
    where did I go wrong?

    Thanks!
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  2. #2
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    I'm not sure what you have done with you derivatives because
    \frac{d}{dx}(8\cos^3x)=-24\sin x \cos^2 x
    but
    \frac{d}{dx}(8\cos (x^3))=-24x^2 \sin (x^3).
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  3. #3
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    I made the mistake of thinking sin^2(x) was the same thing as sin(x^2). oops.

    Let me try this again --

    int_0^2pi sqrt[(-24sin(x)cos^2(x))^2 + (24cos(x)sin^2(x))^2]

    I will multiply through (I'm not sure if I'm distributing the exponent correctly)

    int_0^2pi sqrt[576sin^2(x)cos^4(x) + 576cos^2(x)sin^4(x)]

    I can factor out the 576sin^2cos^2 and move out of the square root to get:

    24 int_0^2pi sin(x)cos(x) sqrt[cos^2(x) + sin^2(x)]

    I integrated to get:

    24 [-cos(x)sin(x) sqrt(sin^2(x) - cos^2(x))

    I plugged in 2pi and got the answer 0 (because of the sin(2pi)) and this is still the wrong answer. Is my problem with integrating cos^2(x) and sin^2(x)? I have very little idea how to integrate these. If anybody wants to help me out with that I'd be very grateful!!
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  4. #4
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    using symmetry ...

    S = 96 \int_0^{\frac{\pi}{2}} \sqrt{\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t}} \, dt

    note the following ...

    \cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t} =

    \cos^2{t}\sin^2{t}(cos^2{t} + \sin^2{t}) = \cos^2{t}\sin^2{t}

    S = 96 \int_0^{\frac{\pi}{2}} \sqrt{\cos^2{t}\sin^2{t}} \, dt

    S = 48 \int_0^{\frac{\pi}{2}} 2\cos{t}\sin{t} \, dt

    S = 48 \int_0^{\frac{\pi}{2}} \sin(2t) \, dt

    S = 48 \left[-\frac{\cos(2t)}{2}\right]_0^{\frac{\pi}{2}}

    S = 48 \left[\frac{1}{2} - \left(-\frac{1}{2}\right)\right] = 48
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  5. #5
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    Thank you very much for the answer!

    I haven't seen this use of symmetry before. Is there a specific formula? where did the 96 and pi/2 come from?


    Is this something that you just have memorized? or how is it found?


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  6. #6
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    look at the graph of the astroid ... it has the same shape and length in each quadrant.

    0 to pi/2 finds the arc length in quadrant I ... I just multiplied by 4 to get the entire length.


    as far as your 2nd question, I just factored and used the Pythagorean identity.
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