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Math Help - Constant circumference of tear drop

  1. #1
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    Constant circumference of tear drop

    A tear drop can be represented by the parametric equations:

    x(t)=r \cos(t)

    y(t)=r \sin(t)\sin^{n-1}(t/2)

    The arc length is:

    A(n)=\int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt

    Find r(n) such that A(n)=K. That is, how must I adjust the value of r so that the arc length remains constant as I vary the parameter n?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shawsend View Post
    A tear drop can be represented by the parametric equations:

    x(t)=r \cos(t)

    y(t)=r \sin(t)\sin^{n-1}(t/2)

    The arc length is:

    A(n)=\int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt

    Find r(n) such that A(n)=K. That is, how must I adjust the value of r so that the arc length remains constant as I vary the parameter n?
    Your definition of A(n) is independent of n. Do you mean find r such that A=K?

    The way A is defined it is proportional to r so if you know A for the case r=1 it is simple to find A for any other value of r, or find the r corresponding to a desired value of A..


    RonL
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  3. #3
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    I don't understand then. I thought the arc length is a function of n. When I simplify the radical I get (Mathematica-generated latex):

    \surd \left(r^2 \text{Sin}[t]^2+\left(r \text{Cos}[t] \text{Sin}\left[\frac{t}{2}\right]^{-1+n}+\frac{1}{2} (-1+n) r \text{Cos}\left[\frac{t}{2}\right] \text{Sin}\left[\frac{t}{2}\right]^{-2+n} \text{Sin}[t]\right)^2\right)

    I'll work on it today.
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  4. #4
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    . . . wonderful. I get it now:


    r(n)=\frac{K}{\int_0^{2\pi} f(n,t)dt}
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