Constant circumference of tear drop

**shawsend** · October 3rd 2008, 01:57 PM

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 $n$ ?

**CaptainBlack** · October 4th 2008, 11:57 PM

Originally Posted by shawsend

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

**shawsend** · October 5th 2008, 07:04 AM

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.

**shawsend** · October 5th 2008, 11:35 AM

. . . wonderful. I get it now:

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

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