That equation again is: ((1/k)')^2 + T^2(1/k)^2 - T^2 = 0(adsbygoogle = window.adsbygoogle || []).push({});

where k is a function of s, T is a constant, and ' denotes differentiation with respect to s.

If you're wondering, it arises when you try to describe the family of curves with constant torsion that lie on the unit sphere. Here, T is the torsion, and k is the curvature. And boy do I wish I remembered my differential equations class a little better.

My attempt:

Let 1/k = R.

Then (R')^2 + T^2*R^2 -T^2 = 0

Suppose R = A*cos(Ts) + B*sin(Ts)

Plugging it in, it is found that A^2+B^2=1

So, R = cos(a)*sin(Ts)+sin(a)+cos(Ts)

So K = R^-1...

Which does not satisfy the equation :(

I also tried setting it up like this:

K' +Sqrt[T^2*K^2*(K^4-K^2)]

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# How to solve: ((1/k)')^2 + T^2(1/k)^2 - T^2 = 0

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