- #1
Pwantar
- 5
- 0
That equation again is: ((1/k)')^2 + T^2(1/k)^2 - T^2 = 0
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)]
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)]