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

  • Thread starter Pwantar
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In summary, the conversation discussed a non-linear equation that arises when trying to describe the family of curves with constant torsion on the unit sphere. The equation involves functions of k and T, where k is the curvature and T is the torsion. Different approaches were discussed, including setting 1/k equal to R and using the characteristic equation. A solution was eventually found, but there were some errors in the process.
  • #1
Pwantar
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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)]
 
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  • #2
Hi Pwantar! :smile:
Pwantar said:
Then (R')^2 + T^2*R^2 -T^2 = 0

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

Nooo …

what are the roots of the characteristic equation? :wink:
 
  • #3
That's a non-linear equation. Neither Acos(Ts)+ Bsin(Ts), nor the concept of "characteristic equation" apply.
 
  • #4
oops!

HallsofIvy said:
That's a non-linear equation. Neither Acos(Ts)+ Bsin(Ts), nor the concept of "characteristic equation" apply.

oops! thanks, HallsofIvy! :redface:

(all those ^2s confusing me)

better write it R'/√(1 - R2) = ±T :smile:
 
  • #5
As I'm too lazy to recheck what I've written now that I'm already at this fancy text box thing that I don't understand, but love, I would like to point out an error that may or may not be present, as well as a possible solution.

((1/k)')^2 + T^2*(1/k)^2 - T^2 = 0

(-k'/k^2)^2 + T^2*(1/k)^2 - T^2 = 0

(k')^2/k^4 + T^2*(1/k)^2 - T^2 = 0

(k')^2 = T^2*k^2*(k^2-1)

dk/ds = T*k*Sqrt(k^2-1)

-atan(1/sqrt(k^2-1) = Ts + C, where C is an arbitrary constant

k = sqrt(1/tan(-Ts-C)^2 + 1)

k = 1/cos(Ts + C)

which happily satisfies my equation... the thing I don't like is taking the square root. Perhaps I should consider both the negative and positive sqrts... however, just looking at it, I don't think the solution will change.One more thing...
That Solution to the equation for R that I mentioned earlier actually does work... I guessed it based off of me forseeing a few sin^2 + cos^2 dealies cancelling out.

the knowledge that k=1/R does not appear to be sufficient information to find k given R... however, should a = Pi/2, it works, save for the phase angle C. It could be that I made a mistake somewhere. Perhaps I'll actually try to find my work for it, examine it, and notify you of anything interesting.Whoops, I made some mistakes in what I typed, but got the answer... I'll leave it as an excercise to you to figure out what they are
 
Last edited:

1. What is the first step in solving this equation?

The first step in solving this equation is to simplify the expression by expanding the square terms and combining like terms.

2. How do I isolate the variable "k" in this equation?

To isolate the variable "k", you will need to use inverse operations to move all terms containing "k" to one side of the equation and all constant terms to the other side.

3. Can this equation be solved without using calculus?

Yes, this equation can be solved without using calculus. You can use algebraic techniques to simplify the expression and isolate the variable "k".

4. What is the significance of the "T" value in this equation?

The "T" value represents a constant in the equation and does not affect the process of solving for "k". It is important to keep the "T" value in the equation until the final step.

5. How many solutions does this equation have?

This equation has two solutions for "k". However, it is important to check for any extraneous solutions that may arise during the solving process.

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