I'm trying to prove the following:
If the tangent and binormal at a point of a curve make angles and φ respectively with a fixed direction, then prove that $\frac{\sin{\theta} \ d\theta}{\sin{\phi} \ d\phi} = -\frac{\kappa}{\tau}$ where $\kappa$ is the curvature, and $\tau$ is the torsion of the curve.
So far, I've tried simplifying it from both the RHS and the LHS.
First I drew this diagram:
To help you visualize, it's a downward opening parabola-like curve. It lies flat on the xy plane.
$\vec{X}$ is a unit vector pointing in the fixed direction.
My approach from the LHS was as follows:
- Draw a diagram with $\vec{T}$, $\vec{T} + \vec{dT}$, $\vec{X}$, and $\theta$. The angle between $\vec{T} + \vec{dT}$ and $\vec{X}$ is $\theta + d\theta$.
- $d\theta = (\theta + d\theta) - \theta$, so I find the angle between $\vec{T} + \vec{dT}$ and $\vec{X}$ using projections, hoping to end up with some kind of trigonometric function and then I can go from there to the RHS.
- Repeat same for $\phi$.
- Simplify fraction to RHS.
However, this didn't work out since I ended up with a huge $arccos$ subtraction, and I stopped there since I was clearly doing something wrong.
My approach from the RHS:
- Use relations between $\kappa$, $\tau$, $\vec{T}$, $\vec{B}$ and $\vec{N}$ to arrive at a simpler expression in terms of vectors.
- Somehow introduce $\theta$ and $\phi$ into the expression and take it from there to the LHS.
I simplified it to $$\frac{\left|\frac{d\vec{T}}{dt}\right|^2}{\frac{d\vec{B}}{dt} \cdot \frac{d\vec{T}}{dt}}$$
But that was about it. I'm not sure how I can introduce $\theta$ or $\phi$ into this.
How do I proceed?
Edit:
A friend suggested a much simpler idea: implicit differentiation.
We know,
$$\vec{T} \cdot \vec{X} = \cos \theta$$ $$\vec{B} \cdot \vec{X} = \cos \phi$$
Differentiating the two implicitly,
$$d\vec{T} \cdot \vec{X} = - \sin \theta d\theta$$ $$d\vec{B} \cdot \vec{X} = - \sin \phi d\phi$$
Simplifying LHS a little,
$$dT \cdot X \cdot \cos(d\vec{T}, \vec{X}) = - \sin \theta d\theta$$ $$dB \cdot X \cdot \cos(d\vec{B}, \vec{X}) = - \sin \phi d\phi$$
Now I want to divide these, but I don't know the angle $(d\vec{T}, \vec{X})$ and $(d\vec{B}, \vec{X})$. Did I go wrong somewhere?
This might help you. To be clear, I divided the dot product of the vectors, not the vector X.
https://i.stack.imgur.com/9gRo9.jpg