1. ## some differential geometry

part b seems pretty confusing for me..

2. ## Re: some differential geometry

What are you finding confusing?

You are given an explicit expression for your smooth curve x, so you just need to compute the Frenet data as per the definitions.

You will end up with something like:

T_2=B_1
N_2=-N_1
B_2=T_1
k_2=t_1
t_2=k_1.

Where T,N,B,k,t denote the tangent, normal, binormal, curvature and torsion respectively.

(Carry out these calculations carefully yourself, there could easily be a small mistake in my answer. Haven't done this kind of stuff in a while.)

3. ## Re: some differential geometry

Originally Posted by seanieg89
What are you finding confusing?

You are given an explicit expression for your smooth curve x, so you just need to compute the Frenet data as per the definitions.

You will end up with something like:

T_2=B_1
N_2=-N_1
B_2=T_1
k_2=t_1
t_2=k_1.

Where T,N,B,k,t denote the tangent, normal, binormal, curvature and torsion respectively.

(Carry out these calculations carefully yourself, there could easily be a small mistake in my answer. Haven't done this kind of stuff in a while.)
well since the curve is parametrised by binormal instead of the usual arc length im used to, but otherwise its still the same?
Overall im just really bad at understanding pure expression I guess, I understand better with a numerical example

4. ## Re: some differential geometry

Fair enough, it is pretty much just application of the definitions:

$T_2(t)=x'(t)=B_1(t)$ and hence x moves at unit speed in the direction of the Binormal vector of r at time t.

To compute the curvature of x, as usual we differentiate T.

So $k_2(t)N_2(t)=T_2'(t)=B_1'(t)=-t_1(t)N_1(t)$ where $N_1$, and $N_2$ have unit length, and the first and last equalities follow from the definition of curvature and torsion respectively.

Hence $k_2(t)=|k_2(t)|=|-t_1(t)|=|t_1(t)|$ (recall that curvature of a curve is a positive quantity).

This also implies that $N_2(t)=-N_1(t)$ by normalisation.

Now that we know the tangent and normal vectors of x, we can compute its binormal.

$B_2=T_2\times N_2=-B_1\times N_1=T_1$

(the last equality can be justified in multiple ways, eg Jacobi's identity for the cross product or by noting that it is clearly a multiple of $T_1$ since it is orthogonal to both $B_1$ and $N_1$, and also that it has length one since $B_1,N_1$ are unit length and orthogonal. Then it is just a matter of checking signs.)

Hence $-t_2N_2=B_2'=T_1'=k_1N_1.$

Since we have already established $N_1=-N_2$, this implies $t_2=k_1$.

This completes the calculation.

5. ## Re: some differential geometry

mostly b and c

struggling quite alot with the geodesic curvature, answer and method dont seem to match up compared to my lecture notes...

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