How Can I Express γ(s) Using Frenet Types t(s), n(s), and b(s)?

[ParisSpart]

even γ: I-> R ^ 2 curve parameterized as to arc length (single speed) with curvature k (s)> 0 and torsion τ(s)> 0. I want to write the γ(s) as a combination of n(s), t(s), b(s). these are the types of Frenet.


the only thing i know is that the types of Frenet are t(s)=γ'(s) , b(s)=t(s)Xn(s) and n(s)=t'(s)/k(s)
but how i will write γ(s) with this types? is there any way or type to do this?

 

[Quesadilla]

Maybe it could help if you told us what you are trying to do. Are you trying to solve a specific problem? Also, did you intend to write $\gamma : I \rightarrow \mathbb{R}^2$, as in a planar curve?

 

[ParisSpart]

my problem is:even γ: I-> R ^3 curve parameterized as to arc length (single speed) with curvature k (s)> 0 and torsion τ(s)>0. we assume that the γ is at the surface sphere with center the origin. Show that for any s we have:
γ(s)=-(1/k(s))*n(s) + (k'(s)/(k^2(s)*τ(s)))*b(s)

and he gives us a hint: write the γ(s) as a combination of n(s), t(s), b(s). these are the types of Frenet
and γ:I->R^3

 

[ParisSpart]

i think that if i find a way to combine γ with the types of frenet and derivative them i will find the problem

 

[Quesadilla]

You are meant to write $\gamma(s) = a_1(s)t(s) + a_2(s)n(s) + a_3(s)b(s)$, for unknown scalar functions $a_i(s)$. You can immediately conclude that one of the $a_i(s)$ is zero identically. Which one and why? The other two can be determined by differentiating the remaining expression.

 

[ParisSpart]

a1(s) will be zero? and after that with differentiation i will conclude to the the expression i wrote?

 

[Quesadilla]

Don't ask. Try it! And if you run into more trouble, come back here, show us what you have done and we will give you more help.

 

[ParisSpart]

i tried to the derivative and i found this :
γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) , we know that γ'(s)=0 beacuse we are on the surface of sphere , after that i tried to replace n'(s)=-k(s)t(s)+τ(s)b(s) and the other types?

 

[micromass]

a1(s) will be zero?

Why will $a_1(s)$ be zero?

i tried to the derivative and i found this :
γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) , we know that γ'(s)=0 beacuse we are on the surface of sphere

Why would $\gamma^\prime(s) = 0$? I don't see what the surface of the sphere has to do with this.

Do you know the Frenet formulas?

 

[ParisSpart]

because γ(s)*γ(s)=c where c is a constant and then 2γ(s)γ'(s)=0

 

[micromass]

because γ(s)*γ(s)=c where c is a constant and then 2γ(s)γ'(s)=0

OK, how does that answer my two questions?

 

[Quesadilla]

$\gamma'(s) = 0$ identically would mean that there is no motion and the curve consists of just one point. Clearly this is not true in our case, since we are allowed to move around as long as we stay on the surface of the sphere.

 

[ParisSpart]

i think that a1(s)=0 because i supposed that for γ'(s)=0 we know that t(s)=γ'(s) and then t(s)=0 and a1(s)=0

 

[ParisSpart]

but if we don't have γ'(s)=0 if we differentiate the γ(s)=a1(s)t(s)+a2(s)n(s)+a3(s)b(s) with what γ'(s) will be equal?

 

[Quesadilla]

i think that a1(s)=0 because i supposed that for γ'(s)=0 we know that t(s)=γ'(s) and then t(s)=0 and a1(s)=0

That is not correct and it is not the reason $a_1(s)$ is zero. You wrote above that $\gamma(s) \cdot \gamma'(s) = 0$, which is correct. Do you see how you could use this to conclude that $a_1(s)$ must be zero?

 

[ParisSpart]

no... can you tell me why? i focused on this that i said prior

 

[Quesadilla]

no... can you tell me why? i focused on this that i said prior

So $\gamma(s) = a_1(s)t(s) + a_2(s)n(s) + a_3(s)b(s)$ and $\gamma'(s) \cdot \gamma(s) = 0$, where $\gamma'(s) = t(s)$ by definition.

By the way, I hope you understand that $\gamma(s)$ and $\gamma'(s)$ are vector-valued: $\gamma(s) = (\gamma_1(s), \gamma_2(s), \gamma_3(s)), \, \gamma'(s) = (\gamma'_1(s), \gamma'_2(s), \gamma'_3(s))$ and that by $\gamma(s) \cdot \gamma'(s)$ I mean the scalar product of these vectors.

 

[ParisSpart]

if i differentiate i find this γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) beacuse γ'(s)=t(s) i will replace this in this equation? and after that i am trying to replace the followings:
b(s)=t(s)xn(s) , n(s)=t'(s)/k(s), t'(s)=k(s)n(s), n'(s)=-k(s)t(s)+τ(s)b(s), b'(s)=-τ(s)n(s) ?

 

[Quesadilla]

if i differentiate i find this γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) beacuse γ'(s)=t(s) i will replace this in this equation? and after that i am trying to replace the followings:
b(s)=t(s)xn(s) , n(s)=t'(s)/k(s), t'(s)=k(s)n(s), n'(s)=-k(s)t(s)+τ(s)b(s), b'(s)=-τ(s)n(s) ?

Use the last two of these and then scalar multiply the expression $\gamma'(s) = ...$ with some suitable vector.

 

[ParisSpart]

i replaced the n'(s) and b'(s) but why we must multiply with a suitable vector , for what reason? i am a little confused

 

[Quesadilla]

i replaced the n'(s) and b'(s) but why we must multiply with a suitable vector , for what reason? i am a little confused

Well, you want to determine $a_2(s)$ and $a_3(s)$, no? I suggested doing scalar multiplication with some well-chosen vector(s) would enable you to do that.

 

[ParisSpart]

i have this
γ'(s)=-a2(s)k(s)t(s)+a2(s)τ(s)b(s)+a2'(s)n(s)+a3'(s)b(s)-a3(s)τ(s)n(s)

we don't know γ'(s) i think that its difficult to find a2(s) and a3 .

 

[Quesadilla]

i have this
γ'(s)=-a2(s)k(s)t(s)+a2(s)τ(s)b(s)+a2'(s)n(s)+a3'(s)b(s)-a3(s)τ(s)n(s)

we don't know γ'(s) i think that its difficult to find a2(s) and a3 .

I literally just told you to scalar multiply both sides with some well-chosen vector(s)! HINT: Try each of $t(s), n(s), b(s)$.

 

[ParisSpart]

with γ(s)=t'(s) and by scalar multiplying i found a2(s)=-1/k(s) a2'(s)=k'(s)/k^2(s) and (k'(s)/k^2(s))-a3(s)τ(s)=0
where a3(s)=k'(s)/k^2(s)*τ(s) and i concluded to this that we wanted