Asymptotic Lines and related Theorems








Let S:r=r(u,v) be a surface. Then curves on the surface whose directions are self-conjugate, is called asymptotic lines. The direction of asymptotic lines are called asymptotic direction.


Equation of Asymptotic lines

Let S:r=r(u,v) be a surface. If (du,dv) and (δu,δv) be conjugate direction, then
Lduδu+M(duδv+δudv)+Nδudv=0
In asymptotic line, the directions (du,dv) and (δu,δv) are self-conjugate.
or(du,dv)=(δu,δv)
Hence, differential equation of asymptotic line is
Ldu2+2Mdudv+Ndv2=0
This completes the proof.


Prove that normal curvature in a direction perpendicular to an asymptotic line is twice the mean curvature.

Solution
Let S:r=r(u,v) be a surface and C be an asymptotic line on it. Then, by Euler’s theorem, normal curvature for asymptotic line is
κacos2ψ+κbsin2ψ=0 [In asymptotic line , κn=0 (i)
Let, C1 be a normal section perpendicular to asymptotic line, then
κacos2(90+ψ)+κbsin2(90+ψ)=κn
orκasin2ψ+κbcos2ψ=κn (ii)
Adding (i) and (ii), we get
κn=κa+κb
orκn=2(κa+κb2)
orκn=2μ
This completes the solution.


Show that asymptotic lines of general surface of revolution is f11du2+uf1dv2=0

Solution
Let r=r(u,v) be general surface of revolution, then position of arbitrary point on the surface is
r=(ucosv,usinv,f(u))
Also, fundamental coefficients of the surface are
L=uf11H,M=0,N=u2f1H,H2=u2(1+f12)
Now, the differential equation of asymptotic lines is
Ldu2+2Mdudv+Ndv2=0
oruf11Hdu2+u2f1Hdv2=0
orf11du2+uf1dv2=0
This completes the proof.


Necessary and sufficient condition for r=r(s) on r=r(u,v) be asymptotic line is dNdr=0

Proof
Let r=r(u,v) be a surface then
dN=N1du+N2dv
dr=r1du+r2dv
Now, necessary and sufficient condition for the curve to be asymptotic is
Ldu2+2Mdudv+Ndv2=0
(N1du+N2dv)(r1du+r2dv)=0
dNdr=0
This completes the proof.


Condition for Asymptotic Lines to be Orthogonal

Let r=r(u,v) be a surface and C be an asymptotic line on it. Then equation of asymptotic line is
Ldu2+2Mdudv+Ndv2=0 (i)
Also, equation of double family of curves on the surface is
Pdu2+2Qdudv+Rdv2=0 (ii)
Comparing (i) with (ii), we get
P=L,Q=M,R=N
Now, condition for the asymptotic line to be orthogonal is
ER2FQ+GP=0
orEN2FM+GL=0
This completes the proof.


Show that asymptotic directions are orthogonal if and only if the surface is minimal.

Solution
Let r=r(u,v) be a surface. Assume that the surface is minimal. Then,
μ=0
orEN2FM+GL=0
This shows that, asymptotic directions are orthogonal.


The necessary and sufficient condition for parametric curves to be asymptotic lines is L=N=0,M0

Proof
Let r=r(u,v) be a surface and C be an asymptotic line on it. Then differential equation of asymptotic line is
Ldu2+2Mdudv+Ndv2=0 (i)
Also, equation of parametric curves on the surface is
dudv=0 (ii)
Now, necessary and sufficient condition for parametric curves to be asymptotic lines is, (i) and (ii) must be identical
L=0,M0,N=0
This completes the proof.


Show that parametric curves of right helicoids are asymptotic lines.

Solution
The position of arbitrary point on the right helicoids is
r=(ucosv,usinv,cv) (i)
Then, the fundamental coefficients of the surface is
E=1,F=0,G=u2+c2,L=0,M=cu2+c2,N=0
In this case, we have
L=0,M=cH,N=0
Hence, on right helicoids, the parametric curves are asymptotic lines.


Show that osculating plane on asymptotic line is tangent plane to the surface.

Proof
Let r=r(u,v)be a surface and C be an asymptotic line on it.
Then
equation of osculating plane on asymptotic line is
(Rr)b=0 (i)
Also, equation of tangent plane to the surface is
(Rr)N=0 (ii)
Here
N.t=0 (iii)
Differentiating of (i) w. r. to. s, we get
dNds.t+N.κN=0
or dNds.drds+N.κN=0
In asymptotic line
dNds.drds=0
Thus, we have
dNds.drds+N.κN=0
or N.κN=0
or N.N=0 (iv)
From (iii) and (iv), we can write
N=b
Thus, (i) and (ii), are same.
This completes the proof.


Show that all straight lines on a surface are asymptotic lines.

Solution
Let r=r(u,v) be a surface and C be a curve on it.
Then, C is trraight line if and only if
κ=0 (i)
Also
Nt=0 (ii)
Differentiating of (ii) w. r. to. s, we get
dNds.t+N.κN=0
or dNds.drds+N.κN=0
In straight line
κ=0
Thus, we have
dNds.drds+N.κN=0
or dNds.drds=0
or dN.dr=0
or (N1du+N2dv).(r1du+r2dv)=0
or Ldu2+2Mdudv+Ndv2=0
This is equation of asymptotic line.
Thus, all straight lines on a surface are asymptotic lines.


Curvature and Torsions of Asymptotic Lines

Let r=r(u,v) be a surface and r=r(s) be asymptotic line on it.
Then expression of curvature for asymptotic line C is
t=κN
Operating dot product on both side by N, we get
κ=t.N
or κ=t.(b×t)
or κ=[t,b,t]
or κ=[t,t,b]
For asymptotic line, we have
N=b.v Thus, the curvature of asymptotic line is,
κ=[t,t,N]
Next, the torsion of asymptotic line C is
b=τN
Operating dot product on both side by N, we get
τ=b.N
or τ=b.(b×t)
or τ=[b,b,t]
or τ=[b,b,t]
For asymptotic line, we have
N=b.
Thus, torsion of asymptotic line is,
τ=[N,N,t]
Hence, curvature and torsion of asymptotic lines are
κ=[t,t,N] and τ=[N,N,t]



Theorems of Beltrami and Ennper

Torsion of asymptotic line is τ=±K, where K is the Gaussian Curvature.
Proof
Let r=r(u,v) be a surface and r=r(s) be asymptotic line on it.
Now, torsion of C is
τ=[N,N,t]
or τ=[N,dN,dr]1ds2
or τ={N.(N1du+N2dv)×(r1du+r2dv)}1ds2
or τ={[N,N1,r1]du2+[N,N1,r2]dudv+[N,N2,r1]dudv+[N,N2,r2]dv2}1ds2
or τ=[N,N1,r1](duds)2+[N,N1,r2]dudsdvds+[N,N2,r1]dudv+[N,N2,r2](dvds)2
or τ=EMFLH(duds)2+(FMGLH+ENFMH)dudsdvds+FNGMH(dvds)2
Without loss of generality, we take asymptotic line along parametric curve, then
L=0,N=0,M0
Thus
K=LNM2H2=M2H2
or M2H2=K
or MH=K (A)
Next, torsion of asymptotic line is
τ=EMH(duds)2+GMH(dvds)2=MH(E(duds)2G(dvds)2)
  1. Case 1: For asymptotic line along v curve, we have du=0
    Thus
    τ=MHG(dvds)2
    Here

    Edu2+2Fdudv+Gdv2=ds2
    or Gdv2=ds2
    or G(dvds)2=1
    Thus
    τ=MH (B)
  2. Case 2: For asymptotic line along u curve, we have dv=0
    Thus
    τ=MHE(duds)2
    Here
    Edu2+2Fdudv+Gdv2=ds2
    or Edu2=ds2
    or E(duds)2=1
    Thus
    τ=MH (C)
Combining both, we get
τ=±MH (D)
Using (A), we get,
τ=±K
This completes the proof.

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