# Differentiable Function - Differential Geometry - Solved Exam, Exams for Computational Geometry. Aligarh Muslim University

## Computational Geometry

Description: This is the Solved Exam of Differential Geometry which includes Normal Vector, Normal Vector, Binormal Vector, Curvature, Torsion, Binormal Vector, Speed Space Curve etc. Key important points are: Differentiable Function, Plane Curve, Curvature, Parabolas, Parametrized, Arclength, Curvature, Tangent Space, Derivatives of Maps, General Second Fundamental
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SOLUTIONS TO FINAL EXAM
QUESTIONS FROM PREVIOUS YEARS.
Curves.
1. Let sbe the arclength parameter.
γ0(t) = (1, f0).
ds
dt =|γ0(t)|=q1 + f02.
ds =1
p1 + f02(1, f0)
d
dt
ds =1
p1 + f02(0, f00 )f0f00
(1 + f02)3/2(1, f0)
=1
(1 + f02)3/2(f0f00 , f00 )
d2γ
ds2=f00
(1 + f02)2(f0,1)
curvature = ¯
¯
¯
¯
d2γ
ds2¯
¯
¯
¯
=f00
(1 + f02)3/2
Curvature of the parabola y=kx2is
2k
(1 + 4k2x2)3/2.
2. Let tbe the arclength parameter for αand let sbe the arclength parameter for β(t) =
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2 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.
α0(t). Let t,n,bbe the Frenet frame for α.
dt =d2α
dt2=kn.
ds
dt =¯
¯
¯
¯
dt ¯
¯
¯
¯
=k.
ds =n.
d
dt
ds =dn
dt =ktτb.
d2β
ds2=ktτb
k=tτ
kb.
curvature of β=¯
¯
¯
¯
d2β
ds2¯
¯
¯
¯
=r1 + τ2
k2.
Tangent space and derivatives of maps.
1. There are misprints in this question. Let us assume that the cylinder is x2+y2= 1 and
that the tangent vector in T(1,0,0) is (0,1,1). In the given coordinates (a coordinate map is
just the inverse of a parameterization), nhas matrix
µ
θ
z
z
θ
z
z =µn0
0 1 .
Now (1,0,0) corresponds to (θ, z) = (0,0). We see that ϕn(1,0,0) = (1,0,0). The basis for
the tangent space associated to the coordinates is
(x, y, z)
θ = (sin θ, cos θ, 0),(x, y, z)
z = (0,0,1).
At (θ, z) = (0,0) this basis is
(0,1,0),(0,0,1).
Then at (1,0,0),
n(0,1,0) = (0, n, 0), dϕ(0,0,1) = (0,0,1).
Hence
n(0,1,1) = (0, n.1).
General second fundamental form questions.
SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS. 3
1. Following Do Carmo, for the graph z=f(x, y), the matrix of the second fundamental
form with respect to the basis
(1,0, fx),(0,1, fy),
is µfxx fxy
fxy fyy .
In our case, the basis at (0,0,0) is
(1,0,0),(0,1,0)
and the matrix of the second fundamental form is
µe f
f g =µ2A B
B2C.
The matrix of the derivative of the Gauss map is
µE F
F G 1µe f
f g =µ1 + f2
xfxfy
fxfy1 + f2
y1µfxx fxy
fxy fyy ,
which gives
K=fxxfyy f2
xy
1 + f2
x+f2
y
, H =(1 + f2
y)fxx + (1 + f2
x)fyy 2fxfyfxy
2(1 + f2
x+f2
y).
In our case,
K= 4AC B2, H = 2A+ 2C.
The unit vector in direction (x, y, 0) is (cos θ, sin θ, 0) = cos θ(1,0,0) + sin θ(0,1,0). The
normal curvature in this direction is
(cos θ, sin θ)µ2A B
B2Cµcos θ
sin θ= 2Acos2θ+ 2Bcos θsin θ+ 2Csin2θ.
2.
elliptic points = {q:K(q)(0,)}.
4 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.
Since Kis a continuous (indeed a smooth) function on Sand (0,) is open, the set of
elliptic points of Sis open, and is hence an open neighborhood of each of its points. The
same is true for the hyperbolic points, but not for parabolic or planar points. For example,
the graph
z=x4+y2
is parabolic at (0,0,0) and elliptic elsewhere, while the graph
z=x4+y4
is planar at (0,0,0) and elliptic elsewhere.
Gauss curvature questions.
1. (a).
yu=xu+cNu⇒ hyu,Ni=hxu,Ni+chNu,Ni=hNu,Ni=1
2
u hN,Ni= 0.
Similarly, Nis perpendicular to yv, and hence is normal to the surface parameterized by y.
(b). Since xuand xvare lines of curvature, we have
Nu=µ1xu,Nv=µ2xv
for smooth scalar functions µ1, µ1(we don’t specify which is bigger.)
yu×yv= (xu+cNu)×(xv+cNv) = xu×xv+cxu×Nv+cNu×xv+c2Nu×Nv
=xu×xvc(µ1+µ2)xu×xv+c2µ1µ2xu×xv= (1 2cH +c2K)xu+xv.
(c). We think of Nas being a function of (u, v). Since xand yhave the same normal,
Ngives the Gauss map for either surface. Consider (u0, v0) and set x(u0, v0) = x0and
y(u0, v0) = y0. Consider a small disc Raround (u0, v0). Then
curvature of xat x0= lim
diam R0
area N(R)
area x(R),
curvature of yat y0= lim
diam R0
area N(R)
area y(R).
Hence
curvature of y
curvature of x= lim
diam R0
area x(R)
area y(R)= lim
diam R0RR|xu×xv|dudv
RR|yu×yv|dudv =1
12cH +c2K.
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