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) =
Typeset by A
M
S-T
E
X
1
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.
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