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.

dγ

ds =1

p1 + f02(1, f0)

d

dt

dγ

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.

dβ

dt =d2α

dt2=kn.

ds

dt =¯

¯

¯

¯

dβ

dt ¯

¯

¯

¯

=k.

dβ

ds =n.

d

dt

dβ

ds =dn

dt =−kt−τb.

d2β

ds2=−kt−τb

k=−t−τ

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), dϕnhas matrix

µ∂nθ

∂θ

∂nθ

∂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),

dϕn(0,1,0) = (0, n, 0), dϕ(0,0,1) = (0,0,1).

Hence

dϕn(0,1,1) = (0, n.1).

General second fundamental form questions.

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Subject:
Computational Geometry

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18/02/2013