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.

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

y¶−1µfxx 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

B2C¶µcos 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×xv−c(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 R→0

area N(R)

area x(R),

curvature of yat y0= lim

diam R→0

area N(R)

area y(R).

Hence

curvature of y

curvature of x= lim

diam R→0

area x(R)

area y(R)= lim

diam R→0RR|xu×xv|dudv

RR|yu×yv|dudv =1

1−2cH +c2K.

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Aligarh Muslim University

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Computational Geometry

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18/02/2013