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SOLUTIONS TO FINAL EXAM

QUESTIONS FROM PREVIOUS YEARS.

Curves.

1. Let *s *be the arclength parameter.

*γ′*(*t*) = (1*, f ′*)*.
ds
*

*dt
*= *|γ′*(*t*)*| *=

√
1 + *f ′*2*.
*

*dγ
*

*ds
*=

1√
1 + *f ′*2

(1*, f ′*)

*d
*

*dt
*

*dγ
*

*ds
*=

1√
1 + *f ′*2

(0*, f ′′*) *− f
′f ′′
*

(1 + *f ′*2)3*/*2
(1*, f ′*)

= 1

(1 + *f ′*2)3*/*2
(*−f ′f ′′, f ′′*)

*d*2*γ
*

*ds*2
=

*f ′′
*

(1 + *f ′*2)2
(*−f ′, *1)

curvature =
∣∣∣∣
*d*2*γ
*

*ds*2

∣∣∣∣ =
*f ′′
*

(1 + *f ′*2)3*/*2

Curvature of the parabola *y *= *kx*2 is

2*k
*(1 + 4*k*2*x*2)3*/*2

*.
*

2. Let *t *be the arclength parameter for *α *and let *s *be the arclength parameter for *β*(*t*) =

Typeset by *AMS*-TEX
1

2 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

*α′*(*t*). Let t*,*n*,*b be the Frenet frame for *α*.

*dβ
*

*dt
*=

*d*2*α
*

*dt*2
= *k*n*.
*

*ds
*

*dt
*=

∣∣∣∣
*dβ
*

*dt
*

∣∣∣∣ = *k.
dβ
*

*ds
*= n*.
*

*d
*

*dt
*

*dβ
*

*ds
*=

*d*n
*dt
*

= *−k*t*− τ*b*.
d*2*β
*

*ds*2
=
*−k*t*− τ*b

*k
*= *−*t*− τ
*

*k
*b*.
*

curvature of *β *=
∣∣∣∣
*d*2*β
*

*ds*2

∣∣∣∣ = √

1 +
*τ*2

*k*2
*.
*

Tangent space and derivatives of maps.

1. There are misprints in this question. Let us assume that the cylinder is *x*2 + *y*2 = 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), *dϕn *has matrix

(
*∂nθ
∂θ
*

*∂nθ
∂z
*

*∂z
∂θ
*

*∂z
∂z
*

) =

(
*n *0
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 (
*fxx 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 (

*e f
f g
*

) =

(
2*A B
B *2*C
*

)
*.
*

The matrix of the derivative of the Gauss map is

*−
*(

*E F
F G
*

)*−*1 (
*e f
f g
*

)
= *−
*

(
1 + *f*2*x fxfy
fxfy *1 + *f*2*y
*

)*−*1 (
*fxx fxy
fxy fyy
*

)
*,
*

which gives

*K *=
*fxxfyy − f*2*xy
*1 + *f*2*x *+ *f*2*y
*

*, H *=
(1 + *f*2*y *)*fxx *+ (1 + *f
*

2
*x*)*fyy − *2*fxfyfxy
*

2(1 + *f*2*x *+ *f*2*y *)
*.
*

In our case,
*K *= 4*AC −B*2*, H *= 2*A *+ 2*C.
*

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*)
(

2*A B
B *2*C
*

)(
cos *θ
*sin *θ
*

)
= 2*A *cos2 *θ *+ 2*B *cos *θ *sin *θ *+ 2*C *sin2 *θ.
*

2.
elliptic points = *{*q : *K*(q) *∈ *(0*,∞*)*}.*

4 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

Since *K *is a continuous (indeed a smooth) function on *S *and (0*,∞*) is open, the set of
elliptic points of *S *is 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 *= *x*4 + *y*2

is parabolic at (0*, *0*, *0) and elliptic elsewhere, while the graph

*z *= *x*4 + *y*4

is planar at (0*, *0*, *0) and elliptic elsewhere.

Gauss curvature questions.

1. (a).

y*u *= x*u *+ *c*N*u ⇒ 〈*y*u,*N*〉 *= *〈*x*u,*N*〉*+ *c〈*N*u,*N*〉 *= *〈*N*u,*N*〉 *= 12
*∂
*

*∂u
〈*N*,*N*〉 *= 0*.
*

Similarly, N is perpendicular to y*v*, and hence is normal to the surface parameterized by y.
(b). Since x*u *and x*v *are lines of curvature, we have

N*u *= *µ*1x*u, *N*v *= *µ*2x*v
*

for smooth scalar functions *µ*1*, µ*1 (we don’t specify which is bigger.)

y*u × *y*v *= (x*u *+ *c*N*u*)*× *(x*v *+ *c*N*v*) = x*u × *x*v *+ *c*x*u ×*N*v *+ *c*N*u × *x*v *+ *c*2N*u ×*N*v
*= x*u × *x*v − c*(*µ*1 + *µ*2)x*u × *x*v *+ *c*2*µ*1*µ*2x*u × *x*v *= (1*− *2*cH *+ *c*2*K*)x*u *+ x*v.
*

(c). We think of N as being a function of (*u, v*). Since x and y have the same normal,
N gives the Gauss map for either surface. Consider (*u*0*, v*0) and set x(*u*0*, v*0) = x0 and
y(*u*0*, v*0) = y0. Consider a small disc *R *around (*u*0*, v*0). Then

curvature of x at x0 = lim
diam *R→*0

areaN(*R*)
areax(*R*)

*,
*

curvature of y at y0 = lim
diam *R→*0

areaN(*R*)
areay(*R*)

*.
*

Hence

curvature of y curvature of x

= lim
diam *R→*0

areax(*R*)
areay(*R*)

= lim
diam *R→*0

∫
*R
|*x*u × *x*v| dudv*∫

*R
|*y*u × *y*v| dudv *=

1
1*− *2*cH *+ *c*2*K .*

SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS. 5

2. *K *= 1: the sphere *x*2 + *y*2 + *z*2 = 1. *K *= 0: the plane *z *= 0. *K *= *−*1: the surface of
revolution obtained by revolving (*e−s, *0*,
*

∫ *s
*0

*√
*1*− e−*2*s ds*), for *s > *0, about the *z*-axis.

We derived the formula for the curvature surface obtained by revolving the curve (*φ*(*s*)*, *0*, ψ*(*s*))
around the *z*-axis in the case when *s *is the arclength - that is *φ′*2 + *ψ′*2 = 1. The Gauss
curvature is

*−φ
′′
*

*φ
.
*

In particular it is zero precisely when *φ′′ *= 0. This implies that

*φ*(*s*) = *as *+ *b, ψ*(*s*) =
√

1*− a*2*s *+ *c.
*

This means that the curve is a straight line and the surface of revolution is a cone or cylinder.

3.
x*s *= *α′*(*s*) + *u*b*′*(*s*) = t(*s*) + *uτ*n(*s*)*, *x*u *= b(*s*)*,
*

so

*E *= 1 + *u*2*τ*2

*F *= 0
*G *= 1

Furthermore

x*s × *x*u *= (t + *uτ*n)*× *b = *−*n + *uτ*t*, *N = *−*n + *uτ*t*√
*1 + *u*2*τ*2

*.
*

*g *= *〈*x*uu,*N*〉 *= *〈*0*,*N*〉 *= 0*.
f *= *〈*x*uv,*N*〉 *= *〈τ*n*, −*n + *uτ*t*√
*

1 + *u*2*τ*2
*〉 *= *−τ√
*

1 + *u*2*τ*2
*.
*

*K *=
*eg − f*2

*EG− F *2 = *−
τ*2

(1 + *u*2*τ*2)2
*.
*

Isometric surfaces.

1. Done in class.

2. Consider the parameterizations x(*u, v*) and

x̄(u, *v*) = r(*u, v *+ *d*) = (*f*(*v *+ *d*) cos *u, f*(*v *+ *d*) sin *u, g*(*v *+ *d*))*.*

6 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

Then
r̄ *◦ *r*−*1

is the map (*u, v*) *→ *(*u, v *+ *d*). This is an isometry if and only if *E *= *Ē, F *= *F̄ *, *G *= *Ḡ.
*But

(
*E F
F G
*

) =

(
*f*2(*v*) 0

0 (*f ′*(*v*))2 + (*g′*(*v*))2

)
*,
*

(
*Ē F̄
F̄ Ḡ
*

) =

(
*f*2(*v *+ *d*) 0

0 (*f ′*(*v *+ *d*))2 + (*g′*(*v *+ *d*))2

)
*.
*

Equating *E *and *Ē *we get *f*(*v *+ *d*) = *±f*(*v*) for all *d*. Assuming *f *is not identically zero,
this gives that *f *is constant and the surface is a cylinder.