课程名称︰几何学导论
课程性质︰数学系大三必修
课程教师︰崔茂培
开课学院：理学院
开课系所︰数学系
考试日期（年月日）︰2016/11/5
考试时限（分钟）：不限(最后改为最多300)
试题 :
Total = 105 points. Enjoy the exam!!!
(1) (15 pts) Suppose α(s) is a curve in R^3 parametrized by arc-length
with ||α''(s)|| > 0. Define T(s) = α'(s), k(s) = ||T'(s)||,
N(s) = T'(s)/||T'(s)||, B(s) = T(s) × N(s), τ(s) = N'(s)．B(s).
Prove that
dT/ds = k(s)N(s)
dN/ds = -k(s)T(s) +τ(s)B(s)
dB/ds = -τ(s)N(s)
(2) (20 pts) A curve is spherical if it lies on some sphere. Assume we
have a unit speed spherical curve satisfying
||γ(s)-c||^2 = R^2 > 0.
Suppose both curvature and torsion are nowhere vanishing.
(a) (10 pts) Show that
(γ-c)．T = 0,
(γ-c)．N = -1/κ,
(γ-c)．B = κ'/(τκ^2).
(b) (4 pts) Show that
1/κ^2 + (1/τ d/ds(1/κ))^2 = R^2.
(c) (3 pts) What kind of equation can one get by differentiating the
third equation
(γ-c)．B = κ'/(τκ^2).
obtained in (a)?
(d) (3 pts) Suppose β(s) is a unit speed spherical curve on the unit
sphere with nonzero curvature and nonzero torsion. Is it possible that
the curvature function of β(s) has the form k(s) = (2+sin(s))/2
(note that k(s) > 0)?
(3) (10 pts) We say that a unit speed curve α(s) is rectifying if, for
some fixed point p, α(s)-p is always contained in the plane spanned
by {T(s),B(s)}. Prove that a unit speed curve α(s) with k(s) > 0 is
rectifying if and only if τ(s)/κ(s) = as + b for constants a and b
with a≠0. (Hint: In one direction,
write α(s)-p = λ(s)T(s)+μ(s)B(s) and differentiate this equation.)
(4) (20 pts) Let I ⊂ R be an interval, and let α(u) = (0,f(u),g(u)),
u∈I, be a regular parametrized plane curve with f > 0 and
f'(u)^2 + g'(u)^2 = 1. Then the surface of revolution obtained by
rotating about the z-axis is parametrized by
X(u,v) = (f(u)cos v, f(u)sin v, g(u)), u∈I; 0 ≦ v ＜ 2π.
(a)(6 pts) Show that E = 1, F = 0, G = f(u)^2 and
l = f'(u)g''(u) - f''(u)g'(u), m = 0, n = f(u)g'(u).
(b)(4 pts) Show that the principal curvatures are
f'(u)g''(u) - f''(u)g'(u) and g'(u)/f(u).
(c)(4 pts) Show that the Gaussian curvature is K = -f''(u)/f(u).
(d)(2 pts) When is the Gaussian curvature positive or negative in
this case?
(e)(4 pts)
(i) Suppose a curve is given by γ(u) = (0, f(u), u) with f(u) > 0.
Recall that the arc-length of the curve is
u ＿＿＿＿_
s(u) = ∫ √1+f'(t)^2dt , and s'(u) = √1+f'(u)^2.
0
We can parametrize the curve γ by arc-length. Note that we have
f(u) = f(s(u)) and f'(u) = df/ds s'(u). Show that d^2/ds^2 f
has the same sign as f''(u).
(ii) If the graph of y = f(u) is given by the following picture.
and consider the surface of revolution X(u,v) = (f(u)cos v, f(u)sin v, u),
where is the region that the surface has positive (Gaussian) curvature,
negative curvature? Please explain your answer.
(5) (10 pts) Suppose we have a regular surface M parametrized by X(u,v)
with unit normal vector N. Let E, F, G and l, m, n be defined as usual
in the class, but suppose that F = 0 everywhere.
Prove that Ev Gu
Xuv = ── Xu + ── Xv + m n
2E 2G
(6) (20 pts) A surface is called umblic if the principal curvature is the
same everywhere, i.e. k_1(u,v) = k_2(u,v) everywhere on the surface.
(a) Show that if a surface is umblic then the principal curvatures
are indeed cosntants (indepedent of the point).
(b) Prove that a surface is umblic then it must be part of a plane
or part of a sphere.
(7) (10 points) Show that Gaussian curvature and the mean curvature of a
regular surface parametrized by X(u,v) satisfies
　　　　　　 (n_u ✕ n_v)．n (Xv ✕ n_u + n_v ✕ Xu)．n
K =　──────── and H = ─────────────
||Xu ✕ Xv|| 2||Xu ✕ Xv||
where
Xu ✕ Xv
n = ──────
||Xu ✕ Xv||
is the unit normal vector field on the surface.