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(MATH321)[2005](f)midterm~ma_yxf^_10505.pdf
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Solutions for Midterm
Dr. Li Wei-Ping
October 27, 2005
1. Given the parametrized curve (25pts)

2
(s) = cos s, s, sin s
2
1.
i.) Show that the parameter s is the arclength parameter.

2.
ii.) Find the tangent vector .t(s), the normal vector .n(s) and the binormal vector .b(s).

3.
iii.) Find the curvature and torsion of the curve .

4.
iv.) Show that the tangent lines to make a constant angle with the y-axis.

5.
v.) Show that the normal lines of meet y-axis under a constant angle.


2.
Let S be a surface parameterized by X.(u, v)=(u, v, u 2 + v 2).

1.
i.) Compute the .rst fundamental form under the parameterization.

2.
ii.) Prove that all normal lines of the surface S intersect z-axis. (20pts)



3.
Let be a regular space curve. Assume the curvature and the torsion of are nowhere vanishing. Prove that all the normal lines of are parallel to a .xed plane if and



only if is a constant. (15pts)

1. Solution: 2
(i)
.(s)= (. sin s, 1, cos s), |.(s)| = 1. Thus s is the arc-length.
2


22

(ii)
.t(s)= .(s)= (. sin s, 1, cos s), .t.(s)= (. cos s, 0, . sin s).


22
.t.(s)
.n(s)= =(. cos s, 0, . sin s).
|.t.(s)|
1

2
.b(s)= .t(s) .n(s)= (. sin s, .1, cos s).
2
2 22
(iii) .b.(s)= . (. cos s, 0, . sin s)= .n(s). Thus (s) = and (s)= |.t.(s)| =
2 22 2
.
2
22
(iv) .t(s) (0, 1, 0)= (. sin s, 1, cos s) (0, 1, 0) = . Thus the angle between the
22 tangent line and the x-axis is /4, a constant.
(v) The parametric equation of the normal line is

2 222
(x, y, z) = (cos s, s, sin s)+ (. cos s, 0, . sin s) = (( . ) cos s, s, ( . ) sin s).
2 222

2
When = , the normal line intersects with the y-axis at (0, c, 0) for some number c.
2 Since (. cos s, 0, . sin s) (0, 1, 0) = 0, the normal line is perpendicular to the y-axis.
2. Solution:
(i) X.u = (1, 0, 2u), X.v = (0, 1, 2v)
22
E = X.u X.u =1+4u, F = X.u X.v =4uv, and G = X.v X.v =1+4v.
The .rst fundamental form is I(aX.u + bX.) = (1+4u2)a2 + 2(4uv)ab + (1 + 4v2)b2 .
(ii) The tangent planes have normal vectors X.u X. =(.2u, .2v, 1).
Normal line has the parametric equation

(x, y, z)=(u, v, u 2 + v 2)+ (.2u, .2v, 1) = (u(1 . 2),v(1 . 2),u 2 + v 2 + ).
1
Thus when = , the normal lines intersect the z-axis.
2
3. Solution: Assume normal lines of are parallel to a .xed plane with the normal vector .v. Since is a regular parametrized curve, let s be an arc length parameter of .
Then .n(s) .v = 0. Therefore .v lies on the plane spanned by .b(s) and .t(s) and we can write .v = a(s).t(s)+ c(s).b(s) where a(s) and b(s) are functions of s.
0= .v = a .(s).t(s)+ a(s).t.(s)+ c .(s).b(s)+ c(s).b.(s)
= a .(s).t(s)+ a(s)(s).n(s)+ c .(s).b(s)+ c(s)(s).n(s)
= a .(s).t(s)+(a(s)(s)+ c(s) (s)).n(s)+ c .(s).b(s).
Since .t(s), .n(s)