Solutions
147a Winter 2012
Homework 4
3.2 Assume that is paramaterized by arclength, then t = ,
n = t + b,
and since and are constant
n = (2 + 2 )n.
This implies that
n = a1 cos(s) + a2 sin(s),
where = 2 + 2 and a1 and a2 are some constant vectors. Since
1.1
1.2
1.4
1.7
Solutions
147a Winter 2012
Homework 1
2 4
(t , t ) is not a paramaterization of y = x2 as it misses the left half of the parabola.
Many dierent paramaterizations are possible.
i. (tan(t), sec(t)
t (/2, /2), t (/2, 3/2)
ii. (2 cos(t), 3 sin
Solutions
Math 147A
Winter 2012
Homework 6
2.7 This surface is the collection of points that are a distance a from the curve .
By the Frenet-Serret equations
s = t + a(t + b) cos n sin ),
= (1 a cos )t a sin n + a cos b.
The theta derivative can be compu
tary Differential Geemetry
surfaces) using, nieth~
on such as 7(t) (as in
ate.
3y t), etc
5 71:721 "'73 Of? is
g/dt3, . exist, for i:
led in this book will 7be
it is cailecl the tangent
2 vector
y(t ~i~ 515) of the image C
+ (it)
1.65 parallel to the tenv
Solutions
147a Winter 2012
Homework 3
2.1 Since ns is a unit vector, ns ns = 1 and 2ns ns = 0. Since is a plane curve, it
follows that ns = at. Taking the derivative of t ns = 0 gives the equation
t ns + t ns = s ns ns + at t = s + a = 0.
3.1
This shows t
Solutions
Math 147A
Winter 2012
Homework 5
2.1 Since u, v, and f (u, v) are all smooth, the surface patch
(u, v) = (u, v, f (u, v),
is smooth. To check regularity compute
u = (1, 0, fu (u, v),
v = (0, 1, fv (u, v),
and
u v = (fu , fv , 1),
which is nev