Math241 D1 Quiz#3 June 22, 2011
Name: Solution Keys
1.[7pts] Find a vector function that represents the curve of intersection of the two surfaces
x2+y2= 1, z =xy.
Sol) x= cos(t), y = sin(t), z = cos(t) sin(t),−→
r(t) =<cos(t),sin(t),cos(t) sin(t)>.
2.[5pts] Find the arc length of the curve. Set up an integral but do not finish computa-
tions. −→
r(t) =<cos(t),sin(t),ln(cos(t)) >, 0≤t≤π
4.
Sol)
−→
r0(t) =<−sin(t),cos(t),−sin(t)
cos(t)>(1)
|−→
r0(t)|=ssin(t)2+ cos(t)2+sin(t)2
cos(t)2=p1 + tan(t)2=|sec(t)|(2)
L=Zπ
4
0|−→
r0(t)|dt =Zπ
4
0
sec(t)dt. (3)
3.[8pts] Find the curvature at t= 0.
−→
r(t) =< t, −t, 1+2t2> .
Hint:
κ(t) = |−→
T0(t)|
|−→
r0(t)|=|−→
r0(t)×−→
r00(t)|
|−→
r0(t)|3.
Sol)
−→
r0(t) =<1,−1,4t > (4)
−→
r00(t) =<0,0,4>(5)
−→
r0(t)×−→
r00(t) = ¯¯¯¯¯¯
−→
i−→
j−→
k
1−1 4t
0 0 4 ¯¯¯¯¯¯
=<−4,−4,0>(6)
κ(0) = √16 + 16
√1 + 1 + 16t23=4√2
(2 + 16t2)3
2
.(7)
1