MATH 251 - QUIZ 2
NAME:
I.D.:
Instruction: Circle your answers and show all your work CLEARLY. Solutions with answer
only and without supporting procedures will have little credit.
1. (Exercise 5 on Page 770) Given a curve with parametric equations x= 3tsin t, y = 3tcos tand
z= 2t2, find the arc length from t= 0 to t= 4/5.
Solution Compute the derivatives to get x0(t) = 3 sin t+ 3tcos t,y0(t) = 3 cos t−3tsin tand
z0(t) = 4t. Thus the arc length Sis
S=Z4/5
0q(3 sin t+ 3tcos t)2+ (3 cos t−3tsin t)2+ (4t)2dt by binomial formula
=Z4/5
0q9 sin2t+ 9t2cos2t+ 9 cos2t+ 9t2sin2t+ 16t2dt apply sin2t+ cos2t= 1
=Z4/5
0p9 + 25t2dt set t=3
5tan θ
=9
5Ztan−1(4/3)
0
sec3θdθ integration by parts
=9
51
2"√9 + 25t2
3
5t
3+ ln √9 + 25t2
3+5t
3!#4/5
0
use sec θ=√9 + 25t2
3and tan θ=5t
3
=9
10 20
9+ ln 3= 2 + 9
10 ln 3.
2. Given a curve r(t) = (t, t2, t3), find the unit tangent vector and unit normal vector of the curve
at (1,1,1).
Solution (Step 1) When t= 1, r(1) = (1,1,1). v= (1,2t, 3t2) and v(t) = √1+4t2+ 9t4.
(Step 2) At (1,1,1), the unit tangent vector is T(1) = (1/√14,2/√14,3/√14).
(Step 3) Note that a(t) = (0,2,6t). At t= 1, a(1) = (0,2,6), and so aT=a·v
v=0+4+18
√14 =22
√14 .
(Step 4) At t= 1, v×a= (1,2,3) ×(0,2,6) = (12 −6,−6,−2). Thus at t= 1,
aN=|v×a|
v=√36+36+4
√14 =r76
14.
It follows by N= (a−aTT)/aNthat
N=√14
√76 (0,2,6) −22
√14(1/√14,2/√14,3/√14)
=√14
√76 (0,2,6) −11
7,22
7,33
7
=√14
√76 −11
7,−8
7,9
7.