Solutions homework 14.
Problem page 190: 1 Solution:
ZF·dR=Z1
0
t2+ 4t+ 6t3·3dt =41
6.
Problem Page 190: 2 Solution:
ZC
F·dR=Z
π
2
0
(−2 cos t)(−sin t) + (2 sint)(cos t)+2etetdt = 2 sin2t+e2t|t=π
2
t=0 =eπ+ 1.
Problem Page 190: 3 Solution:
a. Here R=i+ 4tkwith 0 ≤t≤1. Thus
ZF·dR=Z1
0
4t·4dt = 8.
b. Here dx
dt =−2πsin 2πt,dy
dt = 2πcos 2πt, and dz
dt = 4, so
ZF·dR=Z1
0
4t·4dt = 8.
Problem Page 190: 5 Solution: Here R=i+ 2k+t(2i+ 4j−k) with 0 ≤t≤1. Thus
ZF·dR=Z1
0
8t(1 + 2t)2 + [(1 + 2t)2+ (2 −t)]4 + 4t(−1) dt = 40.
1