Problem Set 1 Solution

0 x i x i x i 1 31 6 a 3 grad 2 ei i

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Unformatted text preview: 2Therefore, 2 (4) CHAPTER 1 from which cos # % % 1-30. or, grad ! )* " grad ! S1 " $ grad ! S2 " 6 9 # % co grad ! S1 "s grad !% 2 "4.2( S7 &1 (5) ! 4e1 & 4e2 ' 2e3 " $ !e1 ' e2 ' 2e3 " 3 - ! )* " 6 6 + -* -) , * %e % e) ' 2 i %1 i 2 -x i cos-*% # % 2 ei ) i from which -x i Thus, i i 4 66 ' 2 ei i . 0 -x i -x i -) * -x i 1-31. 6 a) 3 grad ! )* " % 2 ei i %1 / 1 (4) &1 7 '* grad !# * " %s) grad%* 4.2( grad ) ) % co 9 1-30. (3) (5) + -* - ) , - ! )* " */ % 2 ei .) ' -x i i 0 -x i -x i 1 -* -) 12 n 3 * % 2 ei )-r ' 2 ei - +3 4, x grad r n % 2 ei -n i% 2 ei -xi.5 2 x 2 6 / i i j i %1 Thus, -x i -x i . 7 0 j / 1 8 &1 grad ! )* " % ) grad * ' *2grad ) 4 n3 % 2 ei 2 xi 5 2 x 2 6 j 27 j 8 i 1-31. n a) n &1 3 42 % 2 ei x i n 5 2 x 2 6 j 7j 8+ i 3 4 -r -3 .5 2 x 2 6 grad r % 2 ei n % 2 ei j n & 2 " -x . 8 i7j %i %1 ei-xii n r ! 2x 0 n i Therefore, n n ! n & 2" , / / 1 n (1) &1 42 n3 % 2 ei 2 xi 5 2 x 2 6 j 2 %j 8 i grad r n 7 nr ! n & 2" r 3 42 % 2 ei x i n 5 2 x 2 6 j 7j 8 i 12 &1 (2) 23 MATRICES, VECTORS, AND VECTOR CALCULUS b) 3 grad f ! r " # , ei i #1 %f ! r " 3 % f ! r " % r # , ei %x i % r %x i i #1 ' %& 2 # , ei ( , xj ) %x i * j + i & ' # , ei xi ( , x 2 ) j *j + i # , ei i 12 $1 2 %f ! r " %r %f ! r " %r x i %f r dr (3) Therefore, grad f ! r " # r %f ( r ) r %r (4) c) 5 2 ! ln r " # , i 12 '. % 2 ln r %2 - & # , 2 / ln ( , x 2 ) 0 j %xi2 %x i / * j +0 1 2 $1 2 -1 . & ' / 3 2 xi ( , x 2 ) 0 j 0 *j + % /2 #, / 0 12 & ' i %x i / 0 2 / 0 ( , xj ) *j + / 0 1 2 $1 '. %- & 2 / xi ( , x j ) 0 #, +0 i %x i / 1 *j 2 & ' # , ! $ xi " ! 2 xi " ( , x 2 ) j *j + i ! "! " # , $2x 2 r 2 j i #$ $2 $2 ' %x & 4 , i ( , x2 ) j + i %x i * j $1 -1. 4 3/ 2 0 1r 2 2r 2 3 1 4 2# 2 4 r r r (5) or, 5 2 ! ln r " # 1 r2 (6) 24e have w 1-32. CHAPTER 1 ! ! d (r ) ( r rr ) & * r ' 2 + dt % & dt * - dt Note that the integrand is a perfectrdifferential: r + , ,...
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