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Unformatted text preview: \M'LA ‘ up 20m. Watt... Se 3 , p.31; gattimi. Problem 1  Let F : (4mg + y2 + 2)i + (25:;2 —— 2:63; + y2)j.
Find a potential function for F in two ways: using a a) the algebraic method and
c b) the integration method. Solution. We calculate the potential using both methods.
0 a) We use the algebraic method. We must ﬁnd a. function f where f; = 4.123; + 3,2 + 2
and
fy = 2.162 + 233,": + ya.
Integrating, we see that f = 2922?; + icy2 + 21 + 9(9)
and that 1
f = 222211 + my? + gya + Mm)»
Solvingfor g(y) and {1(3) we obtain h(m) = 23—30 and g(y) = §y3 + 0. So 1
f(.z,y) = 2:32;; + my2 + 2:: + 51,13 + C. We can easily check that V f does indeed equal F.
o b) We use the line integral method. We can arbitr
f, so set f(0, O) = c. Then we set f(a, b) = no, 0) + f C' for each (a, b) where C is a curve from (0,0) to (a, b). We choose 6' to be a. straight line peremeterized as m = at and y = bt for 0 S t S 1. Then the
integral becomes 1 / [(4abt2 + the2 + 2):; + (2a2t2 + 2abt2 + 52:2)b] dt
0
which is arily set some point of (43y + y2 + 2)d$ + (2n2 + 22:3; + 112de 1 [2azht3 + a52t3 + $591? + 2n] = 2:125 + obz + $253 + 20.. Thus, 0 ﬂatt y) = 25ch + wyg + 2:: + $113 + c. . , ____, ._—__ , . , .., 7 __ J _ .. ..  . v _ D Solution. F is not a gradient ﬁeld. o a) If F is a gradient ﬁeld, then $9? = 12% where M = my and N
9%? = 3: 7E 2—2" = 1;, so F is not a gradient ﬁeld. 0 b) As in Part II #1, let f(0,0) = 0. Then (m5)
f(a,b)=f(o,o)+ / wydx+myde
(0.0) = my. But Path 1: Straight line between ( 1 f(a,b) = f(0, 0) + faﬂbﬂadt + at(bt)bdt = émEb + abz) + 0.
0 0,0) and (a,b) : m = at and y = bt. Path 2: First segment to (0, a), second segment to (a, b). a. b
f(a,b) = f({],0) +fmyjy=od$+fxylm=ady = ab.
0 D But if F were a gradient ﬁeld then f0 F  dr should be path
rent answars for two di independent.
ﬁ'erent paths, F cannot be a gradient 1
f; = my => 1” = 5323/ +9ny fy = my e f = in? + he). 2
But it is impossible to ﬁnd a. g(y)
that is only a function of :L' that D WWW? Yum; knew} ”~me an 43.?th gm Coh’n'un" Q9, 9H‘L Yuk} [JIX’HNHJR % Lne]; Doha'm... .‘ C LL Cg CIA
AK
‘ A __ .5 A
g? dr J“ F [xowc 9M Am.
C
0 .u \ i ,1}: Problem ‘1 . Consider F = V(:L'2y + my2). Let C be the semicircle having its
midpoint at the origin and running from (—1,1) to (1,1).
0 a) Write the integral ch  ctr in the fa JVIdre + N dy form. n 5) Evaluate the integral in two diﬁerent but easy ways: using i) the Fan
damcntal Theorem, ii) path—independence. Show the calculations in each €0.58. Solution. FIGURE 1. The curve C ' c a) F=V(m2y+my2) =< 2xy+y2,m2+22:y>
L:a;=t,da:=dt,y=l,dy=0. / (231; + y2)d:z: + (m2 + 23y)dy
c o b) By Fundamental Theorem of Calculus for Line Integrals: (1.1)
2 2 =2— 1—1 :2.
my+wy (—1.1) ( ) Using path independence:
1 1
fFdr=/(2:I:+l)ofz=:eg+:c l=2—(1——1)=2.
L _
—1 .J A l ‘ ﬂ . It
@390“ Z S (”EMVC’JC L r‘iacosfsﬁ  (—Zacw’tt Smtb
1’ + BK M11: @3113) 1.77 M
7.
Z 1Q1§9nmt£05 t Jr Gina—43 3,th 0H:
0 Problem S ' o a) For what simple closed (positively oriented) curve C in
the plane does the line integral jg (way + 113  21W + (3:6 + 23123 + eyidy
C have the lamest positive value? (Hint: use Green’s Theorem).
0 b) What is this maximum value? Solution. We use Green’s Theorem.
3 a.) jt‘(cs2y+y3—y)dm+(3:e+2y2m+ey)dy = ff (3+2y2)"($2+3yg“1)d‘4 = ff (4—372—92M‘4s
c A R R where R is the region enclosed by C. The integrand is positive inside the
circle 1:2 + y2 = 4 and negative outside, so the integral is biggest when G
is the circle :1:2 + y2 = 4. o b) 211' 2 2
/f (4 — $2 — y2)dA = f/(r — 42)rd7'd9 = 271' [2T2 — lard] = 817.
m2+y3<4 U 0 4 0 ...
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 Winter '10
 JohnBush

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