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Unformatted text preview: Engineering Mathematics (ESE 31'?) Exam 3 Apfil 1, 2010 This exam contains six multipieuchoiee prehlems worth two points each, ﬁve truefalse problems worth
one point each, ﬁve short—answer greblems worth one paint each, aud three freeresponse problems worth 18 paints allegether, for an exam total 0f 40 poimsq 9am I. MultigleChoice (two paints each) Clearly ﬁll in the oval en your answer card which correspcmds to the only correct response 3. Let f(x,y, z) = z + sin xy, and note that (Vfﬂm, 91,3) 2 {y ces my, tacos my, I} Find the value of
the following line integral. 37:3
f5? )(ycos xyda: § :rcosxydy’ + (£2) 2. What type of surface is represented by the following parametrization? ?(u,'v) m [Zecosu,222sinu,4} 0 g u < ’21' O S "u 51 (A) circular cylinder (B) half of a circular cylinder (C) onewfourth of a circular cylinder
(D) cone (E) half 0f a cone (F) onefourth of a cone (G) disk (H) half of a disk (i) onefeunh of a disk 3“ What series appmach, if any, shouid be used to soive the foﬂowiog differentiai equation about
mg = 33/’+Sy’+2y=o (A) There is no need to use series to solve this diﬂ‘erentiai equation. (B) The power series method shooid be used.
(C) The Frobenius method should be used, and there will not be a natural logarithm term. (D) The Frobenius method should be used, and there might be or Wiii be a naturak iogarithm term. 4. The point :30 m 0 is a. reguiar point of the diﬁerential equation (1 ~ My” —§~ 3y’ «I» my = 0, so this
DE has power series solutions about 0. Begin the process of solving the DE, and continue until you
have found the pailrot? terms. Exactly what information is given by the pu11~offterms? $126 {32:0 oixﬁaﬁdagmﬁ {D} a? z ""3651 (E) 32 m “261 (F) 92 = “"7331 (G) a2m~§rai (H) «saw—gal (I) omgai (I) The poihoﬁ‘tezms give no information at all. 5. The paint :39 m f} is a singuiar point of the differentiaf eqnation £172ny w 49:31! i~ (3x + 4);; =2 0.
There is at Zeast {me waenius series ssiutien abeut 9, the farm of which is determined by the
indiciai rcots‘ Find the indicial teats. (A) _2:——2
(B) 2,2
{(2) 2,2
{D} —2,—3
(E) 1,«~3
(F) 3,4
(G) 3,:
(H) 4,—1
(1) 4,0
(J) 4,1 6. According to the appiicabie convergence theorem, any series about 0 in the soiution of the following
diﬁ’erentiai equation must converge at ieast on what intervai? 2:33;” + (x +2)y’ + (x2 +1)y = 0 (A) (“"00500)
(B) (4,2)
(C) (*M)
(D) (011)
(E) (0,2)
(F) (0,00)
(G) (1,2)
(H) (1,00) Part H. TrueFaise {one paint each) Mark “A” 011 your answer card: if the statement is true; mark “B” if it is faisei —+ 7” Let F be a centinuous vector ﬁanetion with centinuous partial. deﬁvatives, and iet D be a domain in
' v <  A . ~I+ I
E3, (so everything 13 “nice? If the vaiue ofthe ﬁne integral f F ~ d “1? over every closed path C In
{2 a” l ..
D is zero, then i? IS congervative. 8. The feliowing (2D) differential form is exact, “‘* —l> . .
F~dr m 2ysmxcosa¢dm+51n2$dy 9. A circular cylinder together with its top and bottom “lids” is a smooth surface. It) The ﬁmdamenta] application for the surface integral over a surface S is the totai ﬂux across 3 . 11‘ The Legendre polynomia} P4 (:5) is a soiution of the foﬂ'owing diﬁ‘ereniial equation. (1M may” w ny' 4:“ 293; = 0 Part III. Short Aaswer (one point each) The answer to each of these is right or wrong: no work is required, and no partial credit will be given I2. 13, 14. 15. 16. We}: one of the “big three” theorems from chapter 10 rotates a surface integral to a triple integral? Fiii in the blank: If a ﬁmction f can be represented by a power series about 2: 2 $0 (with R > 0), then the ﬁmction is said to be at 120. The foliowing poiynomiaf solution is obtained for the Legendre equation when n m 3.
y m a; {x w i3.12:3) Traditionaﬁy, a; = —§ is chosen, so that the Legendre poiynomiai of degree 3 is P3(m) m («mgXatw 22:03) 2 31:3 ~ Why is this particaiar choice made for al? In other words, what criterion is being satisﬁed? (The
reason can be stated simply and brieﬂy. Make your answer dear, and don‘t give multiple answers!) What is the value of fijﬂx) da‘ ? What is the value of 17(6)? Pait IV. Free Resgonse (point values as shown) Foilow directions eareﬁiiiy, and Show all the steps aeeded to arrive at your solution. (3} 17. Consider the differentiai equation (a: + My” t 3y’ ~+~ y = 0, and let so = 0. During the solving
process, the foliowing information has aiready been obtained. (You do not need to verify this.) 1 3 2 a ll
a22W§aOW§ai as: “got;ng Use this information to ﬁnd a basis of solutions for the diﬂ’erentia} equation. (7) 38. Let F) m {w y2€$,€§], and cousider the
(timed curve C in the piane which fallews
the parabola x m y? fmm (4, 2) to {1, WE)?
than gaes straight over from (I, «1) to (4, Ml),
and then goes straight up from (4, W1) back to . “‘* «Jr ‘ .
(4, 2} Find the vanilla of f F d r . Simpiﬁ‘y
{, your answer in any reasonable ways such as combining simiiar terms‘ (8) 19. Let if m {3, stigma}, and consider the ci'osed curve C in threespace which gives fmm (0,0, {E} t0
(2, 4, 1) ta (2, 0, 1) back to (3., E}, 0) in. straight Kine segments. Set HQ whatever integrai or integrafe;
wouid be needed to ﬁnd the value of § g ' d ?. Do the comglete setup $0 that the very next step C wculd be integratioa, but do 39;; 60 the integration or ﬁnish the probiem‘ m. These three paints ﬁe in the plane x m 22:. ...
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 Spring '08
 Hastings
 Power Series, Frobenius method, Circular Cylinder, power series method, Apfil

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