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Math 201 — Fall 2004—05 Calculus and Analytic Geometry III, sections 5—8
Final exam, January 29 — Duration: 2.5 hours YOUR _NAME:
YOUR AUB ID#: PLEASE CIRCLE YOUR SECTION: Section 5 Section 6 Section 7 Section 8
Recitation M 1 Recitation Tu 12:30 Recitation Tu 2 Recitation Tu 3:30
Professor Makdisi Mr. Khatchadourian Mr. Khatchadouriau Mr. Khatchadourian READ THESE INSTRUCTIONS”
1. PART I of the exam is quick answers. NO J USTIFI CATION REQUIRED, NO PARTIAL CREDIT.
. PART II of the exam is short answers. No justiﬁcation required, BUT PARTIAL CREDIT IS AVAILABLE.
. PART III of the exam is full problems. FULL JUS TIFICATION AND COMPLETE SOLUTIONS ARE
REQUIRED, and partial credit is available. . Don’t forget to enter your NAME, AUB ID number,
and SECTION! GOOD LUCK! GRADES: TOTAL OUT OF 144: p
f ‘ r
A PART I. QUICK ANSWERS, NO JUSTIFICATION REQUIRED, No PARTIAL CREIBI‘I“ 1 (2 pts/part, total 12 pts). Which of the following series converge and which diverge?
Circle your answer.
DO 2” 1a) Converges Diverges 1b) I ‘ Converges Diverges
1 c) Convorges Diver gee
1d) Converges Diverges
1e) Converges Diverges 1f ) ————I——— Converges Diver ges 2 (2 pts/part, total 6 pts). Fill in the blanks below. (Parts a—c are not related to each
other.)
2a) Fill in the blank: the sum of the following series is 00 4n+1 an
72:0 2b) Fill in the blank: lim 5133;”: 2 can>0 $3 min —————— has radius of convergence
371(712 + 1) 00
2c.) Fill in the blank: the power series 2 11:0 3 (total 14 pts). The following picture shows the gradient vector ﬁeld V7 f for a. certain
function f (a; (Note that the gradient vectors are drawn shorter than their true length,
to make things easier to visualize.) we also know the following values of the gradient at
speciﬁc points:  W = (—5. —3). W W
(0:0) , 33
o: H
T
“ribs i b
at,
\‘I' l
I \ i\\x\\\'\\\\\\\\‘u A \\i\\l\i\'\zl\\i\\lm\\a \\\\\\\f\\\\\\\\
.\\\\\\\\'\\\\q\.\\ \\\\\\\\\{\\\\\\\\‘’\B \\\\\\.\\\'\\\ A
3m.\\
\\\\\\\\\l\\\&\\\b \\\'\\
\\\\\\\\ 3’\\\\\\\\\1i1«.
\\\\\E\\\\\\\\\iua.. \'\\ \\'\\\\\\\\\l\\\1
\'\‘N\\\.\\\\\\\\\\\1u I
/
x
./
./
./
./
/
/
/
/
/ 3a) (2 pts) Which of the following two statements is true? Circle the correct answer. ﬁll,0) > f(2=1) 0R NW) < H231) 3b) (4 pts) Consider a moving point P ( ) = (t2 — 2, 2t — 3). Fill in the blank for the
derivative of f(P(t)) at t = 2: d amt? —2,2t—3)H 2 t:2 3c) (2 pts) The point (62 2) is a critical point for f. Is it a local maximum, a local
minimum. or a saddle point? Circle the correct answer. Local maximum OR Local minimum OR Saddle point. 3d)&3e) (4 pts) Use an approximation to circle the expected answer in the two ques
tions below: 3d) The LARGEST number is: f(0,3) f(0.01,3.01) f(0.01.2.99)
3e)The SMALLEST number is: f(0,3) f(0.01,3.01) f(0.01,2.99) 3f) (2 pts) Draw a. rough sketch of the level curve passing through the point (8. 5).
Please draw ONLY the part of the level curve that is inside the box on the top right Corner= and don’t clutter up the rest of the picture! (Maw—"nvrim.
,AMrJ'Jjw I. .x ..=. LHH:.;11\‘ ]  W
 .ww ——
..... pn' ‘ r
A PART II. SHORT ANSWERS, no justiﬁcation needed, PARTIAL CREDIT AVAILABLE)? 4 (4 pts each part= total 12 pts). Consider the shaded region R below. Fill in the blanks
for the following integrals in rectangular and polar coordinates: 1y I 4e)//1ee=f:*— deep. 4b)//1dA=/—"— /—— dydz
as: y—
+/ / dydm. 4e)//1dA:/——"" /m we
. . 6: 1': “—
R “H”— m"—
+ f / [same]drd9.
6: r: 5 (4 pts each part, total 12 pts). In each of the pictures below, we give a. 3—dimensional
picture and a. cross—section of a. region B (which is always part of the half—ball 332 +312 +32 3
1, z 2 0). In each case, ﬁll in the blanks for integration in spherical coordinates: A
l Ice—cream cone [some] dp (£90 (£6. Holloweduout halféball (cylinder of radius J32 removed) 5c)///1dV=/f;:—— ‘—_ [sameldpdcsdﬁ.
D .... _u___ —_ 6 (4 pts each part, total 16 pts). The parts are not related
6a) Find the tangent plane to the surface 32y + 23;: 2 5 at the point P(1, 1, 2). 6b) Find a potential function for the vector ﬁeld 13 = (y, :L‘ +~ 2, y + 1). 6c) We make a change of variables :1: : mu, y = u + uw to evaluate an integral from
a region R in the avgplane to a region R’ in the urnplane. Fill in the correct value in the blank: ‘
f/ 23: dmdy = dudw.
(may)ER (u,w)ER’ 6d) Given the vector ﬁeld 13 : (372,1:9r yz) and a surface S with oriented boundary C 2 83 as shown. Then Stokes” theorem says that f 13  d7? = ff C3  “Fido for a suitable
C S
vec'tgr ﬁeld Fill in the blank: 1\ C11 :( 7 7 )— OI .J' PART III. FULL SOLUTIONS REQUIRED, PARTIAL CREDIT AVAILABLE. 7 (6 pts each part; total 12 pts).
.01 a) Using power series: express the integral L = f 6“"3 d5: as a certain alternating
1220
series. For full credit, the answer should be written using '2 notation. You can get nearly full credit for just writing out the ﬁrst four (nonzero) terms in the series. b) Find (with justification? of course) a speciﬁc partial sum sn for which the error
satisﬁes Isn —— LI < 10'“. Note: in parts (a) and (b), you may use without proof the fact that your series satisfies
the conditions of the alternating series estimation theorem. WM... HHLA}. l'le'IJIH I '1‘ I iLH’HLch' :
i "W Humu ':
*Mﬂ_h‘ '1 8 (6 pts each part: total 12 pts).
a) Use Taylor’s theorem to ﬁnd a. speciﬁc constant A such that for all x with :1 1 /2,
we have (1+xPﬂ—1—3am 3A3? a) (You do not have to simplify your expression for A.) b) Use the inequality (*) to show that. 1im cc, 2 0, where :12, is deﬁned b7
mmﬁmmﬂ w ﬂ w 3 u+m+yPﬂ—1~3m+yy2 V332 + 3/2 Note: you do not need to know the exact value of A to solve this part. Even if you do
know A, you will have at neater solution if you just write the symbol “A” instead of its
speciﬁc value each time. ﬂaw: n .vv 9 {total 12 pts). Deﬁne the function ﬂay) : m2 + 2112 — y3/3.
a) (4 pts) Find and classify the critical points of f. b) (8 pts) 'We constrain (m, y) to lie in the disk 3:2 + y2 5 1. At What. points of the
disk does f attain its maximum and its minimum values? 10 (6 pts each part, total 12 pts). Let C be the closed curve in the plane starting from
(O, 0), which ﬁrst. goes to (1, 1) along the parabola. y = 3:2, and then returns to (0.; 0) along
‘ the line 3; = 3:. Compute / ydm + 61" dy twice: (.3) directly, (b) using Green’s theorem. C
Solution for (a): w) Solution for (b): 11 (12 pts). Let D be the solid region in Space, shaped like a modern table which is
bounded below by the plane 2 = 0, above by the plane 2 = 1, and on the sides by the
hyperboloid $2 + y: — z2 = 1. The density of D is (Shay, z) = 22. Find the total mass of 12 (3 parts, total 12 pts). Let. D be the solid region in space which is bounded above b}:
the surface 51 : z = 4— yg, below by the plane 32 : z = 0, and on the front and back by the planes S3 : 1? = 1 and S4 : x z 0, respectively. Define the vector ﬁeld F = ($2, 0, z). _‘ (m,y,z}
This is the same as sa in that F : :rzi + zk. The normal vector on the surfaces
3’ g 81, . _ . , S4 is oriented outwards from D.
Z S4 (back) S3 (front)
‘ s2 (bottom) a) (6 pts) Compute the ﬂux integral ff 15  ﬁda.
5: CONTINUED ON NEXT PAGE! Parts (b) and (c) are overleaf. 11 up. Continuation of 12. b) (2 pts) Set up but do not evaluate the ﬂux integral // 13"  77 do.
33 c) (4 pts) Set up but do not evaluate a triple integral over the solid D which is equal bytheDivergenceTheoremtof/ FﬁdU+// ﬁﬁdUif/ FﬁdaO—f/ 1573010.
5; 5'2 53 S4 12 ...
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 Calculus, pts, Gradient, Analytic Geometry III, Professor Makdisi Mr. Khatchadourian Mr. Khatchadouriau Mr. Khatchadourian

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