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Unformatted text preview: Version 070 m Exam03 » Gilbert ~ (57495) 1 This print—out should have 12 questions.
Multiplechoice questions may continue on
the next column or page — ﬁnd all choices before answering. 001 10.0 points Locate and classify all the local extrema of I‘f(a:,y) m 333~y3—322y+2. 1. local min at (m1, 1),
saddle point at (8, 0) 2. local min at (0, 0'),
saddle point at (1, I) 3. local max at (0, 0),
saddle point at (~1', 1) 4. local max at (—1, 1),
saddle point at (9, 0) correct 5. local min at ("1, 1),
local max at (0, 0) Explanation: Since f has derivatives everywhere, the crit— ical points occur at the solutions of
Vflﬂia y) : fssi + .ny “we” 0
But fm m 0 when
. 6f 5; m 3:52m3y = 0, i.€., y = 3:2,
while fy : 0 when
gui = w3y2m3a: = 0, 16,33 2 —y2
Substituting the ﬁrst into the second yields
2: m —(a:2)2 x "334, which can be written as
sc(1+:l:3) = 0, m, a: = 0, m1. '
Thus, the critical points are (0, 0), (—1, l). and to classify these critical points we use the
Second Derivative test. Now Jam: m 633: fyy m “61’: m “3
But then, at (0, 0), A m 'fmm, 0) e 0, s m 11¢,th ~3, ' while 0 == fyy(0,0) = 0. Consequently,
D : $10—82 e —9 < 0. and so there is a saddle point at (0, 0) .' On the other hand, at (~1, 1),
A:f$$(wli 1):“"63 B=fxylmla DEWB: while
C‘ : fyy(—~1,1) = “6. Thus
D = ACWB2 = 27 > c. and so, since A, C < 0, there is a local maximum at (—1, I) . keywords:
002 10.0 points
The paraboloid
1
z = ﬁne? +312)
is shown in Version'O'F'O ~ ExemOB ~ Giibert “— (57495) 3 When
v = 4i—3j therefore, the graph of f has at P in the direction of v. keywords: 7 004 10.0 points Find the J ecobian of the transformation T: (ﬁt?) me (3:,y) when m=efcoséh y H m ~4€2T cos 29 W462?” ll 4. m 482? cos 28 ——3 cos 29 Explanation: For general functions the Jacobisn of the transformation 36'"? sin 6 . 3 cos 26 correct T: (u, U) "+ 0): ylui 0)) is deﬁned by Qt
8(u, ’0) 6y But when
a: : creosﬂ, y = 3eWTsin6.
then by the Product Rule,
823 T 8:1: ,, ,
a? m e c036, 5mg — we 81119,
while
as _ _1« i as _ MT
5—; — ~36 51:19, 5(9 _— 35 0089.
In this case,
am, y) _ ercosﬂ we? sin6
3(7": a) — ——3e”7" sinﬂ 3e” costl m 313Te""7"(cos2 6 w sin2 3) . Consequently,
aim, y)
6&1 9) —— 3cos26
because
ere"? r: 1, c0826 = cosgélwsin26.
keywords: 005 10.0 points  A rectangular box with no top is to be constructeci having a.
volume of 4 cubic inches. What length 3;, in
inches, wiil minimize the amount of material
to be used in its construction. (Assume no
material is wasted.) 1. length :2 2inches correct Version 070 — Exam03 m Gilbert m (57495) 5 we see that I can be written as a repeated
integral 1 m A4(f03(/02($wy)dz)dy)dm. Now
3 2
fowmyldzﬁ [(xwlzloeﬂwwy),
while Thus 4
I :2 f9 (633~—9)d.’13 — l
r—l
0.3
553
M
i
CD
a
:_.1 Consequently, I348—36m12. keywords: integral, triple integral, repeated
integral, linear function,integrai over a speci—
ﬁed volume, limits of integration, polynomial
integrand, exponential integrand, evaluation
of tripie integral 007 10.0 points Use Lagrange Multipliers to determine the
minimum value of ﬁns, y) m 21:2 + 392 +1
subject to the constraint May) = $2+yg~1 e 0 1. min value = 1
2. min value m 2correct 3. min vaiue = 4 ii
ca 4. min value ii
(:0 5. min value Explanation:
The extreme values of f subject to the con—
straint g m 0 occur at the solutions of (WWI, y) e A(V9)(mi y), 9'02, 3;) e 0
NOW ‘
(WW; :1) =1 (411?, 2:1), (WW: 3;) = (2%, 21!)
Thus
43: m ZASC, 2y 2 ZAy,
and so 2:5(Am2) =e, 2y()\~ 1) =0. From the first equation, it follows that either
mm0mX=2 (i) If A = 2, then y z 0 by the second
equation, and so the constraint equation gives g($,0)=m2ml : 0, m, $=i1. i (ii) However, if m m 0, then by the con—
straint equation, 963,31) = y21 ii 0, i.e.,y==i1. Consequently, the critical points are
(i1,0), (ﬂit). Since
f(:l:1, O) m 3, f(0, i1) = 2, we thus see that f has minimum value 2 2 subject to the constraint 9(3):, y) x 0.
keywords: 008 10.0 points Version 070 — Exemﬂ?) — Gilbert M (57495) 7 in the plane bounded by the xeaxis end the
graphs of (not drawn to scale). Integration is taken ﬁrst
with respect to'y‘for ﬁxed :1: along the dashed
vertical line. To change the order of integration, ﬁrst ﬁx
3; in the interval [0, 4] and let x vary along
the solid horizontal line. Then a: varies from
92 to 16. Hence, after changing the order of
integration, keywords: double integral, reverse order inte—
gration, perebole, 810 10.0 points In the contour map below identify the
points P, Q, and R as local minima, local
maxima, or neither. A. local minimum at R,
B. local maximum at Q, C. local minimﬂm at P. 1. all of them 2. A and C only
3. l3 and C only
4. IA and B only
5. C only 6. B only 7. A only correct 8. none of them
Explanation: A. TRUE: the contours near R are closed
curves enclosing R and the contours decrease
in value as we appeoach R. So the surface has
a local minimum at R. B. FALSE: the point Q lies on the 0—
contour and this contour divides the region
near Q into two regions. In one region the
contours have values increasing to 0, while in
the other the contours have values decreasing
to 0. So the surface does not have a local
minimum at Q. C. FALSE: the contours near P are closed
curves enclosing P and the contours increase Version {)70 w Exan103 ~ Gilbert w (57495) by converting to polar coordinates. 1. I ﬂew w 1) correct 2. I = 27r(e4——i)
3. I m Airﬂow“ i)
4. I : 27r(616 1)
5. I = «(ell—l) 6. I : 47r(e4—1) Explanation:
In Cartesian coordinates the region of inte—
gration is {'(aégy) : OSysV16ww2,Osm‘s4}, which is the shaded region in The graphic shows that in polar coordinates
the region of integration is {(r,8):0£r§4,0$05n/2}. Thus in polar coordinates 4 7r/2 2
I 2 46?" ngdi"
0 0
4 16
2
= 27:] re?" d?" = n/ eudu,
0 :0 setting u m: r2. Consequently, I 2 ﬂow m— 1) ...
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This note was uploaded on 03/24/2010 for the course M 57500 taught by Professor Gilbert during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Gilbert

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