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Unformatted text preview: YE let/P . Math 200 Calculus Midterm Exam II March 2, 2007
Naunc:
Lecture IIour: ME— Section Hour: m Guidelines for the test: I No hooks. notes 01' calculators are allowed. 0 You may leave answers in symbolic form: like Jr}. unless they simplify further.
like J?) = 3. e” = l. or eos(37r/J.) = —J§/‘2. « Use the space provided. If necessary, write “see other side” and continue
working 011 the heck of the same sheet. a Circle your ﬁnal answers when relevant. 0 Show all steps in your solutions and make your reasoz'iing clear. Answers with
no explanation will receive no credit. even if they are correct. 0 You have 50 minutes. ('1) (a) Match 1.110 follmving equatiuns (i)—(\‘i) with the. graphs (I)(VI). You need
M explain ymlr choices. The ﬁrst. our has been decided for you. (i) j‘(:r.y) = Hil12:1.'+ 591’
[1 (ii) f(:::.y) = (.n::(::;in(3:r2 + '9'?»
(iii) f(:r.y) = sin(:r) sin(3y) (iv) f(;:‘. y) = ('05(3J') msm) MPH (V) f(?‘. y) = 5iII(IT + lyl) H ‘ (vi) “Iry) = 3:17: Pugl: I (2) Suppose that , , _ if[a.n)#[010)?
m" U) Q {0, + 1mm) = (U: 0.) (a) Is f (one) continuous at (U. D) ? Clearly explain your answer. 1 M u‘s‘m'w
'a u""_—'_'—_ ::
“Mt—7U). a) U14 V1 UL Hw (Qua2 V‘)
M1 2. 0 V12 0 \‘tm (9'1 (11V :0 (X ,\ \ﬁwiul
U7 *Viio l '1 LA 1V —0
d
mm+vlzﬁ OLE{IV lsgl‘n
‘1.
1"?)
L U L 4:! IW
SO OFU—LAVL _ l (3x1 SKI/WQIu' Yet’: '11 “r, cav‘hﬂmOMS (1)) Suppose further that 'u. = 110‘) and n = vhf) are continuous functions of f,
for all t E R.
(1) Write down the chain. rule that. would allow one to compute I‘ll: in
terms of fm ft“ m and 'Ug. You need not compute any derivative ”EL0E Ar explicitly. 4'
OH1 _ “a; ‘llvl H+ +‘Qv‘lVl' / \V l l "f ”l” (ii) Use the two tables of values given below to calculate ﬁg i=1 Page 2 W 3 o: £L°r0\{fﬁj (3) Suppose f(3:,y) = 3:2 + yg, Where (3:,y) E R2. The surface 2 = f(3:._y) is a
paraboloicl.
(a) One of the two contour maps below is for a paraboloid and the other a.
cone. Circle the one that corresponds to a paraboioid. (b) Let P = (2,1,5) is a point on the surface 2 = ﬁst, 3}). Find a. unit vector
11' for which the directional derivative Dﬁf(P] is the largest possible. P=C2,1‘3\ 2:4(Yﬂ1) :xqﬂ'l‘wZ'
'F‘x:2,)( ”CH—'27
"3‘ v¥<¢ﬂ31<4rz>
De—HPPVHPEM \ i P 4.; $7 TEL'zrl7 (0) Consider all possible lines that are tangent. to the surface 2 = f [37,19]
at P. One of these lines has the largest possible slope. What is the magnitude of this slope ".F' WIN—mm : id. two ,__"‘"=..
11W: ”04.4 : ~20 (d)' Find the equation of the plane tangent to z = f(:z:, y) at 13(23 1, 5).
F3 : W 1F”! : 2‘1 FAM " O
Fig : Zr Fug, ’ 2. Ziemzﬂo +¥y(1n7(x_;jipv(2n\(w'l)
: S +‘iL><v7A+ 2044) 7:34 qxg+2ulZ
2 sé‘xrufrlo
**4x+2%'8 Page 3 (4) (a) For each of the following sets. detemliue whether it is closed and whether it is bounded. (i) {[:I:, y) : :32 + y2 >1}
Closed? (Ye. i Bounded? ([email protected] (ii) Hay) = y 2 2}
Closed? e“ Boundei‘l‘? (Yes. {tiac“ 4 Vii—NM + x142ﬁ‘d Jr?
(b) Suppose f(;::1y)=(:r, F 1)2 + (1!; — 1)2 + (:L'+ g)? 2%:qu +ZVI (i) Find all critical points of f[3:, y).
’Qx: 20 :1) + IMHO
We»1 ._ 204 ~13 + “axed : 1 x‘zlLX4'1
ﬁlx O 1 jx+l‘1"7‘ {L 1.1% saw +2 $xx:2+l'r"l pxq‘l
Qviﬂs’ZIT‘l/"L’l :2. '2 4x+7vl L“ +1” ”W: +1):
Gw;O='2FW"L+LK"11W—f Hul'l17C72 1X4h11'2'vl‘l2 0 ”“4””? K11. (ii) Classifyr each of your critical points as local maximum. local mini
mum or saddle point. ”2..
mils3 = more»: spam
= quﬂb’r =W’qt'l>o
\ ‘ 19‘4' \4.
(ER?) \5 mmxmwm / J (iii) Find the shortest distance from the point (1.1.41) to the plane.
a: + y — z = 1. P =mﬁu '1 VH7 + 01 P‘r—(xnH(vi—OIMHO‘ _2_
PX=2L.><—i‘) 'zx—2o 1:2,; x4 ,5
‘szCw—‘l 2‘4'2“) ‘1
vazclH) 11451=O 97'? l1+ l,L 4' (' ll. rd; II Fayed (5) Use the method of Lagrange 111ultipliers to ﬁnd the maximum and minimum
values of the function
[(13 y) = r2  '92
subject. to the constraint. 3:2+'y2=1.
\—V\_/)
_ 650‘!“
@(Vﬁ‘JB = 3(1— ‘41—,
Jazz; / 3)::2): ,
JEN—27 / =3 ./
9w ‘1
V+=AV3 Page .5 [6) Note: You would have to have earned full credit on problem 2 to
qualify for any portion of this extra credit. Consider again the function deﬁned in problem '2
ml!— '1‘ 0.0.
m. w) = T (“ “é (. ’
(3. If (v.1?) = (0.0).
Is f(u.. 1.!) (1il'l'm'clll.ia.hlo at. (l. l) '3' At ((1.0) '3’ Clearl}r and fully explain using the notion of diﬁcrc”mobility. ‘ plan"S
Yes ‘l‘C— Lu N3 ‘ MMMWU‘XA QT (‘9
I n . ‘3
cm «Sm/“d :0 0* 4”“ OIR‘(U‘/(/1 rl K Wu #91:?8‘ . mu‘Ll),
WW ‘“ '  how“ '1 0' 0‘ ‘5 C ME
IQ— PW Z/r’r IS ‘3 _ ‘ﬁl
[MUD "’l'l.ll oxi'l UL Page 6‘ Formula. Sheet, {1} Some surfacme: f(.r. y) = 3‘2 # H2 is a saddle: f{..r y) : 3‘37": — .1":i is n lllOIlkﬂ}:
Huddle; f(3.'. y) 2 Hi — Jr“! — “:3 has at [Jtmk m. ((1.0, 16). {13) 1” is continumm at. (0,3)) if gin: ”ﬂagy) = ﬂan).
(any  In! (3) A: :3 fz AJ? + 1",, Ag: d; =1} :33. 4 f_,, dy. which is the 531110115 LU) = ﬂn} + f“(n){.:r — a) (fur single. variable)
w w L(:1:. y) : f(n.h) + t—(r:.h}(3' — a] + .—{u.b)(y — b).
()3' 0y
(I) To find critical points. solve j". = I] and f." :2 0 simultaneously for 1:. y.
(5) Suppose that. [H.b) is a critical [mint of f. Hllll 10!. D = in if" . Then
9;: .Im (:1) 11' D b (I am! f.,_.r{a,b) > (J. Llauu f(u. b) is a local minimum.
(a) If D ) [l and fn.(u.b) < I]. then ﬁn. h} is a local maximum.
(:1) If D < I). then ﬂab) is not a local maximum or millin'mm.
((5) Suppose z = f[.r. y}. .r 2 arm and y = y“).
(1:: EU rim Bf dy __._+__ E‘mm mm'
(7) Suppose z = f{J',y). ;r := 113.!) nml y = y(s.!].
If): _ ()2 i); (5: By _. _ _ __ + _ _
{is fir (“b. Uy as
62 _ r'): 3);? 02 fly a.  77+Tym (8) Suppose F(J'.y. :} = f} and .. = {.r. 9). Then.
r‘)‘: __ F;
(Jr T F:
(“)3 _ F”
0!; __ F: (9) Suppose I? = (Hub) is a unit vector. Duf[£ y) = AER/(.1 + "b.9111!” — f(.:.',;;) = Vﬂzlr. y}  u {1”} Solve Vﬂjgy] : A v g[.r,y} for .r. 3;. Compute f(.r,y} at. all solutions to find max and min. ...
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