This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 0?
Vincent (3mv777) m HW 10 — Gilbert ~ (57495) lit" 1 This print—out should have 8 questions. and so
Multiple—choice questions may continue on the next columnor page — ﬁnd all choices v f<21 W2)
before answering. 001 . 10.0 points On the other hand, Find the directional derivative, fv, of ﬂay) = V623~3y at the point (2, WZ) in the direction v z i m j .
7
1. = ~—
f" 12
'1
2v fv —— E
1
.3. fv —~ 21:
3
4. fv = a correct . keywords:
5 002 10.0 points
5: fv : '33“
Find the directional derivative, fv, of the
Explanation: function
For an arbitrary vector v,
V f(:c,y) = 4+3m\/§
fV = ' a
at the point P(3, 1) in the direction of the
where we have normalized the direction vector vector
so that it has unit Eength. V = <3, _4) _
Now the partied derivatives of
ﬂay) = 623—39 1, fv = mg
are given by
9
ﬂ = WEWWW 2. fv m —g correct
8:2: 63: m 3y 1
d 3. ' m ~2
83; m 2 a: "— 3y 7 _
Thus 4' fv = “g
3f 3f .
V :13, = ———i+ ——
f( y) 333 3y 5. fv 2 _g m m (ﬂﬁ’ Explanation: 3 '3 iriwr—LE’L'
< 3’»??? I 'ﬁ‘<3g M1) (63, 1’37 Vincent (jaw???) w HW 10 w Gilbert  (57495) ‘ 2 New for an arbitrary vector v, By partial differentiation,
v 8f y 1/2 6f 1 1/2
v :2 v  mm , W = mm (mm ) Mm : mm §
f f 63: 333) 8g (my)
~ where we have normalized so that the direc— so that at PG, 4),
tion vector has unit length. But when
' W; merem <1 _1_>
7 f(:c,y) e": 4+3x¢‘y‘, (4,8) 4 43 4’ 4 _'
then On the other hand, v = (4, 3) which as a
' 3 vector of unit iength becomes
so
Vf = 3 y i~i~w(w)j.
lvi 53 5 ‘ At P(3, 1), therefore, 9
w] = 31+”; P 2 >< > = m
Consequently, when v = (3, we), ,  4 4 5 5 20 5“ ML ampf 004 10.0 points Find the maximum slope on the graph of keywords: flxay) = 25in(%') 003 10.0 points at the pomt P(0, 4). 1. max slope = 8%
Find the directional derivative, fv, of Key) :2(£)1/2 CC 2. max slope r: 271‘ 3. max slope m 4m at P(4, 4) in the direction of Q(8, 7). 4 max Slope _ 7r 3
1 fv = “Eb” ' 5. maxleIpe = 4
2. fv :: wéula correct 6. max Slope “m” 2
1 r 7. max slope = 8correct
3. fv : WEI—(")
8. max slope 2 1
4. fv 2 ——§ Explanation:
20 At PU], 4, 0) the slope in the direction of v
1 is given by '
5 fV = _§ v
Explanation: Vf‘ai‘l). . Vincent (jrnv???) — KW 10 m Gilbert ~ (57495) 3 But when
flat, :9) = 28in(myl,
the gradient of f is
Vf(m, y) = 2ycos(xy)i+2mcos(33y)j, so at P(0, 4) W“ .l m 8i.
(0.4) Consequentiy, the slope at P will be maxi—
mized when v x: i in which case  keywords: slope, gradient, trig function, max—
imum slope 005 10.0 points
Find the gradient of f(m, y) = 333112 + 23:33». 1. W m (23:3 — easy, 3;? + 6%)
2. Vf : (3312 + 63321;, 2st3 — dry)
3. Vf = (Gmy + 2223, 6%; ~ 3292)
4. W e (6563; + 22:3. 392 + 6:62?!) 5. W: rect (3312 + 656231, 6333; + 222:3) cor— 6. Vf = (63323; ~ 31/2, 63:31 + 2m3> Explanation:
Since 8
View) m we see that Vf = (393 + 6.932% 6333.1 + 2w3) . keywords: 006 10.0 points rI‘he contour map given below for a function
f shows also a. path r(t) traversed counter—
clockwise as indicated. Which of the following preperties does the
derivative have?
I positive at R,
II positive at P,
III positive at Q. 1. I and III only
2. II only ‘ 3. none of them
4. all of them 5. I oniy 6. II and III only 7. III only correct 8. I and II only Explanation: Vincent (jmv???) « IIW 10 w Gilbert — (57495) 4 By the multimvariable Chain Ruie, grime) = (Vf)(r(t))r’(t) Thus the sign of d
were» will be the sign of the slope of the surface in
the direction of the tangent to the curve r(t),
and we have to know which way the curve
is being traversed to know the direction the
tangent points. In other words, if we think of
the curve r05) as deﬁning a path on the graph
of f, then we need to know the slope of the
path as we travel around that path — are we
going uphill, downhill, or on the level. That
will depend on which way we are walking!
From the contour map we see that I FALSE: at R we are on the level — we are
following the contour. II FALSE: at P we are descending — the corn
tours are decreasing in the counterwclockwise
direction. III TRUE: at Q we are ascending » the con—
tours are increasing in the counter~cloc1cwise
direction. ' keywords: contour map, contours, slope,
curve on surface, tangent, Chain Rule, militi—
variable Chain Rule, 007 10.0 points Find the directional derivative, 1Dv f , of ﬂex, y, z) = Eixtan‘1 (E) .3 at the point P =2 (1, 1, 1) in the direction of
the vector v = iﬂ2jw2k. 1 1. valp m 17:“ correct 2. m l 3. m 4:. m 5. vaIP = :1 pinice :4th elm 6. nvflp = a Explanation:
For an arbitrary vector v, m = Vf (3—), M where we have normaiized so that the direc»
tion vector has unit length. But when '9 Km, y, z) = 3:6tanmi , then mer QI.. n
— 8x1+8y3+6zk = 3tan‘i(g) ié H 33y
22(1 + (21/le) Vf k. Thus __ _1 y _ 32:2 _ 3mg
Vf—3tan (;)1+22+y23f32+y2k. At P = (I, 1, 1), therefore, it 3 imp: 3(4)'i+%j—§k.
Consequently, when
v = iw23w2k,
weseethat M: \/1+22+22 = 3, and Min: %(3(§)i+§j—gk).(i—2j;2k). Vincent (jmv’???) ~ HW 10 w Gilbert W (57495) 5 Consequently, 1 7T 1
lDVﬂP” 3(1.33333"3+3) “ Z” keywords: directional derivative, gradient,
dot product, unit vector, ‘ .BULV‘Leéﬁ
088 10.0 pomts ERMEM a? 40*?Wt/tgg/M If f(x,y):$2+2y3, use the gradient vector Vf(5,1) to find the
tangent line to the level curve f($, y) z: 27 at
the point (5,1). 1. 2m+4y=27
2.10x+4y327
3.10mm4ym27 4. 553 + 2y 2 27 correct 5. 523 —— 23} = 27
Explanation: (Mi. 4 3:; G
The equation of the tangent line is given by 3; "‘" g M (MUST gammy
vf(5,1)<m—~5,y——1)=0_ W a," < > :0
When H56, 9) = $2 + 292
it follows that Vf = (222,4y) and Vf(5,1) m (10,4). Consequently, the equation of the tangent line 18
 «m «4st
7
W3
1» Ivy—sowv‘t’o
on ¢q1 i g? keywords: TA Sesu‘oa {ﬂint/‘95 __ .Gm‘"°:~+.._of 1% 19%;». _ _
7 M FF? ,<..~’3«..,fr.> ..°{'.VF” <$>9.;;.f>é,f?7.ﬁ . ‘. . Vi? aﬂowﬁ +4.5). . 1;:de Luckv4 'l’V ‘5 5P‘Fﬂ": ..¢(.;V‘?°’.P"’f’\.. . Tar/59¢... Fﬁwcﬂaénﬁgﬁﬁ M 0:935, 55) H I u
5‘ ‘»?DJ’V‘F My») ‘ (x rm .V'i‘XR'.  1%) ..«+.~3w 94/? “ktMW, V (for; #fSihaL7..II ' l I I r (“.11. ' ‘
Fr 1 WE Mfrkrkrdak ‘ 5(0’0?’ ""‘ " if .. W ...... ,_
. . r. I ,7 .. , .‘ .7 m.\;..,.._.w.,imW.,. 4 {swim  ‘ZTsV‘ﬂﬁ/ﬂ... ‘ .. , w “a ‘ Kr..‘>.. ‘..n n ‘
7 .. u , ,, ‘ .. <._ {17’ Cay) :rx73xceg:Qﬁgz ‘ , _ __ f . ‘ . . . .. .. .‘ . .. ' t ' ‘ 47%,.) ; 3114:24 313,914.37  U’Chfa‘?‘ :~ 1' I. . “3’” 3” .. ("373). «4?,7746. k %;,ja.,m_,d.. .. ,7 ' Fairé..:~:.4arw ‘. ..¢.F.fr.+.¢3‘f%ii ‘7 ' ’0“ V" 7 O I V5): ic‘V‘L {3 wk,” cU/ufw “if I ._: \ .. #3:. rt.v.f.._i.:" 7 , [hciﬁrﬂl .Pt39ﬂ3_.”"f . .
.\7_£:,<.26H,%E>
‘ WﬁiffﬂJﬂmi'féz «36 .5 5 W ._
’5 5 . j 5.. ‘ﬂ(?{r.°{).....in.Jihad? .vf‘. (if?) ‘ ‘ . l,..?r‘<fi,%7_.. “ H . M
, M , w, M
w W. a
w M u a
. w ., m a W W
‘ m
.m W m 4
u w
n W
m w u
M _ .V
H. +
m W
m. m i a 13+; 3: ? i <m€fi .. .,. H"
w
m
M 1/3 «*(aijx‘wf cafe"), {{>\;}/)*??7 .'.kP[6;:D  44;va ' ' VHsz,“9'r7/YZII'17 :27
. ‘r "‘7‘. V’F[5?’)r<ww> 470/0" 44M 5 ‘aummksu "AR‘42“.mawhﬁummsm ...
View
Full
Document
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.
 Fall '09
 Gilbert

Click to edit the document details