Homework 10

# Homework 10 - 0 Vincent(3mv777 m HW 10 — Gilbert ~(57495...

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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; mere-m <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- f-V = _§ 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???) « I-IW 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 pin-ice :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) i-é- 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 Luck-v4 'l’V ‘5 5P‘Fﬂ": ..¢(.;V‘?°’.P"’f’\.. . Tar/59¢... 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