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Review Exam 2

# Review Exam 2 - Vincent Univ — Review ExamOZ — Gilbert...

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Unformatted text preview: Vincent Univ???) — Review ExamOZ — Gilbert m {5?495) 1 This print-out should have 18 questions. Multiple—choice questions may continue on the next column or page m ﬁnd 331 choices before answering. 081 10.0 points Find the linearization, L(:z:, y), of H32, 1/) m W17 _ at the point (m2, 9). 1. Mac, 3;) = 3 + 33: m %y correct 1 2. L(a:,'y) w ~3 + ~22 + By 3 3 1 3. L(:c, y) m —6 + ix + ~3—y 1 4. L(;12, y) = —3 + 3:8 + Egg 1 . 5. Has, y) = 3 m 33+ 3y M _ 1 3 6. Many) ~ 6+3}: 2y. Explanation: The Einearizatéon of f rs: f (:13, y) at a point (aa 3)) is given by if, (9113 {(1,6) 3f We 9) e ﬂaw b)+(wwa) Hwﬂw But when If(:z:, y) 2 m/Q, 5f m 31’ W 3.. 6m "” “5’ 8y ”"” Zﬁ’ thus when (a, b) 2 (W2, 9), n m. 3 21: allianbj-w ) 8y while ﬁe, b) = ——6. Consequently, (a,b) = 3 ’ 1 L(m,y) = 3+3m—gy . keywords: 002 10.0 points 63,} (m) ' Which one of the following quadrie surfaces is the graph of the equation Vincent (jaw???) m Review ExamOQ "~— Gilbert —~ (57495) 3 lies on both the cone and the plane. 004 10.0 points Find fm when “53:19) w ln(;c~ 1 . f3 2 correct 2' f3: m 3- fm m 4- fa: m 5me 1 1 _ W l 1 6° f“: Explanation: Differentiating with reSpect to :1: keeping y ﬁxed, we see that SC 1“ 2 2 37 "y fa: 2 2 saw a: my But 005 10.0 points Find an equation in parametric form for the tangent iine to the graph of r(s) m (cos s, 2835, 2665) at the point (3,2, 2). 1. m=t,y=2+6t,zm2—¥6t 2.2: t,ym2—-2t,z 2+6t 9° 2-; Ii 0, y = 2+3t, z: 2m3t 4. :1: = i, y = 2+6t, z m 2~6tcorrect 5. a: m0,'ym 2—313, 3: 2w3t 2m2t, z = 2+3t 6. mmliy Explanation: Note ﬁrst that (1, 2, 2) = r(0). On the other hand, the tangent vector at an arbitrary point r(s) is given by r’(s) : (—sins, 6833, New“), so at 1(0) the tangent vecto:~ is r’(0) m (0, 6, ~6). Thus in vector form. the tangent iine at 1(0) is 1mm r(0)+tr’(0) m (1, 2+6t, 2w6t), which in parametric form becomes = 1, y(t) = 2+6t, 2(t) x 2~6t . keywords: 006 10.0 points The radius of the base of a right circular cone is increasing at a rate of 3 ins/min while its height is increasing at a rate of 5 ins/ min. At what rate is the volume of the cone changing when the radius is 3 inches and the height is 5 inches? H 1. rate 44W (311'. ins/min 2. rate m 477mm. ins/min 3. rate :2 45w cu. ins/min correct Vincent (jmv777) — Review Exa1n02 w Gilbert ~- (57495) 5 l. 4m+ywz+6 = O 2. rr+4ywzm8 = O 3. 4x—y+z—8 m 0 H c: 4. mw4y+z+6 0 correct 5. m—4y—z+8 6. 4\$—y+z+8 m 0 Explanation: The equation of the tangent plane to the graph of z 2 ﬂat, y) at the point P(a, b, ﬁe, 5)) is given by z 2:: f(a, b) 5]“ , 3f +53? ab)“ a” 8y (cablw by Now when f(\$s m V 8+x2‘"2'92: we see that if"; _ a“ 353 x/B—iwmngy21 While if, .. m 2?! 39 8 + 3:2 — 2y2 Thus at P, fﬁ‘: m 11 while if a 1 9i m “4 8:): (1,2) m " 8y {1,2} m ' So at P the tangent plane has equation 2 m 1+(33—1)—4(y—2), which after rearrangement becomes xw4ywz+8 :2 0 . keyworcis: tangent plane, partial derivative, radical function, square root function, 010 10.0 points Find parametric equations for the line passing through the points PM, 3, 3) and Qi5,2,2l- 1. xml—i—Alt, ym~1+3t, zm—l—i’st 2. {Emil-Ft, ym3wt, z=3mtcorrect 3. m=1+4ﬁ, y=——1—3t, 23—1—32? 4. rail—4%, ym~1~3t, zm—1+3t 5. \$=4+t, y=3+t, z=3--t 6. w=4wt, y=3-~—t, z=3+t Explanation: A line passing through a point P(a, b, c) and having direction vector'v is given para— metrically by r(t) = a+tv, a m (a, b, c). Now P'Q’ : (1, m1, m1) is a ciirection vector for the given line, so a 3 (4,3,3), v m (1, ~1,—1). Thus r(t) = (4+t,3-t,3~w~t). Consequently, are parametric equations for the line. 011 10.0 points Vincent (jmv777) w Review Exam02 m Gilbert — (57495) 7 and lim Mt) m liin tint = 0, tv—eU‘l‘ t—-%U+ using L’Hospital’s Rule. Consequently, lim r(t) = (5,6,0) . t—+9+ , keywords: vector function, limit, trig function log function, exponential function 013 10.0 points Find a parameterization of the horizontal circle of radius 4 having center (3, m5, 1). 1. r05) = (3+4sint)i+(5—4cost)j+k 2. = ((3%4c08t)i+5j—(1—4Sint)k 3. r(t) m (3~4cost)i——5j+(l+4sint)k 4. r(t) = (3+4sint) i — (5 ~ 4008293 + k correct 5. r(t) = (3—4sint)i~(5+4cost)jm k 6. r(t) = (3M4cost)i—53—(l+4sint)k Explanation: if the vector function r(t) m i ~i— y(t)j + z(t) k traces out a horizontal circle having center (3, m5, 1), then 2(t) m 1 for all t becauSe the circle must lie in the horizontal plane z(t) : 1. On the other hand, the pro §ection 1‘00?) m x(tli+y(t)j of this circle on the my—plane is a circle of radius 4 having center at (3, —5, 0). Thus, as as circle in the my—piane it has the equation (\$w3)2+(y+5)2 m16. Cerisequently, (at) w 3)2 + are + e2 e 16, and so . r(t) a: (3+4sint)i—(5—éicost)j+k'l is one parametrization of the horizontal circle of radius 4 and center (3, —5, ,1). keywords: vector function, space curve, circle, plane, radius, center circle, 014 10.0 points Find the domain of the function my) = 1. {(23, y) : émgv-t 211:9? __ 1} 2. {(m, y) : \$732+ Egg < i} 3. {(33, y) : 2232+ 2y? > i} 4. {(m, y) : in?” +221? 2; i } correct -5. {(m, y) : %m2+:—yg >1} 6. {(513, y): :32 <1} Explanation: ‘ Since is deﬁned only for .1: 2 0, the function ﬂay) = \/\$2+7y2-4 is deﬁned oniy for 1 2 72 ., :—- +— >1} {(73 y) 4:6 4y .__ keywords: function several variables, square root function, domain 015 10.0 points Vincent (jmv777) —« Review ExamOZ w Gilbert M (57495) Since 1n(my) 2 1mm +111y, we see that f(\$= y) = (356% y)(1n23 HM)- Thus 3mmy m fm W 30112: +1119) + m 3(an+111y) Hm? Consequentiy, after differentiating with re— spect to y we see that ...
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