{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 2 - Version 029-~ ExamOQ w Gilbert ~—(57495 1 rThis...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 029 -~ ExamOQ w Gilbert ~— (57495) . 1 rThis print—out should have 12 questions: Multiple—choice questions may continue on the next column or page — find all choices before answering. 001 10.9 points Which equation has the surface Z as its graph in the first octant? ”(j Z 1. §+Z~é~~§ = lcorrect 3. iii-$3 m 1 4 gut—2+:— = 1 6. g4—%+E m1 Explanation: As the surface is a plane, it must be the graph of a linear function which can be writ— ten in intercept form as x z a; b c But by inSpection we see that the :c—intercept is w = 5, the y—intercept is y = '4 and the z— intercept is z :2 3. Consequently, the surface is the graph in the first octant of the equation m y z m, m + W 2 1 5 + 4 3 002 10.0 points Find a vector function whose graph is the curve of intersection of the paraboloid m 2 93+ng and the parabolic cylinder 2 2: yg. 1.1112) = t2i+tgsintj+étzcostk 2. r(t) = (t4+9t3)i+t23+tk 3. r(t) = t2i+ tcostjtwétsintk 4. r(t) e (t2+9t4)i+t2j+tk 5. r(t) m t2i+ tgsintj+3t2costk e. r(t) (t2 + 9%) i + tj e— 122 k correct Explanation: The graph of the vector function r(t) = Mt) i + y(t)j wlw ZOE) k will be the intersection of the paraboloid at: = y2 wt 9.32 and the parabolic cylinder 2 = y2 when 20?) = W?- Now already 3 of the given vector functions fail to satisfy the first of these conditions; this leaves r05) r(t) m (t2+9t4)i+tj+t2k, roe) m'yefiwzefl (r4+9t2)i+t23+tk, and 1 r(t) = t2i+ tcostj+ gtsintk, Version 029 w ExemOQ w Gilbert — (57495) 3 Consequentiy, 5a: fr=”m' 005 10.0 points The box shown in is the unit cube having one corner at the origin and the coordinate planes for three of its faces. Fingm the cosine of the angle 6 between E and AD. . 2 1. 6 w — cos 3 l 2. cost? = —-——— . 1/3; l 3. coed = m \/§ 4. c056 = fl 2 1 5. coed = 5 correct 6. c089 = O Explanation: To use vectors we shall replace a line seg- ment with the corresponding directed line seg ment. Now the angie 6 between any pair of vectors u, v is given in terms of their dot product by u~v c056 m luilVI‘ On the other hand, since the unit cube has; sidelength 1, A = (1,1,0), B = (0,1,1), while D = (1,0,1). In this case IB’ is a directed line segment determining the vector 11 2 (m1: 0: 1)) whiie 2113’ determines v = (0, W1, 1). For these choices of u and v, u-v m l mW/éK/ficosd. Consequentiy, the cosine of the angie between AB and AD is given by u-v 1 cost? m —————— m — . IUIIVI 2 keywords: vectors, dot product, unit cube, cosine, angle between vectors 006 19.0 points Find the area of the triangle having vertices P(~u~3, W2), en, —2), R(—2, 2), 1. area m 8 correct 17 2. = —— area 2 3. area = 9 4. area m 7 5. area 2:" —— Explanation: Version 029 w Exam02 w— Gilbert —~ (57495) 5 The volume cylinder of a cylinder of height Consequently, h and radius 'r is given by Web) 2 «We. L(t) = (1, 2+6-c, 2min) . Thus the rate of change of V with respect to t 18 6V 6'!" 28h mm = 7r 2 hm» + m» at ( T at T 8t) keywords: But 5?: m .3, 5% 7; 5_ 010 10.0 points at 63 _ . Consequently, ' ' For which of the foltowing quadratic rela— tions is its graph a one—sheetedlhyperboloid? 8V , . . m5? (manna) =. 607tcu.1n./mm. . - 1. $2 ~i~2y2+3z2 = 1 ‘ 2. x2 + 312 — 2’2 m 1 correct 069 10.0 pomts 3 z z: 332 +3;2 Find an equation in vector form for the 2 2 tangent line to the graph of 4- Z = 59 m" y 13(5) 2 (coss, 3623,81103 + 3€~23> 5. 22 - $2 ~ y? m 1 at the point (1, 3, 3). 6. z2 : $2 + y2 1. L(t) = (t, 3 — 2t, 3 — 5t) Explanation: The graphs of each of the given quadratic 2. LOB) 2 (1, 3 + 6t, 3 ~—— 6t) relations is a quadric surface in standard po— sition. We have to decide which graph goes 3. L(t) w (1, 3 — 3t, 3 + 515) with which equation — a good way of doing that is by taking Slices. 4. L(t) = ( 1, 3 + 6t, 3 ~ 5t) correct Now as slicing of- 5. Mt) = (t, 3+2t, 3w 6t) 6. L(t) e (t,3~3t,3+6t) Explanation: ‘ Since (1, 3, 3) w 33(0), the tangent line has vector form Le) = r(0) +tr'(0). r’(s) 2 (Mains, 6e23, coss — 66"”23 ), so at r(0) the tangent vector is shows, the crosswsections of a one»sheeted hy~ r’(0) m (0, 6, 1 —— 6) w (0, 6, —-5). perboioid at ail levels in one direction are Version 029 m EXamOZ ,._ Giibert — (57495) ‘ 7 Consequently, f the surfaces shown, the only one having the given contour map is a keywords: 012 10.0 points From the contour map of f shown below de— cide whether fm and fy are positive, negative, or zero at P. 1. fm>0,fy>9 2. fm<0,fy>0 3. fxz0, fy>0c0rrect 4. fm>0,fy<0 5. f$<0,f3,<0 6. fx=05fy<0 Explanation: ‘ When we walk in the w—direction from P our eievation doesn’t change because we are walking aiong a contour, so f3; = O. Oethe other hand, when we waik in the y—direction from P we are walking uphill, so fy > 0. Consequentiy, at P. fmm8,fy>0. keywords: contour map, contours, partial derivative, slope, ...
View Full Document

{[ snackBarMessage ]}