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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,
uv m l mW/éK/ﬁcosd. Consequentiy, the cosine of the angie between AB and AD is given by
uv 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+6c, 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, ...
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 Fall '09
 Gilbert

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