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Unformatted text preview: Math 230, Makeup Exam 1 October 9, 2006 Name: Student Number: Instructor: Section Number: INSTRUCTIONS
0 Then: are 10 problems on the exam. Some of them haw: multiple parts. 0 Presunt. your work clearly for ALL problems 0 No credit will be given for unsupported answors. 0 Please box your ﬁnal answers. THE USE OF CALCULATORS, BOOKS, NOTES ETC.
DURING THIS EXAMINATION IS PROHIBITED. Cellphones must be turned off during the exam. 1. (8 points) Find thu vulmm: of the parallvlupipttd with mljm't'nt Nigeri
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01’. ()Q and OR, where 0 is theorigiu, P = (1, 2, — 1). Q = (—1,—1.2).
and R = (—1.1.3). VOW/Mg : I 2. )
r: 1 «I 2_
_l I 3 l l
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~C
+ \ ﬁ J. \_
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\1 l‘é 2. (n) (4 points) Find an equation of the sphere with wantur I’ (1, (J, 0) and radius r = '2 in rectangular coordinates. 5L ’
/ (vsIV’r (TL—r 7}; (b) (6 points) Describe and sketch the intersection of the above sphere
with the sphere centered at Q = (—1,0, 0) with radius r = 2. or «4» AM 5/0/1142, @Vl Lx w“+ v 5r 2:? (vi/(sergeath Qﬁuv’FEFf 2‘: y Cl): CNN”? *3" 4‘ at)
who a”) ’a 3"" 0M)“: _ (m)  a 3
2(2X)=0 >f'—0 Wisi 2r=0 we Am far 2‘73.
(Jory/4m is a &, Circle df Y M J3 (eu*6vta( 3. (a) (5 points) Find parmnutriv aquatinns of tlw lint: sugmum. from P: (3,2,1) to Q =(1.—1.2) an (ff (me; x; 3+('2)+:3’1“:
f6: 2—5f:
2': [+47 6 one M {w 7W ~ 0516:) (b) (5 points) Find the intersection point. of the yzplane and the line
through P and Q. if it. exists. 4. (It) points) F‘intl synnmzn'ir nquntions of Mn: tangent line to the ('urvo
given by the parametric equations :1: =e’ y=cost, z = 111(1 —t) at the point P: (1.1.0). 8'6 3 I {:5 O «{:0 Jr ‘
~?’w: <6 ' “W” 115 01 5. (10 points) Givun the vuvtur ﬁlmmm r(l) =< «:()sl+lsj11l.‘ xint — lmsl. —l >. ﬁnd the length of its curve for 0 S I S 3. 5 rm, 0)
:f/ZH; < “5&1th {(05%‘f5M) (Use, 0036 + 1' 1 < {(951%) +9”, D>  {2((03t/ 9%» o) U. If thu sphurival mordinutus of a paint. I’ an: , 1r 31.“
(pI61¢) _ (4v Ev T), (a) (5 points) ﬁnd its rectangular coordinates. 8. (10 points) Find the curvatuu: of thu (:m'vv. C given by
r(t) =< 4f. siut+ 2(zost, 25in! — (1)51 > at P = (02—1). ﬁvcr < m COSﬁ'ls’mt’ awstrswc> “(4/2: 4 0, ~5fm€ 3 lash) ,lg‘wg‘F (03,6) 9 ‘J. (It) points) A moving pm‘tiulc starts at tin: point. I’ = (0. L3) with
initial velocity v = i—j +k. If its acceleration is 1(1) 2 i+1212j+0lk,
iiml its position function. 137W 0c +1) 7; + ((M i)j+ @1534);
F0555“ WWW“ “ ‘pr: fV’Ur/olt: £8 D
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MM, ﬁx»): (0; bio) 27/“? "‘4 0 M 67m; 3' é o; I; bi, Fm; agile, {‘CtH) 14341543) 10. Consider the two lines givoii by HM: pm‘amoti‘ic equations: ;r=4t+'2 y=2i+3 z=10t+6 x=s y=—s+’2 z=s+1 (a) (6 points) Find the point of intersection of these two lines. if it exists.
Lé e + L ; 3 50m 1t +5: —_§+l— _'
“2. t
S h
0 {05+é: Sfi (b) (6 points) Find an equation of the plane containing these two lines.
L J A
if 2. /0 : 6, ll
‘Z(>SD) + aw») + (6)C2—/);0
23+ J—Lw— g+zsc9 ...
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