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Sol-161E2-F05

# Sol-161E2-F05 - MA 161 EXAM II Fall 2005 Name ten—digit...

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Unformatted text preview: MA 161 EXAM II Fall 2005 Name ten—digit Student ID number Division and Section Numbers Recitation Instructor Instructions: 1. Fill in all the information requested above and on the scantron sheet. 2. This booklet contains 14 problems, each worth 7 points. You get 2 points if you fully comply with instruction 1. The maximum score is 100 points. 3. For each problem mark your answer on the scantron sheet and also circle it in this booklet. 4. Work only on the pages of this booklet. 5. Books, notes, calculators are not to be used on this test. 6. At the end turn in your exam and scantron sheet to your recitation instructor. b 2i 7 § —5 Y —3 T .0 9 SD ax+b 2.H : 1~ ’ 9(96) cx+d t10119(1) a+b—c—d 7Q «—M f 9 2 #f 4 (cde w (“W C 3. A box with no top has width twice its height and length four times its height. The material for the sides costs \$6/in2 and for the base \$4/in2. If its height is s in. ﬁnd the rate of change of the cost of the box With respect to s in \$ / in. a.523;§; W:LL‘/ xryl‘ b. 623% C: 5 (12'er 4 c.104si (5,2083% C 2 6(2 24¢. + 2’9A-Z.)VL 9(2L'9‘A) e. 863% 4. If f(x) =secx+tanx then f’(x) = SW W 4. gee/ILA, a. sinx + cosx b. cosx+1 \ WM” 1 1 sin2 x ' (Pi-— 2 2 C24 17 c. tanzxsecx Cw ’r‘ d. tanxsecx , inx+1 “t S In >0 w” a cos2x ‘ (0/7 '247 5. If f(a:) = (1+cos2 x)6 then f’ = f , I ‘3‘ QC (X): C ( (14024») 2&4b(~5/6roj 5" :4; (/ +5) r2 _1 L pug/Cw (—gj‘ 6. D125xe_m = D (Joe‘k) : exk~ x éxk a. —(a: — 124)e—‘” L \ )0 Mb (\$_125)e_z D C}: ~Z P J b? c. (a: — 125)e"z - )0 \k‘ d. (a: — 124)e-m e. none of the above 7. Find the tangent line to y = arcsin(a:) at x = 2 _Z _ / J 1’ a.y—6+a: 2 y 2 y :2 7r 2% 1-1:“ / b.y=g+TIE—2 w W . 7T ,— :/ @‘frh‘ﬁ W / J d.y=%+2x—\/§ \ a IT .M‘ c w + L(>< .1) 2 3 V e.y=g+T\/—x—2 / J zu '7. ‘4 ~u .. lg ” 6 4e) 6 H e. none of the above d cosa: _ 9' da:(\$ )- a. \$1+°°S\$(cos :v — sinzvln :B) b. (— sinzv):v°°S"’ c. (coszv x(°°S‘”‘1 d. :I:(°°”_1)(cos :v — :v sinzv In :13) e. 33°05 "31m cos :3 10. Given 3:11 =y‘”, than (1—1] = dzv a l—lnzv ' 1—1ny b. 3311'”3 lnzv y c. (1 ~ 1n :3)— IL' d.(1——1n Q Q CM )1 m, 6 (kink/r L» i sﬂ/b >C XL? 309 [In/v vl—WW 17’ , _ k x4 XL»XLIA7 J . Ina: I, / —] [A X; 3 11. leen f(a:) —— g, then f _ or, Z X _ L y 1 \ a. —— U 6123’:2 '> QN’] X L 0 \ L, L -L( na: > b F b- ﬂE4 (N, g X , A” 0.1—:4lna: In» ’ x dd.” 1 — 2lna: I I ‘ e. none of the above :— -—- §- 4- i “)7 X 12. A particle moves along the curve y = {711 + 3:4. As it reaches the point (2, 3), the y—coordinate is increasing at a rate of 32 Then, the x—coordinate at that instant '5 inc . _ in at a rate of 5 y b.9m /, I I «1‘— y -;(H—Lx) ."Z/j . /=§U7J c. 13.5 %7‘— V / d. 6.75% .X‘ ‘ / 2‘7 2‘): a; l/ ( \$14M] ‘3 e. None of the above H J ' 2 13. The minute hand on a watch is 9 cm long and the hour hand is 4 cm long. How fast, in Ehﬂ, is the distance between the tips of the hands increasing at ten o’clock? al—sﬁ .11617r b33\/§%r’ ) 7 WV c\/6—1 .. , z z . 6\/6—1 e' 11w £K: “Wisheci’ J Fn 6:2“);0 *: gage: [5 ,8: yi‘th/W/ = OH 6" 277.- 21? 2 77// KC 3€‘§’[%T: 350?” IL é ' Der fa 14. The edge of a cube was found to be 20 cm with a possible error in measurement of 0.1 c . Using diifirrontials, the percentage error in the surface area of the cube is e A z 6 )< b. 2% c. 0.5% MA = /; x 91% d. 5% z; k' 61 e. 2.5% 5113‘ [00 : €00 3 "I; a C)” X Z 5 —~ Ado : // 2,0 ...
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Sol-161E2-F05 - MA 161 EXAM II Fall 2005 Name ten—digit...

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