Sol-161E2-F03

# Sol-161E2-F03 - KEY=bACEB EEABA Eb MA 161 & 161E EXAM 2...

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Unformatted text preview: KEY=bACEB EEABA Eb MA 161 & 161E EXAM 2 FALL 2003 1 1 ﬂ _ ‘du ﬂ _ A 3W? ‘/3 ,‘/ 1 .— (l U/ A, 6 B — 6 7 , . v_,T/ -_ “Kw 3 u ﬁx X/U Z (it‘d 3 ‘4 C. 2% ,{ ‘ /6 1 S -2" [A _66u7 2 E. 3% 2. If the line tangent to the curve y = m2 + 3:5 + 1 at (a,b) passes through the point (1, 4), then the possible values of a are ,6,” 1265+} 0and2 8 / 1 B. iandQ - 2 6‘QJ’=(?QQ+_.,~\$ )[¢“@)) \$3a’*‘3¢+1 C. Zandl ~ - ~ D. Oandl ‘ . W Hf?) 1 3 E. lauda- 4. 53—30, -( 3 (2®+3)((~0/) : —2&2—Q/ +3 2 HWﬁb) Q},Zq/ : 0 Mil Q/a09912 MA 161 & 161E EXAM 2 FALL 2003 3- If M33) = f0”) 9(30), f(0) = 1, f’(0) = 3, 9(0) = 1a and h'(0) (1&0): :él(g)8 +<iO/81(0> ‘ I = 3M + 1x 3W0) _——-—‘ *7gyms5w332 pl 74-) material of the wire, I is the length of the wire and A the area of a cross-section. If speciﬁc resistance and cross-sectional area are ﬁxed, the rate of change of R with respect to length is R 2 61—) Q l 4. The electric resistance of a wire is R = where p is the speciﬁc resistance of the léﬁzzg B; 01L A C_p_, .Az 2l D.—7492— p E; 1) MA 161 & 161E EXAM 2 FALL 2003 ' 2 {XL/v 21C 5. lim sm “5 = 2“ 2x sis—>0 tan3a: «‘7 o 5% 3% A’ :12 aw 2‘ ®=e ‘) 5054.21 1 w. :7: / —_ ~ 2 3’ ‘ :2. D =1 Z MB’X A ' 3 ' 6 . / 3% ~— E. does not exist Z,» web/L 6. If the graplfa’f’fis sketched on the right, which is the graph of f’? MA 161 & 161E EXAM 2 FALL 2003 3 3+1 dy _ 7 Hy: t3_1,then\$ _ 3’ ~. 93 '5 3:2: 2 2‘ A. 25) T a J- si 1?: 0—31“ (ti 4} 3 (t3 — 1>4<t3+1>2 yw‘“ \& a 7 3 2 _ 2t2 i + t/ (2} -5) B. ___———3 (t3 _ 1)2(t3+1)4 :‘21 ~ C. 6152 3 ‘ ‘ '5 AW 3 (t3 — 1)4(t3 + 1)2 Kt-) H) _6t2(t3+1) v3 (t3 - 1)2 ®_ ___2tz___ 3 (t3 _ 1)4(t3 + 1)2 8. An equation of the line tangent to the graph of :1: cos y—I—y cos :3 = 1 at the point (0, 1) is - i ) 1 Ce!) ,..- xm ‘ + _ _ ' : (cos1)m+y__1 1‘ 36 0 B. x+y=1 C. —-(sin1)x+y=1 3’,- D 1 ’ v. 'x—y: WOCF “M3, E. (tan1)\$+y=1 due,” 1% “’9’” how we 3-4 = -m1w-a) MA 161 & 161E EXAM 2 9. Evaluate lirr}r tans 13/3: at SA? 8 _ E E 3:1f 3 2 = 1 z.- ”MJ—m g, .1..- - Q I: f N ’4 4 ﬂ :1 / CM 3 < a) 10. If f(:1;) = cosh 2310, then f”(:c) = 61%)»: W294) (2) €111” (Wﬁx) i?) A. C. D. E. 4 cosh 2x B. —2 cosh 22: C. D. 2 sinh 2x —4 sinh 210 E. 2 sinh 4w FALL 2003 2 4 1 E 1 Z x/E MA 161 & 161E EXAM 2 FALL 2003 11. If f(a:) = «569520162 — 7)9, then f/(w) = 2m 2:2 2 1 Age.) = Emma. "Fm Q» (714) A' W” (x ’"9 i3? +2“ 9&2 —7>i 2:1: igmﬂi‘Xﬁé-‘zqt-sze “29¢ 18\$] |___l \$2 2__ 9 _1_ B. ﬁe (m 7) |:2m+2x+gag—7)8 2 \$2 %-4 C. ﬁe (\$2—7)9[—+1+\$2_7 12. A mini-baseball diamond is a square ABCD with side 9 meters. A batter hits the ball at A and runs toward ﬁrst base B with a speed of 2 m/s. At what rate is his distance from third base D increasing when he is two—thirds of the way to ﬁrst base? oLaL 2 = ’3—‘ms 3 2.) ¢=q 4- 96? A' x/13 / C \UI B.2m/s M = J + C. 4 m/s 3L 24 D<~ﬁn£ E.7_1_—3;m/s ...
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## This note was uploaded on 04/07/2008 for the course MA 161 taught by Professor Gabriel during the Spring '08 term at Purdue University-West Lafayette.

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Sol-161E2-F03 - KEY=bACEB EEABA Eb MA 161 & 161E EXAM 2...

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