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solutions5

# solutions5 - MATH 138 Calculus 2 Fall 2010 Solutions#5...

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Unformatted text preview: MATH 138 Calculus 2 Fall 2010 Solutions #5 iKQ) :l:=e'—l,y=ea'. y=(¢:')2=(:n+l)2and since a: > —1, we have the right side of the parabola y=(m+1)2- oé‘té ZTF oé'L CZTFA‘A W 4/?- 9) (a) :c = 1+ e‘, y = t2. —3 g t 5 3. dI/dt = e" and dyfdt = 2r. sown-10?}: + (@3th = e“ + 4121. Thus,L = f: ./(d:c,rdz)= + {dy/‘thdt = I; V’s"?! +‘4't'2‘dt ' 7; “9 d1: 1 dy 1 da: 2 dy 2 1+1_t2+4t+4 2 = C —:-— _ . _ =__y a: Inty W+1.15t_5. dt tandd_t_2t+1‘so(df) +(dt t? +4(i‘+1} 4t2{t+1)' b 2 2 5 2 5 3 d1 dy t +4t+4 (H2) Th ,L: — — di= = “S f d + dt 4t=(t+1}‘“. 1 {202mm 4m x:e‘+e“‘.y=5—2t.0_<_rgs. drfdt = e' i 9" and rig/d! = —2.50 d 2 d 3 . — -° (5—7) +(ﬁf) =(‘2’-—2+E 2t+4=€va2+e "=(Er4—P—rﬁand L = Eh" 4.» £"'}rﬂ = [1" — 9“]: = e3 — 9—3 _ {1 7 1} : r_,.1 _ E—s_ —————__ 4ND) ’1:Lo\$[1t\) (3.: smbt) (E L -—a W‘WQ, £33;— —<; mum) IL 6‘? \‘Vkﬂmkf = 4‘3“] (11X) 1‘ 4.4.05.1 (2'1\ :1 4 1‘ <5 L Sﬁawgzawm _.__) 6‘ -‘:ir\o\ ‘\:c .SuLl/x W021 PHI—Elie”: (1,0) (8&0, Sinia) : [110) :3) elm: 11. M4 Sink; :0 1‘3"qu 'F/ro)= (i :L) - quqm6:tc_ *e‘bMAi-‘mq of +L¢ hag—cg} IIHC T) ._'Fl4co H M ta [- Ho) 2 (no) +(t*0)‘ (111) =(£+1,t) OIA Car*e:iam etudﬁ‘m 9% 4‘.qu {-HKELW‘f 114% 7 5: iii-xv! ) Us -.~_ 2‘imiw‘1+€ 1: 1) ‘3 '-‘- 3L“ 1L. ml“, Catrfcufam egg/m of 749 79%.“? (am—4. 342—), 7:75 Q x». ﬁxer/(9J4 white amm‘iéaﬂa am VW— J; /E, 7 J, :9 7kg Cadeném +35,“ “(4 +141 fawraag am ms = :2- whom jg ‘F'ivucl, "t gunk *rhcxi P|*\ :— ZOUO) ngU'e .L loo eLgQMLVC : 1WD 63’“— W ______£__;_5_ ,Q/nD‘D __ ...
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