Midterm Exam 2 with Answer Key - Calculus and Analytic Geometry II | MATH 152, Exams of Analytical Geometry and Calculus

Download Midterm Exam 2 with Answer Key - Calculus and Analytic Geometry II | MATH 152 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Math 152 SanePle Second Midterm Name ptuur" l{u"h 1) Complete the foilowing integrais from your repertoire (and a.few more). (Most of these you ' shorrid be able to write"down from memory without straining') [ ,n d., : -!.- y^+l + C-.l * n* l ' | [*-' ls= h' *-F'iff;Lau#ry'' a? ' t ;P* E ' ; ' - ( ' ' v u4 t t t f / 1 / t"(r)d,x: XlL,"6c1 -X rC- T f t I sin(ar)d,r: - eCPsG-x) fC- F 1 I I , I t , a T : I r:, ---:T'* 4J t \ / r ' -o " - t / ry\ sec ra-) (" * -t) I t^^1*1d,x : fT J;"+adr: | ,or(or) d,r : | ,.,(,) d'r -- t-Ldr: I Js'z - az l r"r '(x)dr - +c t A,stei ?to * 4^*"n1 t o v<'t frovt, / %).Ltu tU Ia - /-,..,1 artxl +c - J*" I snoxl + c - l I ' (o'-vY g4t,t(*9 + C r t l t ,Ul snLU)tt'""@) \ - t /& \ z r 5t va lal t \- tt'rr.G) +L 2) compute the derivat ive of /(c) - logz(sin-1 (zr)) + 2" '( lcosrl)" ^ , -- l 4 - | fur. ,p,,.-,("r;' (zx)) + tuet n xgnlc'os'x! 3' , M L.Y.- :. k-, fu) ffi- .z + t ne' r:? !'tr! . !^t u snl - x t""N) l! -f t(j*r*lY(t^'ilr-u,L-lcosnl -xr'"*4 - j/'*&' f' , ' - /..\ r: rz ^.rE I'^<x) t3) Compute I c3 logr(r)dz : JL 'X @4Jz I rz e , , - . , - = TMC) ). x'-L'-@ "V 7*vT; " .w=4Wl o, =,x-/-x = fu, + "-A*c",!: - h" * fnt&tu^ - + dv'7) = * xo' e * *.!-ct _hL@) _h=+*o!: = L AW) 8l='- ,-L^ : gl-- lt-Lnel - Gz- F + !..,^{=1 16!^,,@3T!43) q -TZ,- E@l and | 4 'dx I ,t': f,t.*fu,oq = h .x'a-"-4 li ( f e*& '=-o-{?.) ' t aryl3- a+-( \ oi*t"a'=)tw(i l -1 7 =@1 r i1 =-re\ = G3----- @t 4) Use l'H6spital's Rule to compute the following iimits ,*do) ".il*Ffi*#_ , 'j*l'byr;x!i_(t+s/*) r:gJ*t ' , - '@ = ex^1i - ' X + . o o..,. zsi^ex)y1e(f).s ;,pW ffi,F * Vo>sGiF) Z;^6x7 : lz!.;'^ tre= -+-2il syex_ c e" T-./iffs- | ='Uz*o,;F{ fT='@ 6 t n t / 3 e. l im J^' "'5i' ln(r)-a'l+ n ,'! !r;^ + ,& + !/s 3 7. pa,,', + &+'P o :cs - - - / ) r , , l J - t t l z Ol r ' r . - - c--+n /2 Sln 0 ,l I ,-\ (not a mistake!) dees r"-of So-ti=EQpoi,nscj 1 ,P-'t#trotfu! - l Y ' v I L 4 e .Oa,& F % - -X é J ta0%(2) de ms ee Le ae A es =! 3 = le CE RNG ae ace Ce i wea aed 3 ) 2 te Ovrvs = a= + — SS seraice Sitti aoe eee “Ching = P ee xe AK Ke Yas =| = FRYE IAS CR) OO eal Rot de ee sb Leet aa Bx" tan (x) — £4 FS kn (te ae MOE we er £ sec71(zx) td = Kd y Has -1 . ( RV OF a patter — St cs e ee Gee . Co eor Ce iste — a Sub aw FER (ey, Biro z+1 & ie ce Se In The eet t Ju the second, *« (¢ ee is paamgeede: Fiber Bi betwec= ating EMcet sede a (oe = 4510s (6) dee = 5 ae Cai f F » cee 2ceslhyAt use aiden + a aareen: Roost dt = —J), 72. - sf du +t f (aseeae At ee a eee eae aes a7 ie S (bac2é) +1) Seco Ae — doaamceyy a _ oe subst D= tar) BC i Ce gee Av = sect pdt ed Ce eee te CE art) tanttec As Fe P4134 _—yAt Ze =) His Se ee |» h. [tse ) 43 [4 Ke ata NG aes Pte M + mee! ee ey tale abe 22 4 as ae Rr Kee Xe pan! ( ee 4 4 dye =f \aeaeee've Subst ty cy Pa ok BE 2 fae doe pe eee oe = 58 ae TE 1 ais Seria eave e oF Si 2 S Gay 24x 7 te == Shine pus TERE EU Ot De a) i M the ceeond 2M de hy 9) Compute and simplify fitan-l(tlr) and compare with the derivative of tan-l(z). What does that tell you about the relationship between tan-r (t l") and-tan-r (c). Now substitute x: I to calcuiate the constant. Next substitute r,: -1 to calculate the constant. Explain *W'?tl5':-" i'*::":: :2 Ti' =i-w, + #,:htw'*;er- ' F t I+C/ f ) -Tr,n-^--s +;\ ( /+) J) - tnit,(x) 6c,D e- *t^o- 5a'4"-<-et- Ap-vivoh"te'-t 5o -rr'tz ,ry# +a;t ( t/*) : - i-a;'@) + L (C: ca*',st -*), fn^ K=tt d'(t/D =E: -+A;'(t)+C:-+1c-,to c- +., B,r* lor^ {.=- t +a',4( n/-t )= 4 z - +'^^"tGD +C = -++C25o L--:E 11^.,^4 L rA V;t o- un\st.^*, 5.o dLe-r< L t1^4- oUs"-rre-p*+<fir-Ilrvr- -T1,,-spv-e-,'.- 5%1 Vo frh-9'"7I . , - . - i A)++^",v' bq a- on?t"^^t, ,p roVi dq) t14.- d-e*iu t,a Mr- irtuv"k i6f l,r-r-.: +t's -F t6'*' (n/4 hl|aLgfs i,,t<^rv&t, oGw\tJ,u\ : e*,, o) u (or*c.). 10) write out the FORM of the partial fraction expansion for f r3+5 ) . J f f io , (DoNoTcomputethecoef f ic ientsof the in tegra1s) : (-A 4-B +.c - *PJ ,y-7 ' (X-zY' (X-z)z ' X*, t Q'E|,A +'F-++ 11) Simplify each of the following : tan(sin- l (r lZ)) : L v4_72_ J -a - +' @+,;= ' ry_+ (K-++) - F @-rF v+ - "frz- cos(2 sin-r (r)) : 1 =1 ,l I 25t* ( siwtx) z ( ain ( sr ' i t 'K) )- .>x2^ /h^-bL ^<{L{-"^"*L cr's@>) - lu- z si,'t (*) sin(sin-l (,) - cos-l(")) : sivt(sin' t ) :"Tr-f-6##Fffir"*-*, 4t '"ii# f ,(1 -ix'- = Ay-Lt Gt <.y';1T t tan- l (") + tan-1 ( t l r ) where c ) 0 to;'W) ++c,,'t (a/x) _ E +*^ 'X7o ..-V) ? v'llLr;f tv'io,-,-aee- yt tO"! L""j^'.^) (s*-"o-Qso PT ol&r\^^- 1) -r g/#) f z) U;e the parts s - (lnr)- and d,u = d,x to derive areduction formulaforf J O"o)* dr in terms a lO"r)*-t d,x ry:% )) ^ d/r- - t{-L*F)^-,,+*o Crr t \h^- | , r k )W*)*"\'r = xQL^-.x)*- f x, w.U^ *)^-, i-U f@n)* "V = xU'**)^ - wt- f A_,^_x)^-t & C..r=4 o-K f3) There exist constants I' and B so ihat the foilowing reduction formula holds :f J , "Q - r ' ) * d .x : Aan*tQ - r r )^ * B [ ,n( I - r r )^ - , d , compute the constants '4 and -B and siro#that the reduction formula does holcl.2 t #.- <^^fi' 4 b ot'L ;-;ol,o S x^ ( r - xz) ̂: u+ ) A*; o_x.)\ ; [!rT:;;f:- ) ̂ - ! a- zx) = (w+ r) A Fn (r_ t) zv,,A F"( , _*) L,q-t xz D t',., t dq b otL *i J,--s 4 ?n (,,: U:!i, ;?:;, 1 * K2- = (^+ t) A ( | - X-) - Zr.^.24 y, + )lS5o l-K'= Qt+t)A -t B (z^+w+t) nxz LWWre) l-f :('?^nn+^t)A=-t a'-J A: #(A , r> 21l4+vt* Ia*4 (n+t)A + B *4 B:1- h*(-n-zw\ f *" ( r - f)^.rn = t*iilT x^*t ( p *)^+ ffi*, f x^ (r- fl'-L