Solutions to Problem Set 7

Transcription

1 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 Solutios to Problem Set 7 Lte homework policy. Lte work will be ccepted oly with medicl ote or for other Istitute pproved reso. Coopertio policy. You re ecourged to work with others, but the fil write up must be etirely your ow d bsed o your ow uderstdig. You my ot copy other studet s solutios. Ad you should ot refer to otes from study group while writig up your solutios (if you eed to refer to otes from study group, it is t relly your ow uderstdig ). Prt I. These problems re mostly from the tetbook d reiforce the bsic techiques. Occsiolly the solutio to problem will be i the bck of the tetbook. I tht cse, you should work the problem first d oly use the solutio to check your swer. Prt II. These problems re ot tke from the tetbook. They re more difficult d re worth more poits. Whe you re sked to show some fct, you re ot epected to write rigorous solutio i the mthemtici s sese, or tetbook solutio. However, you should write cler rgumet, usig Eglish words d complete seteces, tht would covice typicl Clculus studet. (Ru your rgumet by clssmte; this is good wy to see if your rgumet is resoble.) Also, for the grder s ske, try to keep your swers s short s possible (but do t leve out importt steps). Prt I(0 poits) () ( poits) p. 38, Sectio 9.5, Problem 6 (b) ( poits) p. 348, Sectio 0.4, Problem 0 (c) ( poits) p. 348, Sectio 0.4, Problem 3 (d) ( poits) p. 348, Sectio 0.4, Problem 4 (e) ( poits) p. 350, Sectio 0.5, Problem 0 (f) ( poits) p. 356, Sectio 0.6, Problem (g) ( poits) p. 356, Sectio 0.6, Problem (h) ( poits) p. 36, Sectio 0.7, Problem 4 (i) ( poits) p. 36, Sectio 0.7, Problem () (j) ( poits) p. 36, Sectio 0.7, Problem (b) Solutio to () Sice t α = d t β = +b, we obti θ = β α = t ( +b ) t ( ).

2 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 To locte whe θ tkes o its etrem, we differetite θ with respect to d set the result to +b d (+b) +b zero: dθ = d t d t = ( +b = + (+b). The gle θ tkes + d d ) + + d (+b) + o its mimum vlue whe tht is = ( + b). dθ = 0, i.e. whe ( + ) = +b, or equivletly, = ( + b), Solutio to (b) The rdicl suggests the chge of vribles = sec θ, d = sec θ t θdθ which yields d sec θ t θ = dθ = ( + cos θ)dθ 3 cos θdθ = 3 sec 3 θ t θ θ θ = ( + si θ) + C = + si θ cos θ) + C () 4 ( = cos ( ) + 3 Solutio to (c) The term suggests the substitutio = si θ,d this leds to itegrtio of sec θ t the ed. While this is oky, s log s you kow tht sec θdθ = l( sec θ + t θ ) + C, there re esier wys: we c use either prtil frctios, or iverse substitutio = th t. Method of prtil frctios proceeds s follows: Solutio to (e) We first eed to work o the rdicl + 3. I ttempt to complete the epressio uder the squre root to perfect squre, we otice tht + 3 = = ( + ) 4. This ispires the substitutio u = +, du = d. The our itegrl becomes + 3 u 4 d = du + u + C A B = ( )( + ) = + + The Heviside method (or multiplyig out d lettig =, = ) yields, respectively, = A d = B. Strightforwrd itegrtig, we obti 4 ( + + ) = + l + C Solutio to (d) The rdicl ( ) 3/ suggests the substitutio = si θ, d = cos θdθ, which leds to ( ) 3/ d = ( cos θ) 3/ cos θdθ = sec θ = dθ () = t θ + C = 3

3 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 which, becuse of the rdicl u is scremig for the chge of vribles u = sec θ, du = sec θ t θdθ. The our itegrl equls t θ sec θ t θdθ = t θdθ = (sec θ )dθ sec θ = t θ θ + C = u 4 cos ( ) (3) u = + 3 cos ( ) + C + Solutio to (f ) As the omitor is secod order polyomil d the deomitor is third order oe, we c directly proceed ito decomposig this rtiol epressio ito its prtil frctios: First we fctor the deomitor The fctor is obvious: = ( + + ) d the remiig prt is just ( + ). Therefore we c write A B C = + + ( + ) + ( + ) The Heviside cover up method goes s follows: Coverig terms without / d ssigig = 0, we obti = = A, i.e A = 3. Similrly, coverig terms without /( + ) d (+) ssigig = yields = = C. The other costt B c ot be determied from ( ) Heviside method remember, Heviside method does ot give ll the coefficiets whe there is repeted fctor, like here we hve ( + ). So how do we fid B? I m gld you sked: the esiest wy is to plug i some rbitrry vlue for. To keep the lgebr simple, we put = i B = ( + ) This yields B =. Therefore d = + + d = ( + ) 3 l l + + C + ( + ) + Solutio to (g) We decompose the give rtiol itegrd ito prtil frctios s A B C = + + ( + 4) Multiplyig out yields = A + 4A + B + C. Equtig the coefficiets of, d respectively, we obti A = 4 + C, = B, d 4 = 4A, hece A =. Therefore C = 3 from the first equtio. Usig the fct tht d t ( ) = d + 3

4 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 to itegrte the secod term d mkig the esy substitutio u = + 4 to itegrte the lst term, we obti d = + + d = l ( 3 ( + 4) t ) + l D Solutio to (h) Let u = si(l ), dv = d, the du = cos(l ) d v =. The itegrtio by prts formul the yields si(l )d = si(l ) cos(l )d = si(l ) cos(l )d We eed oly oe more itegrtio by prts, this time with u = cos(l ), du = dv = d, v =, to obti si(l )d = si(l ) [ cos(l ) ( si(l ))d] = si(l ) cos(l ) si(l )d si(l ), (4) Rerrgig this d dividig by two, we obti si(l )d = [si(l ) cos(l )] + C Solutio to (i) We write cos ()d = cos () cos()d d implemet itegrtio by prts with u = cos (), dv = cos()d, d du = ( ) cos () si()d, v = si(). Therefore, cos ()d = cos () si() + ( ) cos () si ()d = cos () si() + ( ) cos ()( cos ())d (5) = cos () si() + ( ) cos ()d ( ) cos ()d Rerrgig this equtio, we obti the sought recursio formul cos ()d = cos () si( ) + cos ()d 4

5 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 Solutio to (j) Usig our result to (i), we c write π/ π/ cos ()d = ( cos () si() + π/ cos ()d) 0 Observe tht cos () si() π/ 0 is zero for. Let us cll π/ A = cos ()d Combied with this observtio, the recursio formul of (i) gives A = A, = = = π/ = 0 (6) Here the vlues hve bee foud directly for = 0 d =. Therefore, A 7 = A 5 = A 3 = A = /35. Prt II(30 poits) Problem (5 poits) This problem sketches systemtic method for fidig tiderivtives of epressios F (si(θ), cos(θ)), where F is frctio of polyomils. ()(8 poits) With the substitutio t(θ/) = z, verify tht, z dz si(θ) =, cos(θ) = z dθ = (7) + z + z, + z Plese try this o your ow first. However, if you re very stuck, tke look t Problem 5E i the course reder. t( Solutio to () Usig the double gle formul for tgets, we hve t θ = θ ) We the t ( θ ) drw the right trigle with two right sides hvig legth z d z. Pythgore theorem yields the hypoteuse: h = ( z ) + (z) = + z + z 4 = ( + z ).Therefore the hypoteuse hs legth h = + z. Now, we c esily red off the vlues z z si θ =, cos θ = + z + z I dditio, dz = d(t( θ )) = ) dθ sec ( θ, therefore θ dz dθ = cos ( )dz = ( + cos θ)dz = ( z + z )dz = (8) + z + z + z 5

6 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 (b)(7 poits) By Emple 3 o p. 57 of the tetbook, the polr equtio of the coic sectio with eccetricity e d focl prmeter p e is, ep r = f (θ) = e cos(θ) Set up the itegrl for the re of the regio bouded by 0 r f (θ) for θ A. Use () to tur this ito itegrl ivolvig oly z = t(θ/). Use the oegtive costt, e b =, + e to simplify your swer. You eed ot evlute the itegrl. Solutio to (b) Rememberig the re formul for curves i polr coordites, the required re is A r A ep Are = dθ = dθ ( e cos θ) We ow plug i the substitutio give by (7) d obti t(a/) e p dz Are = t(/) ( e z ) + z e p + z = dz ( + e) ( e + z ) +e +z (9) Usig the suggested ottio ±b = e, we seprte the two cses depedig o the sig of the +e epressio e. If e <, the epressig the omitor z + = z + b (b ) we hve +e Are = [ e p t( A/) dz (b ) t( A/) dz if 0 (+e) t( /) b +z t( /) (b +z ) I cse e >, we epress the umertor z + = z b + (b + ), d hece Are = ] e. ] if. [ e p t( A/) dz (b t( A/) + ) dz e > (+e) t( /) z b t( /) (z b ) Problem (5 poits) ()(5 poits) Red through Sectio 0.3 (it is oly 3 d hlf pges). Everybody will get credit for this prt. Solutio to () We first get our hds o the tetbook by Simmos. To locte sectio 0.3, we use the cotets prt of the book. We see tht the sectio strts o pge 340. We ow ope pge 340 d red util the ed of the sectio. 6

7 8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 (b)(5 poits) Use Problem () to rewrite the itegrl, si m (θ) cos (θ)dθ, s itegrl ivolvig oly z = t(θ/). Solutio to (b) Agi, pluggig the substitutios i (7) yields z dz si m θ cos θdθ = ( ( z = + z )m + z ) + z m+ z m ( z ) /( + z ) m++ dz. (c)(5 poits) Assume. Use the followig itegrtio by prts, si m (θ) cos (θ)dθ, u = si m (θ) cos (θ), dv = cos(θ)dθ, to fid reductio formul of the form, si m (θ) cos (θ)dθ = F (si(θ), cos(θ) + A si m+ (θ) cos (θ)dθ. Solutio to (c) Oe requiremet i itegrtio by prts is to choose u d dv so tht we c itegrte dv esily to get v.i this problem, oe such choice would be dv = si m θ cos θdθ,so tht v = si m+ θ d the remiig prt is u = cos θ. The du = ( )cos θ si θ. All m+ these with the itegrtio by prts formul yield si m θ cos = m+ sim+ θ cos θ + m+ si m+ θdθ θ cos θdθ Altertively, you c try dv = cos θdθ d u = si m θ cos θ. It works, but is little bit loger. Not to be tured i: Of these three methods for evlutig the itegrl, which would be esiest to use d fstest o em? 7