THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING

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1 OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING MATHEMATICS MA008 Clculus d Lier Algebr for Egieers Professor J.W. Bruce Professor G.J. Murphy Professor M. Styes Dr. S.J. Wills Aswer Questio d y other two questios from Sectio A. Aswer y three questios from Sectio B. All questios crry equl mrks. Use seprte swer books for ech sectio. Time Allowed: Three hours. Mrks my be lost if ll your work is ot clerly show or if you do ot idicte where clcultor hs bee used. Sectio A This questio is compulsory. ) Solve the iitil-vlue problem dy dx y x = xex for x > 0, y) = e + 2. Express your swer i the form y = fx). Pge of 7 Questio cotiued overlef

2 b) Write dow the first 4 terms i the sequece { }, where = ) for Show tht the sequece is mootoic; use this fct d theorem from the otes to show tht the sequece is coverget. c) Fid the itervl of covergece d rdius of covergece of the series You my ssume tht the series is coverget. x + 2) l 3. ) l 2. ) Solve for x: b) Evlute t 2 x x t + 4)t 2) 4 3t 6) 2. c) Use the itermedite vlue theorem to show tht the equtio x = cos x hs solutio betwee x = 0 d x = π/2. Apply the theorem secod time to show tht the solutio lies i smller itervl. d) Use the quotiet rule to fid the derivtive of the fuctio gx) = Simplify your swer s much s possible. si 2x + cos 2x cos 2x e) Use implicit differetitio to fid y d y if xy + sixy) =.. 3. ) Use differetitio to fid the bsolute mximum d bsolute miimum vlues of fx) = x 2 + 4x + 4 o the itervl 4 x 0, d stte where these vlues re ttied. b) Fid ll poits of iflectio of F x) = x / Pge 2 of 7 Questio 3 cotiued o ext pge

3 c) Use the first fudmetl theorem of clculus to fid the derivtive of the fuctio d) Evlute Gx) = π 2 /4 x=π 2 /9 x 2 t= sit 3 ) dt. si x x e) Suppose tht fx) gx) for ll x [, b]. Stte whether ech of the followig two ssertios is true or flse d give brief justifictio for your swers. b b b b i) fx) dx gx) dx ii) fx) dx gx) dx. 4. ) Roughly sketch the regio eclosed by the prbol y 2 = 4x d the lie 4x 2y = 3. By itegrtig with respect to y, fid the re of the regio eclosed by these two curves. b) Let y = x /x for ll x > 0. Use logrithmic differetitio to fid dy/dx. Determie where dy/dx = 0. c) Use prtil frctio decompositio to evlute the itegrl x x 2 + 3x 4 dx. d) The error betwee the trpezoidl rule formul T d the exct vlue of the itegrl b fx) dx stisfies the iequlity b b )3 fx) dx T mx 2 f x). 3 x [,b] Determie how lrge must be i order tht the trpezoidl rule pproximtio to 0 e2x dx is ccurte to withi dx. Pge 3 of 7

4 Solutios to Sectio A. ) This differetil equtio is lier, i.e. of the form dy/dx + P x)y = Qx), with P x) = /x. Followig the stdrd recipe for solvig this type of ODE, P x) dx = x dx = l x so e P x) dx = e l x l x = e = x. Multiply the ODE by the itegrtig fctor x : x dy dx y x 2 = ex, i.e. d dx y x ) = e x. We c ow itegrte: y x = e x dx = e x + C, so y = xe x + Cx. Use y) = e + 2 here: e + 2 =.e + C. = e + C, so C = 2. Thus the fil swer is y = xe x + 2x. b) 2 = = 3 4 4, 3 = 4 4 = = = 5 8, 5 = ) = = 2 3, = = 3 5. As ech term is multiplied by fctor less th to get the ext term, the sequece is decresig d so is mootoic. It is bouded below by 0 sice it is product of positive umbers; hece by theorem from the otes it is coverget. c) As this is power series we must use the rtio test: + = x + 2) + l + ) 3 + )3 + x + 2) l = x + 2 l + ) 3 l + x + 2 l + ) = 3 l +. Now + = + = + 0 =. Pge 4 of 7

5 By l Hôpitls Rule, l + ) l Cosequetly = + + = + = = x ) ) = + x By the rtio test, the series is bsolutely) coverget whe x = + 0 =. <, i.e. x + 2 < 3, i.e. 3 < x + 2 < 3, i.e. 5 < x <. Check edpoits of this itervl: i) t x = 5 the series is 3) l 3 = d we re told tht this is coverget. ii) t x = the series is 3 l 3 = ) l l ; ow l )/ / for 3 d we kow tht /) is diverget it s the hrmoic series), so by the compriso test l )/ is lso diverget. Coclusio: the origil series is coverget for 5 x <. covergece is 5))/2 = 3. Its rdius of 2. ) The expressio x )/x + 2) c chge sig oly where it equls zero or fils to be defied, i.e. oly where x = 0 or x + 2 = 0, i.e. oly t x = d x = 2. These split poits divide the x-xis ito three itervls:, 2), 2, ) d, ). Choose poit from iside ech of these itervls d compute the vlue of the fuctio there to determie its sig; you ll fid tht x )/x +2) 0 whe x < 2 or x. Note here tht x = is llowed the fuctio equls 0 there) but x = 2 is forbidde becuse it would cuse divisio by zero. b) t 2 t + 4)t 2) 4 t + 4) t 2) 2 t + 4) = = 3t 6) 2 t t 2) 2 t 2 9 = 6 9. c) Let fx) = x cos x. The f0) = 0 = < 0, fπ/2) = π/2 0 > 0. Hece by the Itermedite Vlue Theorem, fx) = 0 for some x 0, π/2). Now fπ/4) = π/4 / > 0, so s f0) < 0, by the sme theorem fx) = 0 for some x 0, π/4). Pge 5 of 7

6 d) By the quotiet rule, g [cos 2x][2 cos 2x 2 si 2x] [si 2x + cos 2x][ 2 si 2x] x) = cos 2 2x = 2cos2 2x + si 2 2x) cos 2 2x 2 = cos 2 2x, where s usul we multiplied out the umertor the ccelled some terms. e) Differetitig xy + sixy) = with respect to x, we get so y + xy + [cosxy)][y + xy ] = 0, y[ + cosxy)] = xy [ + cosxy)] d fter ccellig [... ], filly y = y/x. Differetitig this usig the quotiet rule), y = xy y)) x 2 = x y/x) y x 2 = 2y x ) Now fx) = x 2 + 4x + 4, so f x) = 2x + 4. Cosequetly f x) = 0 oly whe x = 2 d f x) is defied for ll x i the itervl [ 4, 0]. Thus the oly other criticl poits re the edpoits of the itervl. Tht is, the criticl poits re x = 4, 2, 0. To check these for bsolute mxim/miim, simply evlute f there: f 4) = 4, f 2) = 0, f0) = 4. From these vlues we see tht the bsolute mximum is 4 d it is ttied t x = 0 d x = 4, while the bsolute miimum is 0 d it is ttied t x = 2. b) Here F x) = x /3 + 2, so F x) is defied for ll x, F x) = /3)x 2/3 d F x) = 2/9)x 5/3. Thus F x) = 0 is impossible d F x) fils to exist whe x = 0. Tht is, x = 0 is the oly cdidte for poit of iflectio. To check whether it is such poit, ote from the formul for F x) tht for x < 0 we hve F x) > 0 d for x > 0 we hve F x) < 0 tke test poits for x < 0 d x > 0 to see this). Sice F x) chges sig t x = 0, therefore x = 0 is poit of iflectio. c) Gx) = x 2 t= sit 3 ) dt. Let u = x 2. The G = u t= sit3 ) dt d by the fudmetl theorem of clculus dg/du = si u 3. Hece by the chi rule dg dx = dg du du dx = si u3 )2x) = 2x si x 6. Pge 6 of 7

7 d) Mke the substitutio thik of LIATE) u = x = x /2. The du/dx = /2)x /2 = /2 x) so 2 du = / x) dx, which is wht ppers i the itegrl. Chge the its of itegrtio: whe x = π 2 /9, the u = π 2 /9 = π/3, d whe x = π 2 /4, the u = π 2 /4 = π/2. Puttig ll this together, the itegrl equls π/2 2 si u du = 2 cos u) = 2 cosπ/2) + 2 cosπ/3) = 0 + =. u=π/3 π/2 u=π/3 e) i) is true thik of res uder curves. ii) is flse if for exmple fx) < gx) < 0 sice the defiite itegrl will the compute egtive res but the bsolute vlue sigs will remove the egtive sig, chgig true iequlity like 3 < 2 ito flse oe 3 < 2). 4. ) Drw the grph by plottig few poits; the prbol is symmetric bout the x-xis d opes to the right. Solve to fid where the curves cross: y 2 = 4x = 2y + 3, so 0 = y 2 2y + 3 = y 3)y + ), givig y = 3 d y =. The stright lie grph c be writte s x = 2y + 3)/4 d the prbol c be writte s x = y 2 /4. Hece the re see your digrm) is fter some clcultio. 3 y= [ 2y y2 4 ] = = 8 3 b) This problem is solved i the MA008 clss otes. c) x x 2 + 3x 4 = x x + 4)x ) = A x B x. Multiply by x + 4)x ) to get x = Ax ) + Bx + 4). Set x = 4: the 5 = 5A + 0, so A = 3. Set x = : the 0 = 0 + 5B, so B = 2. Hece x x 2 + 3x 4 dx = 3 x ) dx = 3 l x l x + C. x d) fx) = e 2x, so f x) = 2e 2x d f x) = 4e 2x. Thus mx x [0,] f x) = 4e 2. From the give formul, we wt Tht is, 0) e )e2 = , i.e Sice is iteger, the smllest vlue tht we c tke is = 4. Pge 7 of 7