CITY UNIVERSITY LONDON Eg (Hos) Degree i Civil Egieerig Eg (Hos) Degree i Civil Egieerig with Surveyig Eg (Hos) Degree i Civil Egieerig with Architecture PART EXAMINATION SOLUTIONS ENGINEERING MATHEMATICS I EX00 Dte: My 005 Time: 00:00 00:00 Aswer FIVE out of EIGHT Questios A relevt formul sheet is provided
Questio () 5 + y f( ) 5, determie (i) the iverse fuctio f ( ) of f ( ) 5 + y f( ) y 5y 5+ (y 5) 5y+ 5 5y+ 5+ f ( ) y5 5 ( mrks) (ii) the mimum possible domis d the correspodig rges of both f ( ) d f ( ). f( ) : Domi {5 }, Rge {5 } f ( ) : Do mi {5 }, Rge {5 } ( mrks) (b) Ivestigte the cotiuity of the fuctio + if h ( ) if < h ( ) is clerly cotiuous t {}. Ivestigte cotiuity t. lim h ( ) lim( + ) h ( ) lim h ( ) lim h ( ) + lim h ( ) lim( ) + + Therefore, h ( ) is discotiuous t with jump of. ( mrks) (c) Determie the followig limits (i) + + + + lim lim + ( mrks) ( + )( ) 0 + + + (ii) + + + + lim lim ( mrks) ( + )( ) 0 (iii) lim lim + + + + ( mrks) (iv) cos si 0 lim, L'Hospitl rule 0 0 cos si ( cos si ) cos si cos si lim lim lim lim 0 0 0 0 0 ( mrks) of pges
Questio () Fid the derivtives d y e cos of the followig fuctios y + d d d ( e ) e (cos ) cos e cos si cos e d d d cos cos cos cos + d ( ) ( ) d + ( + ) ( +) (5 mrks) (b) Fid the slope of the lie which is tget to the curve + y + y 5 t the poit ( y, ) (,). d + + + + + + + d d d d + y d + y d ( y y) 0 y y 0 ( y) ( y) (,) (5 mrks) (c) Fid d y d if t t d y tt. t y d d t d t t 6() ( )( ) 6 6 + d d d t ( t) ( t) d y t t t t t (5 mrks) of pges
Questio (cot.) (d) A ope cylidricl c is to hve give volume V. Fid the dimesios of the c if the lest mout of mteril is to be used i mkig the c. r h V π + + M πr + πrh V r h h V V π r M πr πr πr π r r dm dr V dm πr πr πh π( r h ). Therefore, 0 for r h r dr dm dr M V π + > 0 r. Therefore M is miimum for r h d r V π π π mi r (5 mrks) Questio () Determie (i) the first order prtil derivtives of f y y y (,, θ ) + cos θ f f f ycos θ, y cos θ, ysi θ, y θ ( mrks) of pges
Questio (cot.) (ii) z if z + ysi z. z ( z + y si z) 0 z + z + y cos z ( z) 0 z z z z + z + z + y z z + y z z zy z z z + zycosz z+ ycosz ( mrks) ( ) cos 0 ( cos ) cos (b) u + y y, d r s y r s +, +, fid u r d u s. u u u y + + r r y r ( y) r (y ) u u u y + y + y s s y s ( ) ( ) s (5 mrks) (c) (i) Determie the totl differetil of the fuctio g y+ y y g g dg d + ( y + y y ) d + ( + y 9 y ) y ( mrks) (ii) The rdius of right circulr coe is mesured s 5 cm with possible error of 0.0 cm, d the ltitude s 8 cm with possible error of 0.05 cm. Fid the mimum possible percetge error i the volume s computed from these mesuremets. (Hit: V π r h). π V V π π + + dv dr + dh r h V r h V r h dv dr dh rhdr r dh Therefore, V r h 0.0 0.05 + +.% V r h 5 8 (8 mrks) of pges
Questio () The dmped oscilltio of vibrtig block is give by Re z, z e + ( 0. 0.5 it ) i terms of the time t. (i) Fid, d determie the vlues of t where is zero. ( 0.+ 0.5 it ) 0.t 0.5it 0.t z e e e e t+ i t 0.t Re z e cos0.5t (cos 0.5 si 0.5 ) π 0 cos(0.5 t) 0 0.5 t + kπ, k 0,, t π + kπ, k 0,, (ii) Fid the velocity of the block s d tht the swers re the sme. d s Re dz d cofirm ( mrks) 0.t d 0.t 0.t ( t) e cos 0.5t 0.e cos 0.5t0.5e si 0.5t dz zt e ie ie t i t dz d e t e t ( mrks) ( 0.+ 0.5 it ) ( 0.+ 0.5 it ) 0.t ( ) ( 0. + 0.5 ) ( 0. + 0.5 ) (cos 0.5 + si 0.5 ) 0.t 0.t Re 0. cos 0.5 0.5 si 0.5 (b) Solve the equtio z + i z + i z + i, z r (-) + d z ϕt 0 0 i 0 + k60 So, z e d z, k 0,,, - Therefore, z 0, z 0, z 0, z 00, (c) Prove tht, if z cosθ + isiθ d y positive iteger, (i) (ii) z z + z cosθ z si i θ iθ ± ± iθ z cosθ + isiθ e z e cos θ ± isi θ So, z + cos θ + isi θ + cos θ isi θ cos θ d z z cos θ + isi θ cos θ + isi θ isi θ z (8 mrks) (5 mrks) 5 of pges
Questio 5 () The positio vectors of t he three poits A,, d C re, b, d +b respectively. Epress the vectors A, C, d CA i terms of d b d prove tht A + C + CA 0. A( ), ( b), C(+ b) A b C + b b + b CA bb A + C + CA b + + b b 0 ( mrks) (b) Show tht the equtios + y+ 8 z+ 5 5 y z 0 6 represet the sme lie. (,5,) 0.5(,0,6), so the lies hve the sme directio. Ivestigte the poit (, y, z ) (, 8, 5) of the first lie with respect to the secod. y 8 y z. 0 0 0 6 z 5 6 6 Therefore (, y, z ) (, 8, 5) is poit of the secod lie s well. So, the two lies hve the sme directio d pss from the sme poit; hece, they coicide. ( mrks) (c) Cosider the three positio vectors ( ), b (0 ), d c ( ). Ivestigte whether the vectors b,, d c re coplr or ot. For the vectors, b, d c to be coplr, ( b c) must be equl to zero. ( b c) 0 + ( ) 6 0. Therefore, b,, d c re ot coplr. (8 mrks) 6 of pges
(d) r cosωt+ bsiωt, where d b re costt vectors, show tht r cosωt+ bsiωt dr ωsiωt + bωcosωt d r ω cosωt bω si ωt ω ( cosωt+ bsi ωt) ω d r Therefore, + ω r 0. r d r + ω r 0 (5 mrks) Questio 6 b () A d b p s d q r. p q r s, show tht A A either whe b 0 or whe b 0, b p q p + br q + bs A b r s bp + r bq + s p q b p+ qb pb+ q A r s b r + sb rb + s p + br p + qb b( r q) 0 q + bs pb + q b( p s) 0 bp + r r + sb b( p s) 0 bq + s rb + s b( q r) 0 So, either b 0, or q r d ps. (5 mrks) (b) Fid the vlues of λ which stisfy the equtio λ 0 λ 0. 0 λ λ 0 λ 0 λ λ λ λ λ λ λ λ λ 0 λ ( ) ( ) ( ) Therefore, λ 0, λ, ±. 0. (5 mrks) 7 of pges
Questio 6 (cot.) (c) Solve, usig mtri iversio, the followig system of equtios + y+ z y+ z y+ z It is of the form A y y b A z z b, where A d b. The cofctors of A re c, 0, c c c, c, c c, c, c 5 Therefore, dja 0 d det A c + c + c. 5 5 So, dj y A 0 A b b det. A z 5 5 (0 mrks) Questio 7 () Determie the followig itegrls (i) d + 0 + d d d d d l + + + + + 0 0 0 0 0 (l l) 0.069 0 0 ( mrks) 8 of pges
(ii) d + A + + ( + ) + A, + 0 So, d d l l c + + + + (iii) cos d (6 mrks) Itegrte by prts. Let u d v cos, therefore du d d v si. Hece cos d si si d si + cos c + ( mrks) (b) Fid the re bouded by the prbol The two curves itersect for The lie y + y d the lie y +. 8 0, + + is bove the prbol y, therefore, 0 ( + ) d + + 8 uits. 6 (6 mrks) Questio 8 () Solve the differetil equtio d y +. d d y d y y c d + y + + + + d For the itegrl we use the followig substitutio + u so d udu. + d Therefore, du u + +. y Hece the solutio to the differetil equtio is + + c. (6 mrks) 9 of pges
() Fid the geerl solutio of the differetil equtio y d +. This is lier differetil equtio with p ( ) d q ( ). Use the itegrtig. p( ) d d fctor µ ( ) e e e. So, yµ ( ) q( ) µ ( ) d ye e d Evlute e dby prts with u d v e, so u d v e. Therefore e d e ed e e + c. Hece ye e e + c y + ce. (6 mrks) (b) Solve the iitil vlue problem d y + 6y 0, y(0) d y (0). d d The chrcteristic equtio is λ + λ 6 0 λ d λ. Therefore y( ) ce + ce d y ( ) ce + ce. So, y(0) c + c d y (0) c + c. Solvig these two equtios gives c d c d the solutio to differetil equtio becomes y ( ) e + e. (8 mrks) Iterl Emier: Dr E Miloidis Eterl Emiers: 0 of pges
FORMULAE IN THE DIFFERENTIAL AND INTEGRAL CALCULUS sic Derivtives y, d y siθ, cosθ d y cosθ, siθ d y tθ, sec θ d y cotθ, cosec θ d y secθ, y cos ecθ, y si, y cos, y t, y cot, y sec, cosec, y siθ t θ secθ d cos θ d d d d d d d cosθ cotθ cosecθ si θ + + y e, y e, y, y l, d e e d log d d iomil Theorem ( ) ( )( ) ± ) ± + ± ( +... of pges
Mcluri s Theorem f ( ) f (0) + f (0) + f (0) + f (0) +...!! Tylor s Theorem ( ) ( ) f ( ) f ( ) + ( ) f ( ) + f ( ) + f ( ) +...!! Product y uv, Quotiet u y, v d d du v d du v d + u u v dv d dv d Formul for Itegrtio by Prts udv uv vdu sic Itegrls + d + ( ) cos θ dθ siθ si θ dθ cosθ sec θ dθ tθ cos ec θ dθ cotθ t θ secθ dθ secθ cot θ cosecθ dθ cosecθ d d d + d + si cos t cot of pges
d d sec cosec e d e e d d d e l l sih d cosh cosh d sih sec h d th d + d d + sih or log + cosh or log th or log + + Trpezoidl Rule 0 h f ( ) d ( f 0 + f + f + f +... + f + f Simpso s Rule 0 h f ( ) d [ f 0 + ( f + f + ) + ( f + f ) + ) + f ) HYPEROLIC FUNCTIONS sih cosh ( e e ( e + e ) ) of pges
SOME TRIGONOMETRIC FORMULAE si( A + ) si Acos + cos Asi si( A ) si Acos cos Asi cos( A + ) cos Acos si Asi cos( A ) cos Acos + si Asi FORMULAE FOR HYPEROLIC FUNCTIONS sih( A + ) sih Acosh + cosh Asih sih( A ) sih Acosh cosh Asih cosh( A + ) cosh Acosh + sih Asih cosh( A ) cosh Acosh sih Asih of pges