Session Trimester 2. Module Code: MATH08001 MATHEMATICS FOR DESIGN

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1 School of Science & Sport Pisley Cmpus Session 05-6 Trimester Module Code: MATH0800 MATHEMATICS FOR DESIGN Dte: 0 th My 06 Time: Instructions to Cndidtes:. Answer ALL questions in Section A. Section A is worth 0 mrks in totl.. Answer ny TWO questions in Section B. Ech question in Section B is worth 0 mrks. If you ttempt ll three questions in Section B, then the two highest mrks will count towrds your Emintion mrk. 3. Clcultors must not be used to store tet nd/or formule nor be cpble of communiction. Invigiltors my require clcultors to be reset. 4. Formule sheets re ttched to the end of this pper. Pge of 7 Continued overlef

2 SECTION A Attempt ALL Questions Q. Consider the line with prmetric equtions = 4 + 3t, y = + 8t, z = 3t, nd the plne 7 5y + z = 57. Obtin the co-ordintes of the point of intersection between the line nd the plne. Q. Given u(, y) = y + sin(3 + y), show tht u u 3 y y is constnt. 3 4y dy d. 0 Q.3 Evlute the double integrl Q.4 Obtin the prticulr solution of the homogeneous liner second order dq dq differentil eqution 99q 0 d d, for which q = 4 nd dq 6 d when = 0. Q.5 The Physics progrmme t University hs 30 second yer students, of whom 8 re direct entry students. The remining students hve progressed from first yer. For survey on student opinion of new ctering fcilities t the University cmpus, two students re selected, t rndom, from the second yer of the Physics progrmme. () (b) Drw tree digrm to show ll possible outcomes, by type of student (i.e. direct entry or otherwise), of the two students selected. Ech brnch, nd ech possible outcome, must be clerly identified, nd ssigned probbility. Clculte, to three deciml plces, the probbility tht ectly one direct entry student ws selected from the second yer Physics cohort. () Pge of 7

3 SECTION B Attempt ny TWO questions Q.6 () Consider the plnes P : + 3y + 4z = 30 nd P : + 6y 5z = 8. Stte norml vector to the plne P, nd hence obtin the eqution of the prllel plne contining the point (3, 7, 6). By first setting = 0 in the equtions of the plnes P nd P, obtin the co-ordintes of point tht lies in both plnes. Similrly, by setting y = 0 in both equtions, obtin the co-ordintes of second such point. Hence, obtin direction vector for the line of intersection of the two plnes. Show tht the line L: = 7 + 3t, y = 4 + 6t, z = 6t does not intersect the plne P. () () (b) Consider the inhomogeneous liner second order differentil eqution dy dy 3t 36y 54e. dt dt Show tht the complementry function for the differentil eqution is given by CF(t) = e 6t (A + Bt), where A nd B re constnts. Obtin the prticulr integrl for the differentil eqution. Further, stte the generl solution of the differentil eqution. () Hence, obtin the prticulr solution of the differentil eqution for which y = 9 nd dy 6 dt when t = 0. Pge 3 of 7

4 Q.7 () Concrete blocks re mnufctured ccording to minimum strength requirements. Strength tests re crried out to determine whether such blocks meet these requirements. On the evidence of mny such tests, it hs been observed tht the probbility tht rndomly tested block does not meet the requirements is Assuming this to be the cse, clculte, to three significnt figures the probbility tht, in smple of 0 blocks tested: ectly one block does not meet the requirements; more thn two blocks do not meet the requirements. () (b) Consider surfce S which hs the eqution z = y k 3 y, where k is constnt. Obtin the vlue of k for which the point P(,, 8) lies on such surfce S. Writing f(, y, z) = y k 3 y z, obtin grd f for the vlue of k obtined in. Hence, obtin norml to this surfce S t the point P. () (c) By first seprting vribles, solve the first order differentil eqution dy 3y 4 e sin d, for which y = 0 when = 0. (d) Consider the liner first order differentil eqution dy 4 y t 6. dt t Show tht suitble integrting fctor for this differentil eqution is μ(t) =. 4 t Hence, obtin the prticulr solution of the differentil eqution for which y() = 0. () Pge 4 of 7

5 Q.8 () Consider the double integrl y 6e dy d. 0 Evlute the double integrl, correct to two deciml plces. Sketch the region in the -y plne defined by the limits of integrtion. With reference to your sketch, give reson why reversing the order of integrtion would led to more difficult problem. () () (b) The surfce S hs the eqution z = 3 3y + y Show tht the co-ordintes of the criticl points on the surfce S stisfy the pir of simultneous equtions. = y nd y =. Solve this pir of simultneous equtions, nd hence obtin the co-ordintes of ll criticl points on the surfce S. Obtin the Hessin H, nd hence clssify the nture of these criticl points. END OF EXAM PAPER (Emintion Formule sheet overlef) Pge 5 of 7

6 Emintion formule Binomil distribution P(X = ) = n C p ( p) n = n! p ( p) n, for = 0,,,.., n.! (n )! Integrting fctor differentil equtions dy Py Q d cn be solved using integrting fctor y Qd. e Pd s Criticl points of functions of two vribles z = f(, y) Loction: where both f f 0 nd 0. y f fy y Nture: Evlute the Hessin, H = f y fyy f f t the criticl point. y y If H < 0, criticl point is sddle point f H > 0 nd < 0, criticl point is locl mimum f H > 0 nd > 0, criticl point is locl minimum H = 0, test fils; further tests re required. f f Rules of Differentition If u nd v re ny functions of vrible nd nd b re constnts, then Linerity D(u + bv) = D(u) + bd(v) Product D(uv) = vd(u) + ud(v) Quotient u vd(u) ud(v) D v v Chin Dfu fudu d u bv du b dv d d d d uv v du u dv d d d du dv v u d u d d d v v dy dy du d du d Pge 6 of 7

7 Stndrd Derivtives nd Integrls, b, n nd c re constnts throughout Function f() Derivtive f Integrl f()d 0 + c n n n n c, n n ln c sin cos cos + c cos sin sin + c tn sec not stndrd ep() = e ep() = e ep() + c = e + c ln not stndrd Function f() Derivtive f Integrl f()d 0 + c n b n b n n b n c, n b b b b ln b c sin( + b) cos( + b) cos( + b) + c cos( + b) sin( + b) sin( + b) + c tn( + b) sec ( + b) not stndrd ep b c e c ep( + b) = e + b ep( + b) = e + b b ln( + b) b not stndrd Pge 7 of 7

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