Solved Question Paper 5-6 B.Tech. Theory Eamination (Semester IV) 5-6 Engineering Mathematics III Time : hours] [Maimum Marks : Section-A. Attempt all questions of this section. Each question carry equal marks. ( ) (a) Write the cauchy's Reimaun conditions in polar coordinates system. Ans. C R conditions in polar coordinates are (i) u v r r (ii) u v r r (b) Write the statement of generalized cauchy's integral formula for n th derivative of an analytic function at the point Z Z. Ans. n n! f( z) f ( z ) dz i n ( z z ) C (c) Find the Z transform of U n {a n } Ans. Z-transform of U n {a n } Z [{U n }] n n az n a a a... z z z z a z a z n (sum of geometrical series i a r ) (d) Write the normal equations to fit a curve y a + b by least square method. Ans. Normal equations of y a + b y a + nb y a 4 + b (e) If covariance between and y variable is and the variance of and y are respectively 6 and 9, find the coefficient of correlation. Ans. Covariance (, y) XY X n Y y y Variance of X 6 n Variance of y Y 9 n covariance (, y) Coefficient of correlation variance variance y 6 9 4.8 Ans.
ii Introduction to Engineering Mathematics III (f) The regression equations calculated from a given set of observations for two random variables are:.4y + 6.4 and y.6 + 4.6 calculate mean values of and y. Ans. The regression equations are.4y + 6.4...(i) y.6 + 4.6...(ii) Both lines pass through, y, so.4y 6.4 y.6 4.6 On solving (iii) and (iv), we get 6, y Ans. (g) Write the Netwon's Raphson iterative formula to find the value of N. Ans. N N f () N, f () By Netwon's Raphson formula is f ( n+ ) n n f( n) n N n f( n) n n + N n Ans. (h) Find the missing data in the given table: f() 58 556 465 Solution f() 58 556 465 y y y y 58 4 556 y 5 y 556 y + 55 y y + 465 465 y y y + 55 y 57.67 Missing data 57.67 (i) If f () is given in following table:.5 f().8.5 Ans. then using trapezoidal rule, evaluate f( ) d....(iii)...(iv)
Solved Question Paper 5-6 iii Solution.5 f().8.5 y y y y h f( ) d [( y y ) y ].5 [(.5).8].5 [.5.6].78 Ans. (j) Find the third forward difference with the arguments, 4, 6, 8 of the function f (). Solution f () f () f () f () 4 5 4 56 96 48 48 6 4 44 9 8 496 Third forward difference 48. Section-B. Attempt any five questions from this section. ( 5 5) 7z 9z8 (a) Find the Laurent series for the function f (z), z is comple variable valid z 9z for the regions (i) < z < (ii) z > Solution f (z) 7z 9z 8 z 9z 7z 9z 8 z ( z )( z ) 4 z z z 4 (i) When < z <, f (z) z z z f (z) z 4 z z z z z 4 z z z...... z 9 7 9 7 (By partial fractions)
iv Introduction to Engineering Mathematics III z z z 4 4z 4z 4z...... z 9 7 8 9 7 8 (ii) When z >, 5z z 5z... z 9 7 8 f (z) 4 z z z 4 z z z z z Ans. 4 z z z z z 9 7 4 9 7...... z z z z z z z z z 9 7 4 6 8...... z z 4 z 4 z z z z z z 7 9 45 8 4... z z z z d (b) Using calculus of residue, evaluate the following integral ( a ) d Solution ( a ) Consider the integral f( z) dz where f( z) c ( a z ) Poles of f(z) are given by putting denominator equal to zero. (a + z ) (a + z ) (a + z ) z ± ai, ± ai Since there is no pole on real ais, therefore we may take the closed contour C consisting of the semicircle C R which is the upper half of a large circle z R, and the real ais from R to R. Here by Cauchy's residues theorem, we have R R R Ans. f() z dz f( ) d f() z i (sum of residues) CR dz d i (sum of residues) R ( a ) CR ( a z ) The only pole within the contour C is z ai, and is order. Here f (z) ( z) ( z ai) ( z ai) ( z ai) ai...() c R Where (z) () z ( z ai) ( z ai) R R
Solved Question Paper 5-6 v Residue at the double pole (z ai) i ( ai) ( ai) 4a and dz dz Rd dz ( a z ) a z ( z a ) ( R a ) CR CR C R R i and Rsince z Re ( R a ) Here, when R, relation () becomes. i d i (residue) i ( a ) 4a a d Ans. ( a ) 4a as (c) Find the inverse Fourier sine transform of e Solution f() f()sin s e as s ds sin sds as e. ( cos s asin s) a..cos a. a a. as e sin sds (d) Using least square method, fit a second degree polynomial from the following data: 4 5 6 7 8 y..5. 8. 7. 8. 7.5 8.5 9. Also estimate y at 6.5 Solution Let the equation of the curve be y a + b + c y y y 4..5.5.5.. 4 4. 8 6 8. 4. 9 7. 7 8 4 7. 8. 6. 64 56 5 8. 4. 5. 5 65 6 7.5 45. 6 7 6 96 7 8.5 59.5 49 46.5 4 4 8 9. 7 64 576 5 496 6 y 8.5 y 99 4 y 697 96 4 877
vi Introduction to Engineering Mathematics III Normal equations are y 9a b c y a b c 4 y a b c 8.5 9a + 6b + 4c...() 99 6a + 4b + 96c...() 697 4a + 96b + 877c...() On solving these equations, we get a.4, b, c. Equation of the curve is y.4 +. y(6.5).4 6.5 +. (6.5) 7.9 Ans. (e) For the following data, calculate the finite differences and obtain the forward and backward difference polynomials. Also interpolate at.5 and.5....4.5 f ().4.56.76..8 Solution f() f() f() (f) 4 (f)..4.6..56.4...76.4.4.4..4.8.5.8 By forward difference formula a + nh, a., h.. + n(.) n n f (a + nh) f () n( n ) f( a) nf( a) f( a)...!.. () ().4 ( )(.6) (.4).4 +.6.6 + 6 +.4 f () +.8 f (.5) (.5) + (.5) +.8.66
Solved Question Paper 5-6 vii By Backward difference formula a.5, a nh.5 n (.) n.5 5. n( n ) f (a nh) f( a) nf( a) f( a)...! (5 )(4 ) f ().8 (5 )(.8) (.4) 9 f ().8.4.8 (.4).8.4 +. 8 +.4.8 + f (). + +.8 f (.5) (.5) +.5 +.8.88 (f) Construct the divided difference table for the data..5.5. 5. 6.5 8. f ().6 5.87.. 8. 5. Hence find the interpolating polynomial and an approimation to the value of f (). Solution f() f() f() (f) 4 (f) 5 (f).5.6 4.5.5 5.87 5. 6.75.9995.. 9.499. 5.. 5.. 4.499..7467.6 6.5 8. 9.5 59.5 8. 5 By Newton's divided difference formula f () f ( ) + ( ) f (, ) + ( ) ( ) f (,, ) + ( ) ( ) ( ). f (,,, ) + ( ) ( ) ( ) ( ) f (,,,, 4 ).6 + (.5) (4.5) + (.5) (.5) (5.) + (.5) (.5) ( ) (.9995) + (.5) (.5) ( ) ( 5) (.).6 + 4.5.5 + (.5.5 +.75) (5.) + ( +.75) ( ) (.9995) +.6 + 4.5.5 + 5. +.75 + Negligible ( +.75 + 6.5) (.9995).4 5.75 + 5. +.9995 4.558 + 6.747.49
viii Introduction to Engineering Mathematics III.9995 + + + Ans. + + (app.) f () () + + Ans. 4 8 (g) Solve the system of equations AX B, where A, B 7 5 using the LU decomposition method. Take all the diagonal elements of L as. Solution + y + z w 4 + z + w 8 + y + z 7 + y + z w 5 LU A u u u u4 l u u u 4 l l u u4 l l 4 4 l4 u44 u u u u4 lu lu u lu u lu4 u4 lu lu lu lu lu u lu4 lu4 u4 l4u l4u l4u l4u l4u l4u l4u4 l4u4 l4u4 u44 On comparing, we get 4 u u u l. u 4 l. u u l. u u l() 4 () () u ()() u l u u lu l u l u l u l u u l() () l( ) () ( 4)() u l l 4 u l4u l4u l4 u l4 u l4 u l4 u l4() () l4( ) ( 5 )() ( 4) l4( ) l 4 l 5 4 4 l4 u4 l u4 u4 () ( ) u4 u4 5 l u4 l u4 u4 ( ) ( 4)5 u4 u 7 4 4 l4 u4 l4 u4 l4 u4 u44 5 7 ( ) ( 4)5 ( 4) u44 u 44
Solved Question Paper 5-6 i A LU 4 5 7 4 4 5 4 A X B LUX B LV B (Let UX V) v v 8 4 v 7 5 4 v 4 5 v v + v 8 + v 8 v 8 UX V v vv 7 4 57v 7 v 9 5 v vvv4 5 4 55 87 v 5 4 4 47 v 5 v 4 5 5 8 7 4 9 4 5 From backward 4 5 4 8 7 4 9 4 7 8 9 4
Introduction to Engineering Mathematics III 5 + 5 4 8 + 4 8 6 + + 4 + 6 6 5 (,,, 4 ) (5, 6,, 8) (h) Solve the initial value problem Use the fourth order Runge-Kutta method. Solution f(, y) y, y,, h. st iteration h.. k h f(, y ).. f(,). ( ) dy y, y() d with h. on the interval [,.]. h k k h f, y. f(.5,). (.5). h k k h f, y. f(.5,.995). (.5.995 ). k4 h f( hy, k). f(.,.99). (..99 ).96 According to Runge-Kutta (for thorder) formula yy ( ) y ( kk kk4) 6 (...96) 6 + (.596).99.99 6 nd iteration., y.99 k h f(, y ). f(.,.99). (..99 ).96
Solved Question Paper 5-6 i h k k h f, y. f(.5,.98). (.5.98 ).88 h k k h f, y. f(.5,.97569).(.5.97569 ).856 k h( h, y k ). f (.,.9654).(..9654 ).698 4 According to Runge-Kutta (for thorder) formula y y ( k k kk4 ) 6.99 (.96.88.856.698.698).9654 6 nd iteration., y.9654, k h f(, y ). f(.,.9654).698. h k k h f, y. f(.5,.945).44467 h k k h f, y. f.5,.99.445 k 4 h f ( + h, y + k ). f (.,.974).55 According to Runge-Kutta (for thorder) formula y y ( k k kk4 ) 6.9654 (.698.4447.44.55) 6.974 y(.).97. Section-C Note : Attempt any two questions from this section. Each question carry equal marks. (5 ). (a) Show that for the function give as - y( iy) if f() z z y if z The C-R conditions are satisfied at origin but derivative of f (z) at origin does not eist. y( iy) if z f() z y Solution if z For, (, ) y y z u y, vy (, ) y y
ii Introduction to Engineering Mathematics III u u ( y, ) uy (, ) u uy (, y) u( y.) lim lim y y y u at(,) u(,) u(,) lim y y lim y y u u(, y) u.) y lim y y at(,) y y lim y y y y lim y y v v ( y, ) vy (, ) v vy (, y) vy (, ) lim lim y y y v v(,) v (,) v v(, y) v(, ) lim lim at(,) y at(,) y y lim lim y y u v y u v y Both C R equations Satisfy at origin. Now, f(z) f() f( y, y) f( y, ) lim, y iy f(, y) f(, ) lim, y iy (. y).( iy) y lim, y iy. m( im) Put y m, f() lim ( m )( im ) m( im) m( im) lim ( )( ) ( m im m )( im ) depends on m, So, f () does not eist (b) Verify that the function on u(, y) f (,y) is harmonic and find its conjugate harmonic function. Epress u + iv as an analytic function f(z). u y y Solution u y y u u y y u u y
Solved Question Paper 5-6 iii u u Satisfies laplace equation. y So, u(,y) f (,y) is harmonic. for conjugate harmonic function dv v d v dy y (c) Solution : u u d dy y d (By C R equations) (y + ) d + dy Eact differential equation v (y) d dy Treat y as const. Ignoring terms of v y + + c f(z) u + iv y y + iy + i + c i + i y + + i y + iy + c i( + iy) + ( + iy) + c iz + z + c Find the Fourier transform of Block function f(t) of height and duration a defined by a for t f() t otherwise ist f s fte () dt a ist le. dt a a/ ist e is a/ isa/ isa/ e e is isin s a is s in as/ s
iv Introduction to Engineering Mathematics III 4. (a) Using Z - transform, solve the difference equation u 4u u 5 n n n n with u u l Solution: un4unu n 5 n...(i) Taking z transform of (), we get zun 4 unun 5z zu ( n) 4 z( un) z( un) 5z zuz () z u zu 4( zuz () zu ) uz () 5z U() z z 4z z z4z 5z z Uz () z 4z z z z 5 z zz ( ) U() z ( z5)( z 4z) z 4z Uz () z z ( z5)( z)( z) ( z)( z ) 8 8 4 z5 z z z (Partial Fractions) 9 z z z Uz () 8 8 4 z5 z z 9 8 8 4 5z z z n 9 n Un.5.. Ans. 8 8 4 (b) The first four moments of a distribution about 4 are, 4,, 45. Comment on the skewness and Kurtosis of the distribution. Solution:, 4,, 4 45 4 () ( ) 4 + () 4 4 4 6 4 45 4 + 6 4 4 45 4 + 4 6
Solved Question Paper 5-6 v Measure of skewness is given by Measure of Kurtosis is given by 6 6 9 4 (c) For observations on price () and supply (y) the following data were obtained, y, 88 y 556 and y 467 Obtain the two lines of regression. Solution:, y, 88 y 556 and y 467 y r. y 467.98 88 556 y y 556 r. r.98.5 88 r y r y Equation of regression of y on is y y y r ( ) y.5 ( ) y.5 9.76 88.98.6 558 y.5.4...() Similarly the regression line of on y is r y y y.6 (y ).6 y.86.6y.86...() Equations () and () are equations of the required lines. 5. (a) Find the root of the equation e by regula falsi method correct up to two decimal places in the interval (,.5). Sol. f () e f () e.8 f (.5) (.5) e.5.7 The root lies between and.5 By Regula Falsi Method a, b.5
vi Introduction to Engineering Mathematics III afb () b f( a) f(.5).5 f() f( b) f( a) f(.5) f().7.5 (.8).4.7.8 f (.4) (.4) e.4.6 The root lies between.4 and.5 a.4, b.5.4 (.7).5 (.6).5.7.6 f (.5) (.5) e.5.5 Hence the root of the given equation is.5. (b) Prove the following identities : Sol. (i) (i) E E E ( E ) E Ans. E E E (E + E ) + +...() E ( E ) E Equation () equation () (ii) E ( E E ) E E Ee ( ) e. e E e Proved....() Sol. (ii) E ( e ) e. E e e e E. E ( E ) e e ( E E ) e. E ( ) E E e
Solved Question Paper 5-6 vii e ( E E ) e. e e e ( e e e )( e ) e e e (c) e e e e e e e ( e e ) e ( e e ) e. Proved. The velocity v of a particle at distance s from a point on its path is given by the following table: s (m.) 4 5 6 v (m./s.) 47 58 64 65 6 5 8 Estimate the time taken to travel 6 m. Using Simpson s one-third rule. Sol. Using Simpson s rd rule. s (m.) 4 5 6 v (m./s.) 47 58 64 65 6 5 8 v 47 58 64 65 6 5 8 y y y y 4 y 5 y 6 y 7 d yd ( y y7) ( y y5) 4( y y4 y6) 4 47 8 64 6 58 65 5 (..6) (.56.64) 4(.7.54.9).476.64.7.88.6 sec. Ans.