By Rajesh Pandey

ISBN-10: 144165772X

ISBN-13: 9781441657725

ISBN-10: 9380250371

ISBN-13: 9789380250373

ISBN-10: 9380257031

ISBN-13: 9789380257037

Quantity i of this sequence serves as a textbook for semester i of thesubject engineering arithmetic. appropriate figures and diagrams havebeen used to make sure a simple figuring out of the recommendations concerned. to stress program of the subjects mentioned, compatible examplesare integrated through the booklet. Solved examples, within the bookinclude options of questions from earlier u. P. T. U. Examinations. earlier years query papers were incorporated as to exposestudents to the trend and sort of questions they might face in anexamination. This publication is especially worthwhile for measure, honours andpostgraduate scholars of all indian universities and for ias, pcsand different aggressive examinations. in regards to the writer dr. Rajesh pandey he has greater than fourteen years experiencein instructing scholars of undergraduate, postgraduate and engineeringlevel. He got his b. Sc and m. Sc measure in 1991 and 1993respectively from gorakhpur college and phd in yr 2003 andparticipated in a number of seminars & meetings of nationwide andinternational point. For graduate & postgraduate, the authorhas additionally written books on enhance calculus, vectors, numericalanalysis, summary algebra, mechanics, fluid mechanics and so forth. shortly, he's operating as an assistant professor/reader inmathematics, sherwood collage of engineering, examine andtechnology, lucknow. desk of contents easy effects and ideas successive differentiation and leibnitz's theorem partial differentiation curve tracing growth of functionality jacobian approximation of blunders extrema of services of numerous variables lagranges approach to undetermind multipliers matrices a number of integers beta and gamma services vector differential calculus vector crucial calculus exam papers of uptu from 2001-2009

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**Extra info for A Text Book of Engineering Mathematics. Volume I**

**Example text**

C 2005) 20 Partial Differentiation Solution: Given u = exYZ au= xyexyz ... az 2 a xye XYz ) = x_ye a xyz _a u = _( dyaz dy ay = x[y xz eXYZ + exyz] = exYZ (X2yz + x) au a 3 Hence = - [exyz (X2yZ+X)] axayaz ax =eXYZ (2xyz+1) + yz exYZ (X2yZ+X) = eXYZ [2xyz +1 + x2y2z2 + xyz] = eXYZ (1 +3xyz + X2y 2Z2) Hence Proved. 2003) +2 2 2 Solution: Given + -{-- + -::-- =1 ................. :axc:,. + ax + ax = 0 ~2+ut (~+ut 2x [X2 or a2 +u - (a 2 y2 ~2+ut Z2 +ut + (b2 +ut + (c 2 +ut 2 2x/(a + u) au or ax =[x2 / (a2 + u _ 2xja 2 +u - L[X 2 au and az 2 2 vJ 2 2y/(b +u) L [ x /(a 2 2 +u) 2 J 2z / (c 2 + u) =-=----'---'--= 2 2 L[x /(a au ax =0 t + y2 / (b + U)2 + Z2 / (c2 + U)2 J /(a +utJ ..

52,32, 12. Answer. Example 13 : If y = [log{x+ + x 2) }P, show that J(1 (Yn+2)0 = - n 2 (yn)O, hence find (Yn)O. 2 or (1+x2)Yn+2 + (2n+1) x)'n+t +n2Yn = 0 ........... (iv) putting x =0 in (iv) we have (1 +0) (Yn+2)0 + n 2 (yn)O =0 or (yn+2)0 = - n2(yn)o ................. (v) Putting n = n-2, n-4, n-6 ......... in (v), we get (yn)o = - (n - 2)2 (y n- 2)0; (yn- 2)0 = - (n - 4)2 (yn -4)0; (Yn-4)0 = - (n - 6)2(Yn_6)0; where we have (on multiplying side wise) (yn)O = {-(n - 2)2} {- (n - 4)2} { -(n - 6)2} {(yn -6)0} :.

Az = az ax + az ay ................ (i) au ax au ayau az az ax az ay .. and - = - . - + - . - ........... (11) av ax av ay av Also given that x =e u + e- v and y = e- u -e V ax u ax _y dy -u ay y :. From (i) we get az = aZ(eU)+ az(_e-U) ................ (iii) au ax ay and from (ii) we get az = az (-e- )+ az (_e- v ) ••••••••••••• (iv) av ax ay Subtracting (iv) from (iii) we get Y 36 Partial Differentiation az- -az= (u az- (-u e +e -v) e -e v) -az au av ax ay az az Hence Proved. =x--y-. ax ay Example 21 : If V = f(2x -3y, 3y -4z, 4z -2x), compute the value of 6V x +4Vy +3V z.

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