Download A Text Book of Engineering Mathematics. Volume I by Rajesh Pandey PDF

By Rajesh Pandey

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 realizing of the techniques concerned. to stress program of the themes mentioned, appropriate examplesare integrated during the ebook. Solved examples, within the bookinclude options of questions from prior u. P. T. U. Examinations. earlier years query papers were integrated as to exposestudents to the trend and kind of questions they could face in anexamination. This ebook is particularly important for measure, honours andpostgraduate scholars of all indian universities and for ias, pcsand different aggressive examinations. concerning the writer dr. Rajesh pandey he has greater than fourteen years experiencein educating scholars of undergraduate, postgraduate and engineeringlevel. He obtained his b. Sc and m. Sc measure in 1991 and 1993respectively from gorakhpur college and phd in 12 months 2003 andparticipated in a number of seminars & meetings of nationwide andinternational point. For graduate & postgraduate, the authorhas additionally written books on develop calculus, vectors, numericalanalysis, summary algebra, mechanics, fluid mechanics and so forth. shortly, he's operating as an assistant professor/reader inmathematics, sherwood university of engineering, study andtechnology, lucknow. desk of contents uncomplicated effects and ideas successive differentiation and leibnitz's theorem partial differentiation curve tracing growth of functionality jacobian approximation of mistakes extrema of features of a number of 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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A2 z 1 1 e - - =-~----:.. log e = 1 =- x(loge'+logx) 1 =---- x log (ex) = - {x log (ex)}-l Hence Proved. 2006) . Weave h ' -1 So Iuhon u- sm (x)y + tan (r)x' ............ t. x and y, we get. au Ox = rn + 1 oy1 (-y) 1 1+(~)' 0 -;1 24 Partial Differentiation = x x +-::--7" y ~y2 _ X2 X2 + y2 au or x ay x xy =- ~y2 _ x2 + x2 + y2 ................. (... ) 111 on adding (ii) and (iii), we have au au x-+y-=O Answer. V. V. 2004) Proof: Since f(x,y) is a homogeneous function of degree n, it can be expressed in the form f(x,y) = xn F(y Ix) ...

We know - = - + - - ................... -'dx (y2 +x) Substituting these values in (ii) we get du x[ (X2 +y)] -=(l+logxy)+- (2 ) Answer. dx y y +x Example 15: If f(x, y) =0, (y, z) =0 show that df d dz df d ---=-dy dZ dx dX dy Solution: If f(x, y) =0 then dy = _(df )/(~) ............. (i) dx dX dy if (y, z) =0, then :-; = -( ~~ )/( ~:) ..................... (ii) Multiplying (i) and (ii), we have dy. (d J. Hence Proved dy dZ dx dX dy Example 16 : If the curves f(x,y) =0 and (x, y) =0 touch, show that at the point of contact df.

Ii) ax ar as .. au au ar au as SImIlarly - = - . - + - . (111 ... 2ar as Z2 => Z2 au = au ....................... (iv) az as and Adding (i) (ii) and (iii) we get 2 au 2 au 2 au 0 x -+y ax -+z ay -az= Hence Proved. Example 19 : If u = u (y - z, z - x, x - y) Prove that au + au + au =0 ax ay az Solution: Here given u = u(y - z, z - x, x - y) Let X = Y - z, Y Z - x and Z = x - y ................ (i) Then u = u (X,Y,Z), where X, Y, Z are function of x,y and z. Then :::0 35 A Textbook of Engineering Mathematics Volume - I au au ax au ay au az ..

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