By Rajesh Pandey

Quantity i of this sequence serves as a textbook for semester i of thesubject engineering arithmetic. compatible figures and diagrams havebeen used to make sure a simple realizing of the options concerned. to stress software of the themes mentioned, compatible examplesare integrated in the course of the ebook. Solved examples, within the bookinclude strategies of questions from earlier u. P. T. U. Examinations. past years query papers were integrated as to exposestudents to the development and sort of questions they might face in anexamination. This booklet 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 collage and phd in yr 2003 andparticipated in a variety 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 on. shortly, he's operating as an assistant professor/reader inmathematics, sherwood collage of engineering, learn andtechnology, lucknow. desk of contents uncomplicated effects and ideas successive differentiation and leibnitz's theorem partial differentiation curve tracing enlargement of functionality jacobian approximation of blunders extrema of capabilities of a number of variables lagranges approach to undetermind multipliers matrices a number of integers beta and gamma capabilities vector differential calculus vector fundamental calculus exam papers of uptu from 2001-2009

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

**Sample text**

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) ... (i) :. af ax or x =~(xnF(Y/x)}=nxn-lF(Y/x)+xnF' ax (r)(-y) x x2 ~: =n xn F (~) - yx n- F{ ~) ................. (ii) 1 Again from (i), we have af a ay = ay (xnF(y/x)} = 1 xn P(y/x). x af -_ yx n-l F'( y I x).............. (111 ... ) or y ay Adding (ii) and (iii), we get af af x-+ y - = nxn F(y/x) ax ay = nf using(i) Hence Proved. Note. In general if f (Xl, X2 ...............

F = ell x ax (e") + y ay (e" ) = e" au au or xe" - + yeO - = e" ax ay au au or x - + y:;- = 1 Hence Proved. ax oy Example 8 : If u ~ sin-' { fx : Jy ),show that au au 1 x-+y-=-tan u ax ay 2 Solution: Here u = sin- l => sin u = x+y JX+fy 1JX+fy x+y ) = {(say) Here f is a homogeneous function in x and y of degree ( 1 :. e ~ Partial Di[ferentiation af af 1 x-+y-=-f ax ay 2 or x -a (. sm u ) + y -a (. sin u ) = -1. sin u ax ay 2 :. f = sin u au au 1. or x cos u - + Y cos u - = - sin u ax ay 2 au au 1 Hence Proved or x-+y-=-tanu.

A )(av . 0] ayax oX 38 Partial Differentiation 'iiv Hence Proved. P. 2003, Delhi college of Engg. U. r=~x2+y2, O=tan-l(~) ar = x = r cos 0 = cos 0, ar = y = r sin 0 = sin 0 ax ~X2 + y2 r ay ~X2 + y2 r ao 1 ( y) y sinO ax = 1 + y2 -7 = - x2 + y2 = --rx2 ao 1 1 x cosO and - = - - . - = ? =-ay 1 + y2 X x- + yr 2 x Here u is a composite function of x and y au = au ar + au ao = cos 0 au _sin 0 au ax ar ax ao ax . ar r ao a a sinO a . ~ - == cosO- - - - - ......................... (1) ax ar r ao Also au = au .