Download PDF by Rajesh Pandey: A Text Book of Engineering Mathematics. Volume I
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 innovations concerned. to stress program of the subjects mentioned, compatible examplesare integrated in the course of the booklet. Solved examples, within the bookinclude strategies of questions from past u. P. T. U. Examinations. prior years query papers were incorporated as to exposestudents to the trend and kind of questions they could face in anexamination. This ebook is especially worthwhile 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 bought his b. Sc and m. Sc measure in 1991 and 1993respectively from gorakhpur college and phd in 12 months 2003 andparticipated in a variety of seminars & meetings of nationwide andinternational point. For graduate & postgraduate, the authorhas additionally written books on increase calculus, vectors, numericalanalysis, summary algebra, mechanics, fluid mechanics and so forth. almost immediately, he's operating as an assistant professor/reader inmathematics, sherwood collage of engineering, study andtechnology, lucknow. desk of contents easy effects and ideas successive differentiation and leibnitz's theorem partial differentiation curve tracing enlargement of functionality jacobian approximation of mistakes extrema of services of numerous variables lagranges approach to undetermind multipliers matrices a number of integers beta and gamma services vector differential calculus vector essential calculus exam papers of uptu from 2001-2009
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Extra resources for A Text Book of Engineering Mathematics. Volume I
Ao ay ar ay ao ay . + cosO ~ ................. (ii) ay ar ao Now we shall make use of the equivalence of Cartesian & polar operators as given by (i) and (ii) 2 a u = ~(au) = (cosoi. _ sinO ~)(coso au _sinO au) ax 2 ax ax ar r ao ar r ao r 39 A Textbook of Engineering Mathematics Volume - I au- sin S au) Sa (' cos Sau- sin S au) =cos S-a ( cos S-- -sin --ar ar r as r as ar r as 2 2 2 2 = COSs[COSS a u _ sinS au (-~J sinS a u ]_ sins[-sins au + cosS a u _ coss au _ sinS a u] ar2 r as r2 r aras r ar asar r as r as 2 _ _ 2 S a2u 2cosSsinS au sin 2 S au _ 2cosSsinS a2u sin 2 S a2u ('" - cos ar2 + r2 as + r ar r aras + r2 as 2 ......................
And az = ax (-1) + ay (1) + az (0) = - ax + ay ............... (V11) Adding (v), (vi) and (vii) we get au + au + au = O. Hence Proved. UP 2005) Solution: Here z is a function of x and y, where x and yare functions of u and v. :. 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.
U. 2001, 2003) 42 Partial Differentiation 1 1 log x - log y af af 23. Iff(x,y)= -2 +-+ 2 2 ,showthat x~+y:\+2f(x,y)=0 x xy x +y ux uy (PTV 2004) 24. 2003) ~+I+~ 25. 2003) 2 1t(2X + y2 + xzt2 ) 26. 1991) 27. If u = x2 - y2 + sin yz where y = ex and z = log x; find du dx 1 Ans. 2(x - e2x )+ex cos (ex log x) (log x + -). x au au au 28. If u = f(r, s) and r = x+y, s = x-y; show that - + - = 2 ax ay ar r r 02U 29. If x = e cosS, y = e sinS, show that ax 2 au (a u 02U) + ay2 = e- ar2 + OS2 2 2 2r 30.
A Text Book of Engineering Mathematics. Volume I by Rajesh Pandey