# Engineering Mathematics- II (As per the new syllabus of VTU by A. Ganesh (Author), Balasubramanian G. (Author)

By A. Ganesh (Author), Balasubramanian G. (Author)

Read Online or Download Engineering Mathematics- II (As per the new syllabus of VTU (B. E. , II Semester) PDF

Best engineering books

Trends in Computer Science, Engineering and Information Technology: First International Conference on Computer Science, Engineering and Information Technology, CCSEIT 2011, Tirunelveli, Tamil Nadu, India, September 23-25, 2011. Proceedings

This e-book constitutes the refereed court cases of the 1st overseas convention on laptop technology, Engineering and data expertise, CCSEIT 2011, held in Tirunelveli, India, in September 2011. The seventy three revised complete papers have been rigorously reviewed and chosen from greater than four hundred preliminary submissions.

Strategies and Applications in Quantum Chemistry: From Molecular Astrophysics to Molecular Engineering

On the time whilst expanding numbers of chemists are being attracted through the fascination of supposedly effortless computing and linked vibrant imaging, this e-book seems as a counterpoint. the 1st half makes a speciality of primary thoughts of quantum chemistry, masking MCSCF conception, perturbation remedies, foundation set advancements, density matrices, wave functionality instabilities to correlation results, and momentum area idea.

Extra resources for Engineering Mathematics- II (As per the new syllabus of VTU (B. E. , II Semester)

Sample text

2! 3! n! (3) This is called the Maclaurin’s series for the function f (x). If f (x) = y and f ′ (x), f ′′ (x), .......... are denoted by y1, y2, ...... the Maclaurin’s series can also be written in the form: y = y (0) + xy1 (0) + x2 xn y2 (0) + ...... y (0) + ...... 2! n! n WORKED OUT EXAMPLES 1. Verify Rolle’s theorem for the function f (x) = x2 – 4x + 8 in the interval [1, 3]. Solution f (x) = x2 – 4x + 8 is continuous in [1, 3] and f ′ (x) = 2x – 4 exist for all values in (1, 3) ∴ f (1) = 1 – 4 + 8 = 5; f (3) = 32 – 4 (3) + 8 = 5 ∴ f (1) = f (3) Hence all the three conditions of the theorem are satisfied.

T. (4) x = 1 2y4 (1) + 3 – 3 = 0 y4 (1) = 0 Substituting these values in the expansion, we get tan–1 x = π 1 + 4 2 R| U S|b x – 1g – b x –21g + b x –61g |V| T W 2 3 13. By using Maclaurin’s theorem expand log sec x up to the term containing x6. Solution Let y (x) = log sec x, y (0) = log sec 0 = 0 We have y1 = sec x tan x = tan x sec x y1 (0) = 0 y2 = sec2 x y2 (0) = 1 To find the higher order derivatives Consider y2 = 1 + tan2 x y2 = 1 + y12 This gives, Hence, y3 = 2y1 y2 y3 (0) = 2y1 (0) y2 (0) = 0 y4 = 2 [ y1 y3 + y22] so that and y4 (0) = 2 [ y1 (0) y3 (0) + y22 (0)] = 2 [0·1 + 12] = 2 This yields y5 = 2 [ y1 y4 + y2 y3 + 2y2 y3] = 2 [ y1 y4 + 3y2 y3] y5 (0) = 2 [ y1 (0) y4 (0) + 3y2 (0) y3 (0)] = 2 [0·2 + 3·1·0] = 0 y6 = 2 [ y1 y5 + y2 y4 + 3 {y2 y4 + y32}] 45 DIFFERENTIAL CALCULUS—I = 2 [ y1 y5 + 4y2 y4 + 3y32] This yields y6 (0) = 2 [ y1 (0) y5 (0) + 4y2 (0) y4 (0) + 3y32 (0)] = 2 [0·0 + 4·1·2 + 3·0] = 16 Therefore by Maclaurin’s expansion, we have y = y (0) + xy1 (0) + log sec x = 0 + x · 0 + = x2 y (0) + ........

The given f (x) is continuous in [a, b] since 0 < a < b in LM x + ab OP MN ba + bg x PQ 2 f (x) = log The given function can be written in the form f (x) = log (x2 + ab) – log (a + b) – log x f ′ (x) = 2x 1 –0– x x + ab 2 d 2 x 2 – x 2 + ab = dx 2 i + ab x i= x 2 – ab dx 2 f ′ (x) exists in (a, b) L a + ab OP = log 1 = 0 f (a) = log M MN ba + bg a PQ L b + ab OP = log 1 = 0 f (b) = log M MN ba + bg b PQ 2 Also 2 ∴ f (a) = f (b) = 0 Hence all the conditions of the theorem are satisfied. , ∴ that c = ± ab c = + ab ∈ (a, b) and we know ab is the geometric mean of a and b.