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

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

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

Example text

5! x 2 x 3 2 x 4 5x 5 + + + ........ – 2 ! 3! 4! 5! 15. Expand log (1 + ex) in ascending powers of x up to the term containing x4. Solution Let y = log (1 + ex), y (0) = log 2 Now y1 = y1 (0) = y2 = y2 (0) = ∴ ∴ ex 1 + ex 1 2 ex d1 + e i 1F 1I 1– J G 2H 2K x 2 b = ex 1 . = y1 1 – y1 x 1 + e 1 + ex = 1 4 y3 = y1 (–y2) + (1 – y1) y2 = y2 – 2y1 y2 y3 (0) = 1 1 1 – 2⋅ ⋅ = 0 4 2 4 y4 = y3 – 2 ( y1 y3 + y22) y4 (0) = 0 – 2 FG 1 ⋅ 0 + 1 IJ H 2 16 K = –1 8 g 47 DIFFERENTIAL CALCULUS—I Therefore from Maclaurin’s theorem, we get y = y (0) + xy1 (0) + log (1 + ex) = log 2 + x .

F ′ (c) = 0 Now consider From (1), we have, 2 e sin c = 0 (Œ ec ≠ 0) c ∴ sin c = 0 = sin nπ c = nπ where n = 0, 1, 2, ..... c = π∈ But FG π , 5π IJ H4 4 K Thus Rolle’s theorem is satisfied. LM x + ab OP MN ba + bg x PQ 2 5. Show that the constant c of Rolle’s theorem for the function f (x) = log a ≤ x ≤ b where 0 < a < b is the geometric mean of a and b. Solution. 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.

We get r (– sin θ) + (1 + cos θ) · dr =0 dθ dr r sin θ = dθ 1 + cos θ or We have, 1 p2 1 1 = 2 + 4 r r = FG dr IJ H dθ K 2 1 1 r 2 sin 2 θ + ⋅ r 2 r 4 1 + cos θ 2 b g 1 L sin θ O M P 1+ r M b1 + cos θg P N Q 1 L b1 + cos θg + sin M r M N b1 + cos θg 2 = 2 2 2 = 2 2 2 θ OP PQ 25 DIFFERENTIAL CALCULUS—I = = where 1 + cos θ = LM 2 b1 + cos θg OP MN b1 + cos θg PQ 1 r2 r 2 2 1 + cos θ b 2 g a r 1 p2 = 2 a r2 ⋅ r = 2 ar ar which is the pedal equation of the curve. t. p, we get Hence, p2 = 2p = a dr ⋅ 2 dp ⇒ 4p dr = a dp ∴ ρ = r⋅ dr dp = r⋅ 4p where p = a = r.

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