Download e-book for iPad: Engineering Mathematics- II (As per the new syllabus of VTU by A. Ganesh (Author), Balasubramanian G. (Author)

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

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

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2 ar OP a (vi) r = b1 − cos θg 3 PQ 2 MN LM Ans. na OP (viii) r = a sin nθ N 2Q n rn an n −1 3. If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode 2 2 r = a (1 + cos θ) which passes through the pole. 1 Rolle’s Theorem If a function f (x) is 1. continuous in a closed interval [a, b], 2. differentiable in the open interval (a, b) and 3. f (a) = f (b). Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0 (No proof). 2 Lagrange’s Mean Value Theorem Suppose a function f (x) satisfies the following two conditions.

F (x) is differentiable in the open interval (a, b). Then there exists at least one value c of x in the open interval (a, b), such that bg f (b) – f a b–a = f ′ (c) 28 ENGINEERING MATHEMATICS—II Proof. (1) where k is a constant. Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b). (2) Hence, if k is chosen as given by (2), then φ (x) satisfy all the conditions of Rolle’s theorem. Therefore, by Rolle’s theorem there exists at least one point c in (a, b) such that φ′(c) = 0.

7. Prove that b–a 1 – a2 < sin–1 b – sin–1 a < b–a 1 – b2 where a < b < 1. (3) 1 – b2 Combining (2) and (3), we get 1 1– a 2 1 or 1 – a2 1 < 1– c < 2 < 1 1 – b2 sin –1 b – sin –1 a < b–a 1 1 – b2 On multiplying by (b – a) which is positive, we have b–a 1– a 2 < sin–1 b – sin–1 a < b–a 1 – b2 by using (1) 39 DIFFERENTIAL CALCULUS—I 8. Show that b–a b–a π 3 4 π 1 –1 –1 , if 0 < a < b and deduce that + < tan –1 < + · 2 < tan b – tan a < 1 + a2 1–b 4 25 3 4 6 Solution Let f (x) = tan–1 x is continuous in [a, b], a > 0.

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