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Note for Applied Mathematics - 2 - M-2 By JNTU Heroes

  • Applied Mathematics - 2 - M-2
  • Note
  • Jawaharlal Nehru Technological University Anantapur (JNTU) College of Engineering (CEP), Pulivendula, Pulivendula, Andhra Pradesh, India - JNTUACEP
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VECTOR CALCULAS UNIT STRUCTURE 5.0 Objectives 5.1 Introduction 5.2 Vector differentiation 5.3 Vector operator 5.3.1 Gradient 5.3.2 Geometric meaning of gradient 5.3.3 Divergence 5.3.4 Solenoidal function 5.3.5 Curl 5.3.6 Irrational field 5.4 Properties of gradient, divergence and curl 5.5 Let Us Sum Up 5.6 Unit End Exercise 5.0 OBJECTIVES After going through this unit, you will be able to Learn vector differentiation. Operators, del, grad and curl. Properties of operators 5.1 INTRODUCTION Vector algebra deals with addition, subtraction and multiplication of vertex. In vector calculus we shall study differentiation of vectors functions, gradient, divergence and curl. Vector: Vector is a physical quantity which required magnitude and direction both. Unit Vector:

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Unit Vector is a vector which has magnitude 1. Unit vectors along coordinate axis are i and j , k respectively. i = j = k =1 Scalar Triple Vector: Scalar triple product of three vectors is defined as a. b c or a b c . Geometrical meaning of a b c is volume of parallelepiped with cotter minus edges a, b and c . We have, abc = abc =- bca = cab bac Vector Triple Product: Vector triple product of a b and c is cross product of a and b a c or cross product of b a b a c b = a.c b b and c a a.b c= a.c b c i.e. b.c c a Remark : Vector triple product is not associative in general i.e. a a b Coplanar Vectors: Three vectors a, b and c a 0 0,b b c 0,c c are coplanar if abc = 0 for 5.2 VECTORS DIFFERENTIATION Let v be a vector function of a scalar t. Let a corresponding to the increment t in t. Then, v be the small increment in

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v v t + t - v(t) v t + t - v(t) v = t t Taking limit t 0 we get, v t + t - v(t) v = lim t 0 t 0 t t v t + t - v(t) dv v = lim = lim t 0 t 0 dt t t v t + t - v(t) dv = lim t 0 dt t lim Formulas of vector differentiation: (i) d dv = k v =k dt dt k is a constant (ii) d dt u +v = dv dt (iii) d dt u .v =u . (iv) d dt u (v) If v du dt v =u dv dt v. dv dt du dt du dt v v1i + v2 j + v3k dv Then, dt dv dv1 dv 2 i+ j+ 3 k dt dt dt Note: If r xi + yj + zk then r = r = x 2 y2 z2 Example 1: If r t + 1 i + t 2 + t - 1 j + t 2 - t + 1 k find dr d2 r and dt dt

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Solution: t + 1 i + t2 + t - 1 j + t2 - t + 1 k r dr dt d2 r dt 2 i + 2t + 1 j + 2t - 1 k 2j + 2 k Example 2: If r a cos wt + b sin wt where w is constant show that dr d2 r r = w a b and 2 = -w r dt dt Solution: r a cos wt + b sin wt------------ (i) dr a cos wt + b sin wt------------ (ii) dt dr r a cos wt + b sin wt -aw sin wt + bw cos wt dt a a=0 a b w cos 2 wt b a w sin 2 wt b b=0 a b w cos 2 wt a a b w cos 2 wt sin 2 wt a b w 1 w a b w sin 2 wt b = -a b Again differentiating eqn d2 r dt 2 -a w 2 cos wt - b w 2sin wt = -w 2 a cos wt + b sin wt -w 2 r from (i) Example 3. Evaluate the following: i) d dt a b c ii) d dt a da dt d2a dt 2 a=0 b

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