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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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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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Solution: = d dt = a. i) a. d dt d dt b b a b c c c + b c . da dt = a. b dc db + c + b dt dt = a. b dc dt = dc dt a b Solution: + d ii) dt +a. a db dt c . c + b db c + dt da a dt b c c . da dt da dt d2 a dt 2 da d 3 a d 2 a d 2a = a + a c 2 dt dt 3 dt dt 2 (From Result i) da d 3 a = a +0+0 dt dt 3 = a. da dt da dt d 2 a da dt 2 dt d = dt a b b c + da d 3 a dt dt 3 Example 4. Evaluate the following: d = a b c dt dc d b + a b c dt dt dc db da b + a + dt dt dt c Solution: = a = a = a b dc + a dt Example 5. Show that r db dt dr = dt c+ r da dt b c dr dt , where r = r 2 r r

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