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30) The vector o = t ( ~ c2, l , c3) is called the angular velocity, which depends on t and is determined by the movement of 0. We now take a fixed point in 0. Let z(t) be its position at time t. l(O>i(t) + x2(0)j(t)+ 23(O)k(t) Basic Notions of Vector Analysis 23 because 0 is a rigid body. In the use o f the representation of components by { i ( O ) , j ( O ) , k(O)},this implies that d(0) = q(o)i’(o) + z a ( O ) f ( o ) -I- za(O)k’(o) =( = - c3( 0 ) X Z (0) c3(0)21(0) - Cl(o)z3(0) c1(0)z2(0) - c2(0)z1(0) CZ (o)z3(0) wx 1 Because the above relation holds at any time t , the velocity v is given by the position x and the angular velocity w at that moment in such a way as dx dt v=-= wx 2.

K ( t ) = k ( t ). i ( t )= 0 and i ( t ). i ( t )= j ( t )* j ( t )= k ( t ) k ( t ) = 1. This implies i'. j ' = j ' . k + j . k ' = k ' . i + k . i ' = 0 i'. i = j ' . j = k ' . 29) by the differentiation in t. 29) that + + c23 c32 = c31 c13 = c12 c11 = c22 = c33 = 0. 30) The vector o = t ( ~ c2, l , c3) is called the angular velocity, which depends on t and is determined by the movement of 0. We now take a fixed point in 0. Let z(t) be its position at time t. l(O>i(t) + x2(0)j(t)+ 23(O)k(t) Basic Notions of Vector Analysis 23 because 0 is a rigid body.

Then, maximum and minimum values are in the boundary values, f ( z , y ) for z = a,b with c I y I d , and f ( z , y ) for y = c , d with a 5 z I b, and the critical values f(zj,yj)’s. The third problem is to seek maximum and minimum values of z = g ( x , y), under the constraint that f ( z , y ) = 0. Then, solving f ( z , y ) = 0 as y = h ( z ) ,for example, we may obtain them by z = g(z, h ( z ) )defined on a 5 z 5 b. Those rough answers are justified in the following way. In the first problem, Weierstrass’ theorem guarantees that if y = f(z) is continuous, then its maximum and minimum are attained.

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