... thatddt((x 1 (t) 1) 2 +(x 2 ) 2 (t)) =2( x 1 (t) 1) ˙x 1 (t)+2x 2 (t)˙x 2 (t) =2( x 1 (t) 1) x 2 (t)+2x 2 (t)(−x 1 (t) +1) =0.Consequently, the motion satisfies (x 1 (t) − 1) 2 +(x 2 ) 2 (t) ≡ r 2 1 , for ... solve? We compute˙d = 2 ¨bb− 2( ˙b) 2 b 2 = 2 ¨bb−d 2 2.Hence (R) gives 2 ¨bb=˙d +d 2 2 =2 2d =2 2 2˙bb;55Hencevt+ |vx|−|x| = x − 1+ t + |t 1| −|x| = 0 in Region III,because ... .6.4 .2 APPLYING DYNAMIC PROGRAMMING. First, we check the minimaxcondition, for n =2, p =(p 1 ,p 2 ):f(x, a, b) ·p + r(x, a, b)=(m 1 − c 1 bx 2 )p 1 +(m 2 − c 2 ax 1 )p 2 + (1 a)x 1 − (1 −b)x 2 =...