134Chapter 22.Solutions: Case Study: Robot Control: Swinging Like a Pendulumso the solution to the diferential equation isy(t)=α1·1λ1¸eλ1t+α2·1λ2¸eλ2t,whereα1andα2are constants determined by two additional conditions. IF thediscriminant satis±esc2/(4m2±2)−g/± >0, then the solution decays; otherwise itcan have an oscillatory component in addition to a decaying one.CHALLENGE 22.2.We note thatv(0,0) = 0, and it is easy to see thatv>0For all other values oF its arguments.We diferentiate:ddtv(y(t)) =gsinθ(t)±dθ(t)dt+dθ(t)dtd2θ(t)dt2=gsinθ(t)±dθ(t)dt−dθ(t)dt1m±(cdθ(t)dt+mgsin(θ(t)))=−cm±µdθ(t)dt¶2≤0.ThereFore, we can now conclude that the pointθ=0,dθ/dt= 0 is stable For both thedamped (c>0) and undamped (c= 0) cases. ²or the undamped case, dv(y(t))/dtis identically zero, and we cannot conclude that we have asymptotic stability. ²orthe damped case, we note that the set de±ned by dv(y(t))/dt= 0 contains all points(θ,ddt= 0), and the only invariant set is the one containing the single point (0
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.