0 1 00127 2 ℎ 1 000497 2 ℎ 2 000995 2 ℎ 3

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= 0 , 𝜕𝑓 1 𝜕𝑞 = 0.0127 , 𝜕𝑓 2 𝜕ℎ 1 = 0.00497 , 𝜕𝑓 2 𝜕ℎ 2 = −0.00995 , 𝜕𝑓 2 𝜕ℎ 3 = 0.00497 , 𝜕𝑓 3 𝜕ℎ 1 = 0 , 𝜕𝑓 3 𝜕ℎ 2 = 0.00884 , 𝜕𝑓 3 𝜕ℎ 3 = −0.018 Hence the linear approximation 𝑥̇(?) = 𝑓(𝑥 0 , ? 0 ) + ?𝑥(?) + ??(?) , where 𝑓(𝑥 0 , ? 0 ) = 0 at equilibrium equal to: 𝑥̇(?) = [ −0.00318 0.00318 0 0.00497 −0.00995 0.00497 0 0.00884 −0.018 ] [ ℎ1 − 3 ℎ2 − 2 ℎ3 − 1 ] + [ 0.013 0 0 ] [?1 − 0.5]
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Question4 zID: z5081559. Z = 100+559 =659. C3A=0.5-659/4000=0.33525 Step1: ? 1 (?) − 0.5√ℎ 1 − ℎ 2 ? 1 = 0 0.5√ℎ 1 − ℎ 2 − 0.5√ℎ 2 − ℎ 3 ? 2 = 0 0.5√ℎ 2 − ℎ 3 − 0.33525√ℎ 3 ? 3 = 0 Step2: ? 1 (?) − 0.5√ℎ 1 − ℎ 2 = 0 0.5√ℎ 1 − ℎ 2 = 0.5√ℎ 2 − ℎ 3 0.5√ℎ 2 − ℎ 3 = 0.33525√ℎ 3 ? 1 (?) − 0.33525√ℎ 3 = 0 0.5 − 0.33525√ℎ 3 = 0 3 = 2.2201 0.5√ℎ 2 − ℎ 3 = 0.33525√ℎ 3 0.5√ℎ 2 − 2.2201 = 0.33525√2.2201 h2=5.2 0.5√ℎ 1 − ℎ 2 = 0.5√ℎ 2 − ℎ 3 0.5√ℎ 1 − 5.2 = 0.5√5.2 − 2.2201 h1=8.18
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