MAT 1332 Lecture Notes - Lecture 4: Product Rule

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lemonchinchilla454

9 Dec 2018

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This session covers sections 8. 3, 8. 4 and part of complex numbers: find the equilibria of the equation y" = y3 5y2 + 6y, and determine the stability of each equilibrium. Let y3 5y2 + 6y = y(y 2)(y 3) = 0. Then y = 0, y = 2, y = 3. Let f (y) = y3 5y2 + 6y. Since f "(0) = 6 > 0, and f "(3) = 3 > 0, equilibria y = 0 and y = 3 are unstable. Since f "(2) = 2 < 0, equilibrium y = 2 is stable. Recall the stability theorem: let y = y0 be an equilibrium of autonomous equation y" = f (y). It is stable if f "(y0) < 0; it is unstable if f "(y0) > 0: find the equilibria of the equation y" = y(e stability of each equilibrium. The equilibria of this equation are y = 0 and y =

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[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

MAT 1332 Lecture Notes - Lecture 4: Product Rule

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