Problem Set #5Chapter7, Part1: Ex. 1; Part2: Ex. 1-6;October 22, 2004Ex. 1, page 125A trajectory starting at the point (110,110) goes toward the equilibrium point(3, 4). This equilibrium point is a nodal sink, a stable equilibrium point, thusas t → ∞, the trajectory gets arbitrarily close to this equilibrium point.1Ex.1, page 126.Compute v + w and v − w when v and w equal:a) v =55and w =−32v + w =27v − w =83b) v =31and w =−10−31v + w =−7−30v − w =1332c) v =20and w =6−1v + w =8−1v − w =−41d) v =01and w =−10v + w =−11v − w =11Ex. 2, page 168.Compute rv when r and v are givena) r = 3 and v =−32rv =−96b) r = 1 and v =20rv =20c) r = 0 and v =−10−31rv =00d) r = −2 and v =31rv =−6−22Ex. 3, page 126Compute the length of the vectors v in Exercise 2.In order to compute the length of these two vectors, we can use the dotproduct. We know that |v| =√v · v.a) v =−32|v| =q−32·−32=√13b) v =20|v| =q20·20=√4 = 2c) v =−10−31|v| =q−10−31·−10−31=√1061d) v =31|v| =q31·31=√10Note: We can also find the length of these vectors using the PythagoreanTheorem.Ex. 4, page 126Compute v · w for the vectors v and w, in Exercise 1.a) v =55and w =−32v · w =q55·−32= 5 · (−3) + 5 ·2 = −5b) v =31and w =−10−31v · w =q31·−10−31= 3 · (−10) + 1 ·−31 = −61c) v =20and w =6−1v · w =q20·6−1= 2 · 6 + 0 ·−1 = 12d) v =01and w =−10v · w =q01·−10= 0 · (−1) + 1 ·0 = 03Ex.5, page 126.If v(t) is the vector function of time given, computeddtv:a) v =cos(2t)3e−2t.v0=(cos(2t))0(3e−2t)0=−2 sin(2t)−6e−2t.b) v =e2tt2.v0=(e2t)0(t2)0=2e2t2t.c) v =31.v0=(3)0(1)0=00.d) v =− sin(t)2t.v0=(− sin(t))0(2t)0=− cos(t)2.Ex.6, page 126.Compute the antiderivate for the vectors v in Exercise 5.a) v =cos(2t)3e−2t.Rvdt =Rcos(2t)dtR3e−2tdt=12sin(2t)+C−32e−2t+C.b) v =e2tt2.Rvdt =Re2tdtRt2dt=12e2t+C13t3+C.c) v =31.Rvdt =R3dtR1dt=3t+Ct+C.d) v =− sin(t)2t.Rvdt =R−
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