Problem Set #7Chapter 10, Ex. 1,3,4, p.172;Chapter 11: Ex. 1,2,3,4, p.177November 2, 2004Ex. 1, p. 172ddtxy=x − x2− 2xy2y − y2− 3xyEquilibrium points are:00,10,02,0.60.2.D =1 − 2x − 2y −2x−3y 2 − 2y − 3xEquilib.point Matrix Det(D) T r(D) Stability001 00 22(2 > 0)3(3 > 0)Unstable02−3 0−6 −26(6 > 0)−5(−5 < 0)Stable10−1 −20 −11(1 > 0)−2(−2 < 0)Stable0.60.2−0.6 −1.2−0.6 −0.2−0.6(−0.6 < 0)−0.8(−0.8 < 0)UnstableEx. 3, p.172• h = x2y3∂h∂x= 2y3x∂h∂y= 3x2y2∂2h∂x2= 2y3∂2h∂y2= 6x2y∂2h∂y∂x= 6xy21∂2h∂x∂y= 6xy2The last two are equal.• h = x cos(xy)∂h∂x= −xy sin(xy) + cos(xy)∂h∂y= −x2sin(xy)∂2h∂x2= −xy2cos(xy) − 2y sin(xy)∂2h∂y2= −x3cos(xy)∂2h∂y∂x= −2x sin(xy) − x2y cos(xy)∂2h∂x∂y= −2x sin(xy) − x2y cos(xy)The last two are equal.• h = sin(x + y2)∂h∂x= cos(x + y2)∂h∂y= 2y cos(x + y2)∂2h∂x2= − sin(x + y2)∂2h∂y2= 2cos(x + y2) − 4y2sin(x + y2)∂2h∂y∂x= −2y sin(x + y2)∂2h∂x∂y= −2y sin(x + y2)The last two are equal.• h = xey∂h∂x= ey∂h∂y= xey∂2h∂x2= 0∂2h∂y2= xey∂2h∂y∂x= ey∂2h∂x∂y= eyThe last two are equal.Thus, we have seen through these examples that∂2h∂y∂x=∂2h∂x∂y.2Ex. 4, p. 172Integrate h(x, y) over the indicated rectangle.a)Z2−1Z10dxdy =Z2−1x|10dy = y|2−1= 3b)Z10Z1−1x dxdy =Z10(x2/2)|1−1dy = 0c)Z10Z10(x+y) dxdy =Z10(x2/2+yx)|10dy =Z10(1/2+y) dy = (y/2+y2/2)|10= 1/2+1/2 = 1d)Z32Z1−2xy dxdy =Z32(x2y/2)|1−2dy =Z32y2(1−4) dy =Z32−32y dy = (−3y2/4)|32= −15/4e)Z10Z10cos(xy) dxdy =Z10sin(xy)y10dy =Z10sin yydy = 0.946Chapter 11. Ex. 1, p.177a) M =1 20 1→v=11M→v=1·1+2·10·1+1·1=31b) M =3 2−5 4→v=13M→v=97c) M =0.5 0.28 0.4→v=01M→v=0.20.43Ex. 2, p.177a b cMM′4 171 69 92 1115 −218 −15MM′2 91 815 −23 59 542 −9Ex. 3, p. 177a b cdet(M) 1 −27 7det(M) 2 −4 8Ex. 4, p. 177Verify that→v=10is an eigenvector for M =1 20 1.M→v= λ→v1 20 1xy= λxy1 20 1xy− λ1 00 1xy= 01 − λ 20 1 − λxy=→0(1 − λ)x + 2y = 0(1 − λ)y = 0xy6=→0Thus, 1 − λ = 0 ⇒ λ = 1. The eigenvalue is λ = 1. Verify that10is aneigenvector:M10= λ101 20 110= λ10Thus,10is an eigenvector for
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