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preptest3Bsol
Course: MATH 2605, Fall 2008School: Georgia Tech
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to Solutions practice test 3B Problem I: a) 1 1 2 Q= 2 1 5 1 5 7 R= 5 0 9 b) 1 5 1 + 3i U= 35 1 3i 5 U AU = c) eAt = 1 sin(3t) + 3 cos(3t) 5 sin(3t) 3 2 sin(3t) sin(2t) + 3 cos(3t) 3i 119 + 42i 0 105i Problem II: a) x (t) = b) T(t) = sin t cos(t/2) = 1 cos t sin(t/2) 2(1 cos t) N(t) = c) 2 2 sin t 1 cos t 1 sin(t/2) cos(t/2) L= 0 2(1 cos t)dt = 0 sin(t/2)dt = 8 d) (t) = 1 4 sin(t/2) Problem...
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to Solutions practice test 3B Problem I: a) 1 1 2 Q= 2 1 5 1 5 7 R= 5 0 9 b) 1 5 1 + 3i U= 35 1 3i 5 U AU = c) eAt = 1 sin(3t) + 3 cos(3t) 5 sin(3t) 3 2 sin(3t) sin(2t) + 3 cos(3t) 3i 119 + 42i 0 105i
Problem II: a) x (t) = b) T(t) = sin t cos(t/2) = 1 cos t sin(t/2) 2(1 cos t) N(t) = c)
2 2
sin t 1 cos t
1
sin(t/2) cos(t/2)
L=
0
2(1 cos t)dt =
0
sin(t/2)dt = 8
d) (t) = 1 4 sin(t/2)
Problem III: a) = cos1 (1/7) . b) 1 u 1 = 1 3 1 1
c) 4/7I + (1 1 1 1 4/7) 1 1 1 + 1 1 1 0 1 1 3/7 1 0 1 1 1 0
Problem IV: a) x y b) ( c) 0 2+ 0 2 1 1 + 42 0 1 1 + 42 0 1 (1 + 2 1 + 42 ), 0) , ( 1 (1 2 1 + 42 ), 0) = y (1 x2 ) x
e) The rst critical point is always a saddle, and the...
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