Final Exam (Year N:A) Solutions - FINAL EXAM CALC II(BIO...

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FINAL EXAM: CALC II (BIO AND SOC. SCI.) Problem 1. a) Integrate dy = (3 t - 1) dt : y = R (3 t - 1) dt = 3 t 2 / 2 - t + C . Determine the constant: y (2) = 6 - 4 + C = 5 C = 1. So y ( t ) = 3 t 2 / 2 - t + 1. b) Integrate dy 1 - y = 2 dt : - ln(1 - y ) = 2 t + C 1 . Exponentiate: 1 - y = Ce - 2 t , so y ( t ) = 1 - Ce - 2 t . Determine the constant: y (0) = 1 - C = 2 C = - 1. So y ( t ) = 1 + e - 2 t . c) This is the autonomous equation dy dt = g ( y ) with g ( y ) = y 5 . Equilibrium solu- tion: ˆ y = 0. The first derivative criterion is inconclusive since g 0 y ) = g 0 (0) = 0. However g changes sign near ˆ y and since g ( y ) > 0 for y > ˆ y , this means that ˆ y = 0 is an unstable equilibrium. Problem 2. a) det A = 2 and det B = 0, so A is invertible and B is not. A - 1 = 1 2 - 3 - 1 - 1 - 1 = - 3 / 2 - 1 / 2 - 1 / 2 - 1 / 2 . b) AX = - 2 3 X = A - 1 - 2 3 = - 3 / 2 - 1 / 2 - 1 / 2 - 1 / 2 ¶ • - 2 3 = 3 / 2 - 1 / 2 . Problem 3. a) --→ QR = - 1 - 7 ⇒ | --→ QR | = p 1 2 + ( - 7) 2 = 50. b) cos P = --→ P Q · -→ P R k --→ P Q kk -→ P R k . But --→ PQ = 3 1 and -→ PR = 2 - 6 . Hence --→ PQ · -→ PR = 0. Therefore cos P = 0 P = π 2 = 90 . c) The equation of the line passing through (1 , 2) and perpendicular on the vector --→ QR = - 1 - 7 is ( - 1)( x - 1) + ( - 7)( y - 2) = 0 x + 7 y - 15 = 0. Problem 4. a) f x = sin( πxy ) +

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