651 0 70 e 1 0 70 e125t 125t c lim vt 0 t 25

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Unformatted text preview: ) = A0 B0 (ekA0 t − ekB0 t ) A0 ekA0 t − B0 ekB0 t (b) v(t ) = 1 2xy y2 . Then the resulting differential equation is y2 = − + , y y (y )2 , where C is an arbitrary constant. α v0 α v0 e−(a/m)t = (α/m)t − β v ( α + β v0 ) e α + β v0 − β v0 e−(α/m)t 0 (b) v(t ) = 15. 65(1 + 0. 70 e 1 − 0. 70 e−1.25t −1.25t (c) lim v(t ) = 0 t →∞ 25, 000 , y(20) ∼ 1544 = 1 + 249 e−0.1398t (b) 40 days 35. (a) 88. 82 m/sec ) (c) 15.65 m/sec CHAPTER 9 SECTION 9.1 1. (a) 2 13 (b) 29 13 3. (0, 1) is the closest point ( − 1, 1) the farthest away 5. 17 2 7. Adjust the sign of A and B so that the equation reads Ax + By = |C |. Then we have B |C | A +y√ =√ . x√ A2 + B2 A2 + B 2 A2 + B 2 Now set A = cos α , √ A2 + B 2 B = sin α , √ A2 + B2 |C | = p. √ A2 + B2 p is the length of OQ, the distance between the line and the origin; α is the angle from the positive x-axis to the line segment OQ. y Q P α O x 9. y2 = 8x y 11. (x + 1)2 = −12(y − 3) y (–1,3) 13. 4y = (x − 1)2 y 15. ( y − 1)2 = −2(x − 3 ) 2 y ( , 1) 3 2 x x (1,0) x x...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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