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math16B_review_exer2_ans

math16B_review_exer2_ans - 13(a P = 05 P-12 000(b P t = 240...

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Math 16B, Section 1 Fall 2005 Sarason ANSWERS TO REVIEW EXERCISES 2 (USER BEWARE) 1. (a) cos(ln t ) t (b) cot t (c) e t sec( e t ) tan( e t ) (d) sec t tan t sec 2 (sec t ) 2. (a) 4 - 2 e (b) π 2 4 - 2 (c) 1 / 4 (d) 2 / 3 3. By substitution: 2 5 ( x - 1) 5 / 2 + 2 3 ( x - 1) 3 / 2 + C By integration by parts: 2 3 x ( x - 1) 3 / 2 - 4 15 ( x - 1) 5 / 2 + C Reconciliation: 2 5 ( x - 1) 5 / 2 + 2 3 ( x - 1) 3 / 2 = ( x - 1) 3 / 2 2 5 x - 2 5 + 2 3 = ( x - 1) 3 / 2 2 5 x + 4 15 = ( x - 1) 3 / 2 2 3 x - 4 15 x + 4 15 = 2 3 x ( x - 1) 3 / 2 - 4 15 ( x - 1) 5 / 2 . 4. Tangent of adjacent angle: 12 / 5 Tangent of opposite angle: 5 / 12 5. 300 feet 6. 7 π/ 30 . 733 ft/min. 7. 2 . 856 mph 8. 3 / 2 miles/minute (= 90 mph) 9. (a) y = Ce 1 2 sin 2 t (b) y = ± 1 / 2 t 4 4 + C , and y = 0 (c) y = Ce sin t - 1 10. (a) y = sec t (b) y = 0 (c) y = 100 , 000 - 50 , 000 e . 075 t 11. (a) y = 1 / (2 t 3 / 2 - C ), and y = 0 (b) y = 1 / (2 t 3 / 2 - 1) (c) y = 0 12. (a) - 1 / ( t + C ) 2 , C = constant, and the zero function.

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Unformatted text preview: 13. (a) P = . 05 P-12 , 000 (b) P ( t ) = 240 , 000-190 , 000 e . 05 t 14. (a) P = . 045 P-9000 (b) P ( t ) = 200 , 000-Ce . 045 t (c) P ( t ) = 200 , 000-164 , 000 e . 045 t (d) ≈ 4 . 4 years. Corrections 1. In the statement of Exercise 5, an incorrect value is given for cos 1 . 25. The correct value (to three decimals) is . 315. 2. The answer given for Exercise 12(a) is incorrect. The correct answer is y =-1 t + C , and y = 0 ....
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