ISM_T11_C07_B - 452 Chapter 7 Transcendental Functions 10....

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452 Chapter 7 Transcendental Functions 10. (a) There are (60)(60)(24)(365) 31,536,000 seconds in a year. Thus, assuming exponential growth, œ P 257,313,431e and 257,313,432 257,313,431e ln œœ Ê œ kt 14k 31 536 000 ÐÎ ßßÑ Š‹ 257,313,432 257,313,431 31,536,000 14k k 0.0087542 ʸ (b) P 257,313,431e 293,420,847 (to the nearest integer). Answers will vary considerably œ¸ ÐÑ 0.0087542 ab "& with the number of decimal places retained. 11. 0.9P P e k ln 0.9; when the well's output falls to one-fifth of its present value P 0.2P !! ! œÊ œ œ k 0.2P P e 0.2 e ln (0.2) (ln 0.9)t t 15.28 yr Êœ Ê œ Ê œ ¸ ÐÞ Ñ Ñ ln 0 9 t ln 0 9 t ln 0.2 ln 0.9 12. (a) p dx ln p x C p e e e C e ; dp dp dx 100 p 100 100 œ ± ± ±²Êœ œ œ "" " " Ð Þ Ñ Þ Þ 001x C C 001x p(100) 20.09 20.09 C e C 20.09e 54.61 p(x) 54.61e (in dollars) œÊœ Ê œ ¸Ê œ Ð Þ ÑÐ Ñ Þ 001 100 (b) p(10) 54.61e $49.41, and p(90) 54.61e $22.20 Ð Þ ÑÐ Ñ Ð Þ ÑÐ Ñ 0 01 10 0 01 90 (c) r(x) xp(x) r (x) p(x) xp (x); œ ² ww p (x) .5461e r (x) œ ± Ê Þ (54.61 .5461x)e . Thus, r (x) 0 œ ± œ Þ w 54.61 .5461x x 100. Since r 0 Ê œ ³ w for any x 100 and r 0 for x 100, then ´´ ³ w r(x) must be a maximum at x 100. œ 13. (a) A e A e ÐÞ Ñ Þ 0045 02 œ (b) 2A A e ln 2 (0.04)t t 17.33 years; 3A A e ln 3 (0.04)t œ Ê œ ¸ œ 004t ln 2 0.04 t 27.47 years Êœ ¸ ln 3 0.04 14. (a) The amount of money invested A after t years is A(t) A e œ t (b) If A(t) 3A , then 3A A e ln 3 t or t 1.099 years Ê œ ¸ ! t (c) At the beginning of a year the account balance is A e , while at the end of the year the balance is A e . t t1 Ð Ñ The amount earned is A e A e A e (e 1) 1.7 times the beginning amount. ! t t ± œ ± ¸ 15. A(100) 90,000 90,000 1000e 90 e ln 90 100r r 0.0450 or 4.50% Ê œ Ê œ Ê œ ¸ r 100 100r ln 90 100 16. A(100) 131,000 131,000 1000e ln 131 100r r 0.04875 or 4.875% Ê œ Ê œ ¸ 100r ln 131 100 17. y y e represents the decay equation; solving (0.9)y y e t 0.585 days Ê œ ¸ ! ± Þ Þ 0 18t 018 t ln (0.9) 0.18 18. A A e and A A e e k 0.00499; then 0.05A A e Ê œ Ê œ ¸ ± œ ! ## kt 139k 139k 0 00499t ln (0.5) 139 Þ t 600 days ¸ ln 0.05 0.00499 ± 19. y y e y e y e (0.05)(y ) after three mean lifetimes less than 5% remains ´ ! ! ÐÎÑ kt k 3 k 3 yy e2 0 $ 20. (a) A A e e k 0.262 œ ¸ ! " Þ kt 2 645k ln 2 .645 (b) 3.816 years " k ¸ (c) (0.05)A A exp t ln 20 t t 11.431 years œ ± Ê ± œ ± ¸ ˆ‰ ˆ ln 2 ln 2 2.645 ln 20 2.645 2.645 ln # 21. T T (T T ) e , T 90°C, T 20°C, T 60°C 60 20 70e e ± œ ± œ Ê ± œ ss s kt 10k 10k 4 7 k 0.05596 ln 10 7 4
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Section 7.6 Relative Rates of Growth 453 (a) 35 20 70e t 27.5 min is the total time it will take 27.5 10 17.5 minutes longer to reach ± œÊ ¸ Ê ± œ Þ 0 05596t 35°C (b) T T (T T ) e , T 90°C, T 15°C 35 15 105e t 13.26 min ± œ ± œœ ± Ê ² ¸ ss s kt 0 05596t !! Þ 22. T 65° (T 65°) e 35° 65° (T 65°) e and 50° 65° (T 65°) e . Solving ± œ ± Ê ± œ ±± œ ± ! kt 10k 20k 30° (T 65°) e and 15° (T 65°) e simultaneously (T 65°) e 2(T 65°)e ± œ œ ± Ê ± œ ± ! ! 10k 20k 10k 20k e 2 k and 30° 30° e6 5 ° 60° Êœ Ê œ ± ± œ ± œ 10k 10 ln 2 ln 2 10 e T6 ! ± 10k ln 2 10 ± ˆ‰ œ± Ê œ ± 5 ° 5 ° 30° e 23. T T (T T ) e 39 T (46 T ) e and 33 T (46 T ) e e and ± œ ± Ê ± œ œ ± kt 10k 20k 10k ! ± ± 39 T 46 T s s e e (33 T )(46 T ) (39 T ) 1518 79T T 33 T 33 T 39 T 46 T 46 T 46 T ± ± # # ## Ê œ Ê œ ± Ê ±² 20k 10k s s s ab Š‹ s s 1521 78T T T 3 T 3°C œ Ê ± œÊ œ ± s s # 24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
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This note was uploaded on 10/13/2011 for the course MATHEMATIC 103 taught by Professor Thommas during the Spring '11 term at LCC Intl University.

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ISM_T11_C07_B - 452 Chapter 7 Transcendental Functions 10....

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