140Bquiz11asols

# 140Bquiz11asols - b 85 = 70 28 6 e-Nk 80 = 70 28 6 e N 1 k...

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1 MATH 140B QUIZ#11 Solutions Due: Nov, 2007 1. P ( t ) = 100 e kt , so 560 = 100 e 3 k . Therefore, 5 . 6 = e 3 k or k = 1 3 ln 5 . 6 = 0 . 5743 We get, P ( t ) = 100 e 0 . 5743 t . The generation time is ln 2 k = ln 2 0 . 5743 = 1 . 2. Therefore, B) is the answer. 2. At depth 16 meters, the intensity is 1% of that at the surface, so I (16) = 0 . 01 · I 0 . That is, I 0 · a 16 = 0 . 01 · I 0 . Canceling I 0 gives a 16 = 0 . 01 . Taking the sixteenth root of both sides, we get a = (0 . 01) 1 16 = 0 . 75 . Therefore, D) is the answer. 3. If I 0 denotes the initial amount stored, then the amount remaining after t days is I ( t ) = I 0 p 1 2 P t/ 60 . 1 . For some value of t (to be determined) that I ( t ) = 0 . 75 · I 0 , or equivalently, I 0 p 1 2 P t/ 60 . 1 = 0 . 75 · I 0 . Canceling I 0 gives 0 . 75 = p 1 2 P t/ 60 . 1 . Taking ln of both sides and using the power rule for logs, gives ln(0 . 75) = t 60 . 1 ln(0 . 5) , or ±nally, t = 60 . 1 · ln(0 . 75) ln(0 . 5) = 24 . 9 days. Therefore, A) is the answer. 4. We know that T ( t ) = C + ( T 0 - C ) e - kt = 70 + (98 . 6 - 70) e - kt = 70 + 28 . 6 e - kt . Let t = N be the time the body was discovered. Then we get two equations:
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Unformatted text preview: b 85 = 70 + 28 . 6 e-Nk 80 = 70 + 28 . 6 e-( N +1) k or equivalently, b 15 = 28 . 6 e-Nk 10 = 28 . 6 e-( N +1) k Dividing these equations gives 3 2 = e k . Putting this back into the ±rst of the two equations above, gives 15 = 28 . 6 ( 2 3 ) N . Taking ln of both sides and using the power rule for logs, gives N = ln p 15 28 . 6 P ln 2 3 = 1 . 6 hours. Therefore, C) is the answer. 5. Let S be the initial amount of St 90 . We want the value of t such that S ( t ) = 1 3 S . That is, S ( 1 2 ) t/ 29 = 1 3 S . Canceling S gives ( 1 2 ) t/ 29 = 1 3 . Taking logs and using the power rule, gives t = 29 · ln 3 ln 2 ≈ 46 . Therefore, D) is the answer....
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