a 500 gallon tank initially contains 50 gallons of brine solution in which 28 pounds of salt have been dissolved. Beginning at time zero，brine containning 2 pound of salt per gallon is added at the rate of 3 gallons per minute， and the mixture is poured out of the tank at the rate of 2 gallons per minute，how much salt is in the tank when it contains 100 gallons of brine ???
題目如上
答案:176 pounds of salt
請好心人幫忙解惑吧!!!!
2013-03-13 5:33 pm
回答 (4)
2013-03-16 9:05 am
✔ 最佳答案
老怪物題目看錯了，請重算一遍
500gallon容量的桶子在一開始時內中有50gallons的鹽水溶液, 其中含鹽28pounds. 自時間 0 開始, 以每分鐘 3gallons, 其中每gallons含食鹽 2pounds(每分鐘 3gallons, 含食鹽6pounds), 的速度注入; 同時又把混合的溶液以每分鐘2pounds的速度排出. 在桶中溶液是 100gallons 時, 含鹽量多少?
2013-03-17 13:40:19 補充：
dS/dt = 6-2S(t)/(52+t)
注意：6
注意：52
2013-03-16 7:26 pm
嗯
Beginning at time zero，brine containning 2 pound of salt per gallon
應該是
第一分鐘
有6 lb的鹽進來
Beginning at time zero，brine containning 2 pound of salt per gallon
應該是
第一分鐘
有6 lb的鹽進來
2013-03-14 9:41 pm
500gallon容量的桶子在一開始時內中有50gallons的鹽水溶液, 其中
含鹽28pounds. 自時間 0 開始, 以每分鐘 3gallons, 其中含食鹽 2
pounds, 的速度注入; 同時又把混合的溶液以每分鐘2pounds的速度
排出. 在桶中溶液是 100gallons 時, 含鹽量多少?[解]溶液: V(0) = 50(gallons), dV/dt = 3-2 = 1 (gallon/min).
食鹽: S(0) = 28(pounds), dS/dt = 2 - 2*S(t)/V(t) (pounds/min).
問: S(t) when V(t)=100(gallons).V(t) = 50+t, t≧0.dS/dt = 2-2S(t)/(50+t)
dS/dt+S[2/(50+t)] = 2
解 S(t) = e^{-∫ 2dt/(50+t)} ∫ 2 e^{∫2dt/(50+t)} dt = (2/3)(50+t)+C/(50+t)^2
S(0)=28 代入, 得 S(t) = 2(50+t)/3-40000/[3(50+t)^2]V(t) = 100 ==> t = 50 ==> S(t) = 200/3-40000/30000 = 196/3 (pounds)我不知為何你的答案是 176 pounds,
以每一分鐘為考量的離散型計算之一:
(排出之含鹽量是以期初濃度計算)
t V(t) S(t) S(t)/V(t)
0 50 28.00 0.56000
1 51 28.88 0.56627
2 52 29.75 0.57207
3 53 30.60 0.57742
4 54 31.45 0.58238
5 55 32.28 0.58698
6 56 33.11 0.59125
7 57 33.93 0.59522
8 58 34.74 0.59891
9 59 35.54 0.60236
10 60 36.33 0.60557
11 61 37.12 0.60858
12 62 37.91 0.61139
13 63 38.68 0.61402
14 64 39.46 0.61649
15 65 40.22 0.61880
16 66 40.98 0.62098
17 67 41.74 0.62302
18 68 42.50 0.62495
19 69 43.25 0.62676
20 70 43.99 0.62847
21 71 44.74 0.63009
22 72 45.48 0.63161
23 73 46.21 0.63305
24 74 46.95 0.63442
25 75 47.68 0.63571
26 76 48.41 0.63693
27 77 49.13 0.63809
28 78 49.86 0.63919
29 79 50.58 0.64023
30 80 51.30 0.64122
31 81 52.02 0.64216
32 82 52.73 0.64306
33 83 53.44 0.64391
34 84 54.16 0.64473
35 85 54.87 0.64550
36 86 55.58 0.64624
37 87 56.28 0.64694
38 88 56.99 0.64761
39 89 57.69 0.64826
40 90 58.40 0.64887
41 91 59.10 0.64946
42 92 59.80 0.65002
43 93 60.50 0.65056
44 94 61.20 0.65107
45 95 61.90 0.65156
46 96 62.60 0.65203
47 97 63.29 0.65249
48 98 63.99 0.65292
49 99 64.68 0.65334
50 100 65.37 0.65374
2013-03-14 13:46:09 補充：
若排出之鹽量以 "已注入3gallons鹽水溶液後之結果" 計算,
最後含鹽量雖有少許不同, 但相差不大:
5010065.29 0.65294
兩種結果都與前面連續型計算結果(196/3=65.333)相近.
含鹽28pounds. 自時間 0 開始, 以每分鐘 3gallons, 其中含食鹽 2
pounds, 的速度注入; 同時又把混合的溶液以每分鐘2pounds的速度
排出. 在桶中溶液是 100gallons 時, 含鹽量多少?[解]溶液: V(0) = 50(gallons), dV/dt = 3-2 = 1 (gallon/min).
食鹽: S(0) = 28(pounds), dS/dt = 2 - 2*S(t)/V(t) (pounds/min).
問: S(t) when V(t)=100(gallons).V(t) = 50+t, t≧0.dS/dt = 2-2S(t)/(50+t)
dS/dt+S[2/(50+t)] = 2
解 S(t) = e^{-∫ 2dt/(50+t)} ∫ 2 e^{∫2dt/(50+t)} dt = (2/3)(50+t)+C/(50+t)^2
S(0)=28 代入, 得 S(t) = 2(50+t)/3-40000/[3(50+t)^2]V(t) = 100 ==> t = 50 ==> S(t) = 200/3-40000/30000 = 196/3 (pounds)我不知為何你的答案是 176 pounds,
以每一分鐘為考量的離散型計算之一:
(排出之含鹽量是以期初濃度計算)
t V(t) S(t) S(t)/V(t)
0 50 28.00 0.56000
1 51 28.88 0.56627
2 52 29.75 0.57207
3 53 30.60 0.57742
4 54 31.45 0.58238
5 55 32.28 0.58698
6 56 33.11 0.59125
7 57 33.93 0.59522
8 58 34.74 0.59891
9 59 35.54 0.60236
10 60 36.33 0.60557
11 61 37.12 0.60858
12 62 37.91 0.61139
13 63 38.68 0.61402
14 64 39.46 0.61649
15 65 40.22 0.61880
16 66 40.98 0.62098
17 67 41.74 0.62302
18 68 42.50 0.62495
19 69 43.25 0.62676
20 70 43.99 0.62847
21 71 44.74 0.63009
22 72 45.48 0.63161
23 73 46.21 0.63305
24 74 46.95 0.63442
25 75 47.68 0.63571
26 76 48.41 0.63693
27 77 49.13 0.63809
28 78 49.86 0.63919
29 79 50.58 0.64023
30 80 51.30 0.64122
31 81 52.02 0.64216
32 82 52.73 0.64306
33 83 53.44 0.64391
34 84 54.16 0.64473
35 85 54.87 0.64550
36 86 55.58 0.64624
37 87 56.28 0.64694
38 88 56.99 0.64761
39 89 57.69 0.64826
40 90 58.40 0.64887
41 91 59.10 0.64946
42 92 59.80 0.65002
43 93 60.50 0.65056
44 94 61.20 0.65107
45 95 61.90 0.65156
46 96 62.60 0.65203
47 97 63.29 0.65249
48 98 63.99 0.65292
49 99 64.68 0.65334
50 100 65.37 0.65374
2013-03-14 13:46:09 補充：
若排出之鹽量以 "已注入3gallons鹽水溶液後之結果" 計算,
最後含鹽量雖有少許不同, 但相差不大:
5010065.29 0.65294
兩種結果都與前面連續型計算結果(196/3=65.333)相近.
2013-03-14 1:28 am
一個500加侖的油箱，最初包含50加侖的鹽水中，28磅的鹽被溶解。在零時刻開始，鹽水洗滌用於含鹽每加侖2磅是3加侖每分鐘的速率加入，並將該混合物傾滿分罐在2加侖率每分鐘多少鹽時，水箱中它包含了100加侖的鹽水？
收錄日期: 2021-05-04 01:51:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130313000016KK01151