2. Tidal Flows 2. Tidal Flows You are a team of consulting engineers studying th
ID: 2889833 • Letter: 2
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2. Tidal Flows 2. Tidal Flows You are a team of consulting engineers studying the fow of water from a certain river into a data in the fors of a graph desailing the amount of waber that has flowed into the lake from the river over a f period. The graph is given in Figure I tocal amount (100 million galoss) days) Figure I Your job is to repont several results to your client as follows l. First, determine the total volume of waer that has flowed from the river in the first five days of the observation period: in the first sen days; in the entire fourteen-day pericd Questions 2 and 3 are individual parts. That is, each group member is to answer the questions for a different day. Member 1: Day 4 Member 2: Day6 Member 3: Day 8 Member 4: Day 10 according to the following list 2. Estimate the average rate of fow from three days before your day until your day; from two days before your day util your day; from one day before your day unil your day. Also estimate the average rate of flow from your day until three days after your day: from your day until two days afher your day; from your day umil one day afher your day 3. Estimate the indantaneoss rale of fow on your day.Explanation / Answer
We are given the Volume Vs time graph for the flow of water.
1> The total volume of water that has flowed in the first five days:
We check the grapg ans see where the graph is when time is = 5 days
From the graph we could say total volume of water is = 23 million gallon (approximately)
2. Lets would this problem for Member 1 rest would be done in the same manner.
Average rate of flow on an time interval t E [a,b] for a function f(t) is = [f(b) - f(a)]/(b-a)
Member 1: Day: 4
Three days before your day is = Day 1
Here a = 1 and b = 4
From the graph we could see that flow on day 1 is = 0 million gallons
flow on day 4 is = 20 million gallons
=> Average rate of flow = (20-0)/(4-1)= 20/3 million gallons per day
from two days before your day that is day 2
flow on day 2 = million 11 gallons (approximate value)
flow on day 4 = 20 million gallon
=> Average rate of flow = (20-11)/(4-2) = 4.5 million gallons per day
From 1 day before your day is day 3
flow on day 3 = 17 million gallons (approximate value)
flow on day 4 = 20 million gallons
=> Average rate of flow = (20-17)/(4-3) = 3 million gallons per day
day three days after your day is , day day 7
flow on day 7 = 26 million gallons (approximate value)
flow on day 4 = 20 million gallons
=> average rate of flow = (26-20)/(7-4) = 2 million gallon per day
two days after your day is, day 6
flow on day 6 = 25 million gallon
flow on day 4 = 20 million gallons
=> average rate of flow = (25-20)/(6-4) = 2.5 million gallon per day
one day after your day is, day 5
flow on day 5 = 23 million gallon (approximate value)
flow on day 4 = 20 million gallons
average rate of flow = (23-20)/(5-4) = 3 million gallon per day
4>From the graph we could see that the flow is always increasing from day 1 till day 14.Hence the interval of increasing flow is t E [1, 10]
5> As the flow is always increasing hence there is no interval of decreasing flow.
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