Yes, notice that $\displaystyle R(t)$ is a differenciable function, and $\displaystyle R(0)=R(24)$ thus, by Rolle's Theorem (or Mean-Value theorem) there exists a real number, $\displaystyle 0<t<24$ such as,Originally Posted by FrozenFlames
$\displaystyle R'(t)=0$
This is my 6th Post!!
Mid-point Riemann Sum, avoiding too-technical stuff, is about computing the area between a curve and the horizontal axis by subdividing the interval [a,b] into n number of equal sub-divisions, and then by multiplying the midpoint height or depth of a subdivision by the length or width of the subdivision, thereby assuming the area of a subdivision is a rectangle, and then summing all the individual areas of the n number of subdivisions.Originally Posted by frozenflames
Zeez, if you understand that, then it is not my fault.
I am just avoiding using the summation symbol and all those a(i), x(i), a(i-1), etc. With LaTex there'd be no problem showing those, but I'd shun LaTex as long as I can express myself good enough with the old reliable ASCI. I am a simple guy. :-)
So, the interval is [0,24] in hours.
The n is 4.
Hence, length or width of a subdivision is (24-0)/4 = 6 hrs.
It so happens that in the given data, the midpoint heights of the 4 subdivisions are already shown:
Subdiv. [0,6]....midpoint height = reading at t=3, = 10.4
Subdiv. [6,12]....midpoint height = reading at t=9, = 11.2
Subdiv. [12,18]....midpoint height = reading at t=15 = 11.3
Subdiv. [18,24]....midpoint height = reading at t=21, = 10.2
Hence,
INT.(0-->24)[R(t)]dt
= 6*10.4 +6*11.2 +6*11.3 +6*10.2
= 6(10.4 +11.2 +11.3 +10.2)
= 6(43.1)
= 258.6 gallons per 24hrs, or per day, or gpd.
Or,
= 260 gpd. --------answer. Approximate, remember.
Explain that answer in terms of waterflow?
Easy. The waterflow of the pipe is about 260 gpd. Or, the pipe dislodges water at the rate of about 260 gallon per day.
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Is there a time between 0 and 24 hours where R'(t)=0?
Where the 1st derivative of R(t) is zero?
Yes. Per the given data, the slope of the rate of flow curve is increasing t=from t=0 to t=12, and then decreasing from t=15 to t=24.
That means somewhere between t=12 and t=15, the slope of the rate of flow curve became horizontal---where R'(t) is zero.
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Q(t) = (1/79)(768 +23t -t^2) will approximate the average rate of waterflow.
Solve for R(t) by using Q(t).
This portion is not clear to me.
Are R(t) and Q(t) not one and the same?
If not, then R refers to rate and Q refers to quantity or volume?
Then Q = R*A
where A refers to the effective cross-sectional area of the pipe.
We do not know the size of the pipe, so let me pass on this portion of your question.