The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. The table below gives the rate as measured at various times in an 8-hour time period.
t---------0-----2------3-------7----8
(hours)
R(t)--1.95---2.5---2.8----4.00--4.26
(gallons per
hour)
Use a trapezoidal sum with the four sub-intervals indicated by the data in the table to estimate Using correct units, explain the meaning of your answer in terms of water flow.
So I did this and found out
1. 4[1.95 + 2(2.5 + 2.8 + 4.0) + 4.26] = 99.24 gallons per 8 hours
Now im just trying to figure out the second one .
Is there some time t, 0 < t < 8, such that R′(t) = 0? Justify your answer.
I thought no because the function is always increasing.
Avery, you may have a typo, though I may be late in answering, haha. The trapezoidal sum yields 24.83
its 24.83
for the trapezoidal sum, you cannot factor out the width as 4, because the intervals are not even. you have to do the width of each individual interval. The actual answer is around 23.83
Nice
To estimate the total amount of water that flows into the tank in an 8-hour time period using the trapezoidal sum, you first need to divide the time interval into sub-intervals. In this case, you have four sub-intervals: [0, 2], [2, 3], [3, 7], and [7, 8].
Next, you calculate the average rate of flow on each sub-interval by adding the rates at the beginning and end of the sub-interval and dividing by 2. Then, you multiply each average rate by the width of the corresponding sub-interval.
For the given data, the trapezoidal sum can be calculated as follows:
Trapezoidal sum = (2 - 0)*[(1.95 + 2.5)/2] + (3 - 2)*[(2.5 + 2.8)/2] + (7 - 3)*[(2.8 + 4.0)/2] + (8 - 7)*[(4.0 + 4.26)/2]
Simplifying this expression gives:
Trapezoidal sum = 2.45 + 1.65 + 12.4 + 2.13 = 18.63 gallons
Therefore, the estimated amount of water that flows into the tank in an 8-hour time period is approximately 18.63 gallons.
As for your second question, you are asked to find out if there is any time t between 0 and 8 (exclusive) such that the derivative of the flow rate R(t) is equal to 0, denoted as R'(t) = 0.
To justify your answer, you need to examine the behavior of the function R(t) and determine if it is always increasing or not. From the given table values, you can observe that the flow rate is actually increasing from 1.95 gallons per hour at t = 0 to 4.26 gallons per hour at t = 8. This means that R(t) is increasing over the entire interval [0, 8], and therefore it does not equal zero at any point within that interval.
Hence, your initial understanding is correct - there is no time t between 0 and 8 (exclusive) such that R'(t) = 0. The rate of water flow is continuously increasing within the given time period.