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.

Question

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.

Answer 1

unless the function has some pathological behavior hidden in one of the intervals, you are likely correct.

Answer 2

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

Answer 3

Avery, you may have a typo, though I may be late in answering, haha. The trapezoidal sum yields 24.83

Answer 4

Even still, the sum does not come out to 24.83, but 24.81 instead. Unless there is something I missed, This thread has an unfortunate series of typos. Quite amusing to be honest

Answer 5

its 24.83

Answer 6

Nice