Q. 48 - Calculus

Question

The National Oceanic and Atmospheric Administration tabulates flow data from many American rivers. From this data they compute and plot median annual flows. The flows are given by functions whose closed form is not known, but for which we can read off values for any day of the year we like. The following table describes the flow f(t) in cubic feet per second for Idaho’s Lochsa River, t days after January 1: 

t	0	60	120	180	240	300	360
f(t)	700	1000	6300	4000	500	650	700

(a) Keeping in mind that the data are periodic with period 365, compute the left Riemann sum for the function that these data have sampled. 

(b) What is the relation between the numbers you computed and the total amount of water that flows down the Lochsa annually? Estimate the total amount of water that flows down the Lochsa each year.

(c) Most of the flow down the river takes place from April to July. We can get a better idea of the total flow if we add a few data points. Recompute the left Riemann sum, adding the data points (90, 2100), (150, 11000), and (210, 1000). 

(d) What would you need to do to get an even better estimate of the total flow? 

Step-by-Step Solution

Answer

Total amount of water that flows down the Riverannually is 7,89,000.

left -Riemann sum is 8,73,000.

Step 1. Given information is.

$\mathrm{The} \mathrm{following} \mathrm{table} \mathrm{shows} \mathrm{the} \mathrm{flow} \mathrm{f}\left(\mathrm{n}\right) \mathrm{in} \mathrm{cubic} \mathrm{feet} \mathrm{per} \mathrm{second} \mathrm{for} \mathrm{River}, \mathrm{t} \mathrm{days} \mathrm{after} \mathrm{January} 1.$

t	0	60	120	180	240	300	360
f(t )	700	1000	6300	4000	500	650	700

Step 2. Calculating Left Reimann sum

$$\begin{gathered} \mathrm{Left} \mathrm{Riemann} \mathrm{sum} \mathrm{is} \mathrm{the} \mathrm{approximation} \mathrm{of} \mathrm{the} \mathrm{total} \mathrm{amount} \mathrm{of} \mathrm{water} \mathrm{that} \mathrm{flow} \mathrm{sdown}, \mathrm{the} \mathrm{refore} \mathrm{it} \mathrm{will} \\ \mathrm{give} \mathrm{almost} \mathrm{slightly} \mathrm{more} \mathrm{or} \mathrm{less} \mathrm{than} \mathrm{the} \mathrm{actual} \mathrm{total} \mathrm{amount}. \mathrm{The} \mathrm{actual} \mathrm{water} \mathrm{flow} \mathrm{is} \mathrm{calculated} \mathrm{by} \mathrm{finding} \\ \mathrm{the} \mathrm{average} \mathrm{of} \mathrm{left} \mathrm{Riemann} \mathrm{sum} \mathrm{and} \mathrm{right} \mathrm{Riemann} \mathrm{sum}. \\ \mathrm{The} \mathrm{right} \mathrm{Riemann} \mathrm{sum} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{sum} \sum _{\mathrm{k}=1}^6\mathrm{f}\left({\mathrm{t}}_{\mathrm{k}+1}\right)∆\mathrm{t}. \\ \sum _{\mathrm{k}=1}^6\mathrm{f}\left({\mathrm{t}}_{\mathrm{k}+1}\right)∆\mathrm{t}= ∆\mathrm{t}\left[\mathrm{f}\right({\mathrm{t}}_2)+\mathrm{f}({\mathrm{t}}_3)+\mathrm{f}({\mathrm{t}}_4)+\mathrm{f}({\mathrm{t}}_5)+\mathrm{f}({\mathrm{t}}_6)+\mathrm{f}({\mathrm{t}}_7\left)\right] \\ =60(1000+6300+4000+500+650+700) \\ =7,89,000 \\ \mathrm{And} \mathrm{the} \mathrm{left} \mathrm{Riemann} \mathrm{sum} \mathrm{for} \mathrm{the} \mathrm{function} \mathrm{that} \mathrm{given} \mathrm{data} \mathrm{have} \mathrm{sampled} \mathrm{is} \mathrm{also} \mathrm{the} \mathrm{same} \mathrm{value} 7,89,000. \\ \mathrm{Thus}, \mathrm{the} \mathrm{actual} \mathrm{water} \mathrm{flow} \mathrm{is} 7,89,000. \\ \mathrm{Hence}, \mathrm{the} \mathrm{total} \mathrm{amount} \mathrm{of} \mathrm{water} \mathrm{that} \mathrm{flows} \mathrm{down} \mathrm{the} \mathrm{Riverannually} \mathrm{is} 7,89,000. \end{gathered}$$

Step 3. Part (c)

$$\begin{gathered} \mathrm{Given} \mathrm{information}: \mathrm{The} \mathrm{following} \mathrm{table} \mathrm{shows} \mathrm{the} \mathrm{flow} \mathrm{f}\left(\mathrm{t}\right) \mathrm{in} \mathrm{cubic} \mathrm{feet} \mathrm{per} \mathrm{second} \mathrm{for} \mathrm{Idaho}\text{'}\mathrm{s} \mathrm{Lochsa} \mathrm{River}, \mathrm{t} \mathrm{days} \mathrm{after} \mathrm{January} 1. \\ \mathrm{t} 0 60 120 180 240 300 360 \\ \mathrm{f}(\mathrm{t} ) 700 1000 6300 4000 500 650 700 \\ \mathrm{The} \mathrm{more} \mathrm{data} \mathrm{points} \mathrm{are} \mathrm{given} \mathrm{by} (90,2100), (150,11000) \mathrm{and} (210,1000). \\ \mathrm{Calculation}:\mathrm{From} \mathrm{the} \mathrm{table},\mathrm{we} \mathrm{can} \mathrm{write} \mathrm{as} \\ ({\mathrm{t}}_1.\mathrm{f}({\mathrm{t}}_1\left)\right)=(0,700) \\ ({\mathrm{t}}_2,\mathrm{f}({\mathrm{t}}_2\left)\right)=(60,1000) \\ ({\mathrm{t}}_3,\mathrm{f}({\mathrm{t}}_3\left)\right)=(90,2100) \\ ({\mathrm{t}}_4.\mathrm{f}({\mathrm{t}}_4\left)\right)=(120,6300) \\ ({\mathrm{t}}_5.\mathrm{f}({\mathrm{t}}_5\left)\right)=(150,11000) \\ ({\mathrm{t}}_6,\mathrm{f}({\mathrm{t}}_6\left)\right)=(180,4000) \\ ({\mathrm{t}}_7.\mathrm{f}({\mathrm{t}}_7\left)\right)=(210,1000) \\ ({\mathrm{t}}_8.\mathrm{f}({\mathrm{t}}_8\left)\right)=(240,500) \\ ({\mathrm{t}}_9.\mathrm{f}({\mathrm{t}}_9\left)\right)=(300,650) \\ ({\mathrm{t}}_{10}.\mathrm{f}({\mathrm{t}}_{10}\left)\right)=(360,700) \\ \mathrm{Here} \mathrm{we} \mathrm{use} \mathrm{the} \mathrm{left} \mathrm{end} \mathrm{points} \mathrm{of} 9 \mathrm{rectangles} \mathrm{on} \mathrm{the} \mathrm{interval} [0,360]. \\ \mathrm{The} \mathrm{width} \mathrm{of} \mathrm{each} \mathrm{rectangle} \mathrm{is} \mathrm{differ} \mathrm{by} 30 \mathrm{or} 60. \\ \mathrm{The} \mathrm{right} \mathrm{Riemann} \mathrm{sum} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{sum} \sum _{\mathrm{k}=1}^9\mathrm{f}\left({\mathrm{t}}_{\mathrm{k}}\right)∆\mathrm{t}. \\ \sum _{\mathrm{k}=1}^9\mathrm{f}\left({\mathrm{t}}_{\mathrm{k}}\right)∆{\mathrm{t}}_{\mathrm{k}}= [∆{\mathrm{t}}_1\mathrm{f}({\mathrm{t}}_1)+∆{\mathrm{t}}_2\mathrm{f}({\mathrm{t}}_2)+∆{\mathrm{t}}_3\mathrm{f}({\mathrm{t}}_3)+∆{\mathrm{t}}_4\mathrm{f}({\mathrm{t}}_4)+∆{\mathrm{t}}_5\mathrm{f}({\mathrm{t}}_5)+∆{\mathrm{t}}_6\mathrm{f}({\mathrm{t}}_6)+∆{\mathrm{t}}_7\mathrm{f}({\mathrm{t}}_7)+∆{\mathrm{t}}_8\mathrm{f}({\mathrm{t}}_8)+∆{\mathrm{t}}_9\mathrm{f}({\mathrm{t}}_9\left)\right] \\ =60\left(700\right)+30\left(1000\right)+30\left(2100\right)+30\left(6300\right)+30\left(11000\right)+30\left(4000\right) +30 \left(1000\right) 60\left(500\right) + 60\left(650\right) \\ =8,73,000 \\ \mathrm{Hence}, \mathrm{the} \mathrm{left} \mathrm{Riemann} \mathrm{sum} \mathrm{for} \mathrm{the} \mathrm{function} \mathrm{that} \mathrm{given} \mathrm{data} \mathrm{have} \mathrm{sampled} \mathrm{is} 8, 73, 000. \end{gathered}$$

Step 4. The way to get a better estimation .

$$\begin{gathered} \mathrm{For} \mathrm{a} \mathrm{better} \mathrm{estimation} \mathrm{of} \mathrm{the} \mathrm{total} \mathrm{flow}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{subdivide} \\ \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{into} \mathrm{more} \mathrm{number} \mathrm{of} \mathrm{intervals}. \\ \mathrm{So} \mathrm{that} \mathrm{we} \mathrm{can} \mathrm{add} \mathrm{few} \mathrm{more} \mathrm{data} \mathrm{points} \mathrm{that} \mathrm{leads} \mathrm{to} \mathrm{an} \mathrm{even} \mathrm{better} \mathrm{approximation}. \end{gathered}$$