Chapter 6, Problem 58RE

(a)

To determine

To calculate:The value of the water flow.

Answer to Problem 58RE

The volume of water in gallons that flowed through the pipe during the 24 hour period.

Explanation of Solution

Given information:

The given table is shown below,

  

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of f on [a, b] can be approximated by a straight line segment at each subinterval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases".

That is,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=h\frac{y_0+y_1}{2}+h\frac{y_1+y_2}{2}+\mathrm{.....}+h\frac{y_{n-1}+y_n}{2}\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

  $y_0=f\left(a\right),y_1=f\left(x_1\right)\mathrm{....}{y}_{n-1}=f\left(x_{n-1}\right),y_n=f\left(b\right)$

Now, apply the trapezoidal rule with

  $\begin{array}{l}n=6\\ \left[a,b\right]=\left[0,24\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{24-0}{6}\\ h=4\end{array}$

Now, substitute values in equation,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{24}{\int }}R\left(t\right)}dt=\frac{4}{2}\left(9.6+2\left(10.3\right)+2\left(10.9\right)+2\left(11.1\right)+2\left(10.5\right)+9.6\right)\\ {\displaystyle \underset{0}{\overset{24}{\int }}R\left(t\right)}dt=253.2\end{array}$

This approximation gives volume of water in gallons that flowed through the pipe during the 24 hour period.

Conclusion:

The volume of water in gallons that flowed through the pipe during the 24 hour period.

(b)

To determine

To find:The value varies between 0 to 24.

Answer to Problem 58RE

The value is $f\text{'}\left(c\right)=0$

Explanation of Solution

Given information:

The given table is shown below,

  

Formula used:

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Yes, it is possible.

According to mean value theorem if a function is continuous on [a, b] then there is a point e in the interval satisfy this condition

  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Since,

  $f\left(0\right)=f\left(24\right)$

Hence,

  $\begin{array}{l}f\text{'}\left(c\right)=\frac{f\left(24\right)-f\left(0\right)}{24-0}\\ f\text{'}\left(c\right)=0\end{array}$

Conclusion:

The value is $f\text{'}\left(c\right)=0$

(c)

To determine

To find:The average rate of water flow during 24 hr period.

Answer to Problem 58RE

The average rate of water flow during 24 hr period is $Q\left(t\right)=10.34\text{gal/hr}$

Explanation of Solution

Given information:

The given table is shown below,

  

Formula used:

Substitution method is used.

Calculation:

Since, water flow has been given by,

  $Q\left(t\right)=0.01\left(950+25t-t^2\right)$

Since, average rate of water flow is required

  $\begin{array}{l}t=4\\ Q\left(t\right)=0.01\left(950+25\left(4\right)-4^2\right)\\ =0.01\times 1034\\ Q\left(t\right)=10.34\text{gal/hr}\end{array}$

Conclusion:

The average rate of water flow during 24 hr period is $Q\left(t\right)=10.34\text{gal/hr}$