Chapter 24, Problem 36P

Repeat Prob. 24.35, but use (a) Simpson's $1/3$ rule, (b) Romberg integration to ${\epsilon }_s=0.5\%$, and (c) Gauss quadrature.

To determine

To calculate: The work done for the given equations of $F\left(x\right)$ and $\theta \left(x\right)$ using Simpson’s 1/3 rule.

Answer to Problem 36P

Solution:

The work done byusing 4-segment Simpson 1/3 rule is $I=66.57542124$.

The work done by using 8-segment Simpson 1/3 rule is $I=66.96047$.

The work done by using 16-segment Simpson 1/3 rule is $I=66.92582125$.

Explanation of Solution

Given Information:

The given expressions are as follows,

$\begin{array}{l}F\left(x\right)=1.6x-0.045x^2\\ \theta \left(x\right)=0.8+0.125x-0.009x^2+0.0002x^3\end{array}$

Work done in integral form (Refer Sec. 24.4)

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)dx}$

If the direction between the force and displacement changes between initial and final position, then the work done is written as,

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$ …… (1)

Here, $\theta \left(x\right)$ is the angle between force and displacement.

Formula Used:

Simpson’s 1/3 rule.

$I=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Calculation:

Calculate the work done.

Substitute the value of $F\left(x\right)\text{ and }\theta \left(x\right)$ in equation (1).

Force $F\left(x\right)$ is given by,

$F\left(x\right)=1.6x-0.045x^2$ …… (2)

And,

$\theta \left(x\right)=0.8+0.125x-0.009x^2+0.0002x^3$

4-segment Simpson’s 1/3 rule.

Calculate integral from $x_0=0\text{ to }{x}_n=30$ value of $h$ is 7.5. So, for four segments Simpson’s 1/3 rule the whole interval is divided into four equally spaced interval.

$n$	$x_n$
0	0
1	7.5
2	15
3	22.5
4	30

Calculate $F\left(x_0\right)$ from equation (2),

$\begin{array}{c}F\left(x_0\right)=1.6x_0-0.045{\left(x_0\right)}^2\\ =1.6\left(0\right)-0.045{\left(0\right)}^2\\ =0\end{array}$

Calculate $F\left(x_1\right)$ from equation (2),

$\begin{array}{c}F\left(x_1\right)=1.6\left(7.5\right)-0.045{\left(7.5\right)}^2\\ =9.46875\end{array}$

Similarly, calculate $F\left(x_2\right),F\left(x_3\right)\text{ and }F\left(x_4\right)$

Calculate $\theta \left(x_n\right)$

$\theta \left(x\right)=0.8+0.125x-0.009x^2+0.0002x^3$

Calculate $\theta \left(x_0\right)$ for $x_0=0$.

$\begin{array}{c}\theta \left(x_0\right)=0.8+0.125\left(0\right)-0.009{\left(0\right)}^2+0.0002{\left(0\right)}^3\\ =0.8\end{array}$

Calculate $\theta \left(x_1\right)$ for $x_0=7.5$.

$\begin{array}{c}\theta \left(x_1\right)=0.8+0.125x_1-0.009x_1{}^2+0.0002x_1{}^3\\ =0.8+0.125\left(7.5\right)-0.009{\left(7.5\right)}^2+0.0002{\left(7.5\right)}^3\\ =1.315625\end{array}$

Similarly, calculate $\theta \left(x_2\right)$, $\theta \left(x_3\right)$ and $\theta \left(x_4\right)$.

Calculate $F\left(x_0\right)\cos \left[\left({\theta }_0\right)\right]$

Substitute value of $F\left(x_0\right)\text{=0 and }\theta \left(x_0\right)=0.8$

$\begin{array}{c}F\left(x_0\right)\cos \left[\left({\theta }_0\right)\right]=\left(0\right)cos\left(0.8\right)\\ =0\end{array}$

Calculate $F\left(x_1\right)\cos \left[\left({\theta }_1\right)\right]$

Substitute value of $F\left(x_1\right)\text{=9}\text{.46875 and }\theta \left(x_0\right)=1.315625$

$\begin{array}{c}F\left(x_1\right)\cos \left[\left({\theta }_1\right)\right]=\left(9.46875\right)\cos \left(1.315625\right)\\ =2.39001847\end{array}$

Similarly, calculate all the other values.

All the values which are calculated in are tabulated below,

$n$	$x_n$	$F\left(x_n\right)$	$\theta \left(x_n\right)$	$y_n=F\left(x_n\right)\cos \left[\left({\theta }_n\right)\right]$
0	0	0	0.8	0
1	7.5	9.46875	1.315625	2.39001847
2	15	13.875	1.325	3.376187019
3	22.5	13.21875	1.334375	3.096161858
4	30	7.5	1.85	–2.066926851

Apply Simpson’s 1/3 rule to calculate work.

According to Simpson’s 1/3 rule integral $I$ is written as,

$I=\frac{h}{3}\left[f\left(a\right)+f\left(b\right)+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

$I=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$ …… (3)

Here, $a$ and $b$ are the lower and upper limit of the integral respectively and $h$ is the step size that is $x_n-x_{n-1}$.

Work done is given as,

$W={\displaystyle \underset{0}{\overset{30}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$

Here, $a=0,b=3\text{0 }\text{and }h=7.5$

Calculate $I$ for $n=4$ from equation (3).

$I=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2\left(y_2\right)\right]$ …… (4)

Substitute values of $h,$ $y_0,y_1,y_2,y_3\text{and }{y}_4$ in equation (4).

$\begin{array}{c}I=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2\left(y_2\right)\right]\\ =\frac{7.5}{3}\left[0+\left(-2.066926851\right)+4\left(2.39001847+3.096161858\right)+2\left(3.376187019\right)\right]\\ =66.5754212475\end{array}$

Hence, the value of integral is $I=66.57542124$.

8-segment Simpson’s 1/3 rule.

For eight segmented rule value of $h=3.75$ .The whole interval from $x_0=0\text{ to }{x}_n\text{=30}$ is divided into 8 intervals. Here, $n=8$ and $h=3.75$.

All the values are tabulated below.

$n$	$x_n$	$F\left(x_n\right)$	$\theta \left(x_n\right)$	$y_n=F\left(x_n\right)\cos \left[\left({\theta }_n\right)\right]$
0	0	0	0.8	0
1	3.75	5.367188	1.152734	2.179025
2	7.5	9.46875	1.315625	2.390018
3	11.25	12.30469	1.351953	2.671355
4	15	13.875	1.325	3.376187
5	18.75	14.17969	1.298047	3.819728
6	22.5	13.21875	1.334375	3.096162
7	26.25	10.99219	1.497266	0.807535
8	30	7.5	1.85	–2.06693

Apply Simpson’s 1/3 rule to calculate work done,

According to Simpson’s 1/3 rule integral $I$ is written as,

$I=\frac{h}{3}\left[f\left(a\right)+f\left(b\right)+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

$I=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

$a$ and $b$ are the lower and upper limit of the integral respectively and $h$ is the step size that is $x_n-x_{n-1}$.

Work done is given by,

$W={\displaystyle \underset{0}{\overset{30}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$

Here, $a=0$, $h=$ 3.75 and $b=30$

Substitute values of $h,$ $y_0,y_1,y_2,\mathrm{...},y_8$ in equation (4)

Calculate $I$ for $n=9$ using equation (3)

$\begin{array}{c}I=\frac{h}{3}\left[y_0+y_8+4\left(y_1+y_3+y_5+y_7\right)+2\left(y_2+y_4+y_6\right)\right]\\ =\frac{3.75}{3}\left[\begin{array}{l}0+\left(-2.06693\right)+4\left(2.179025+2.671355+3.819728+0.807535\right)\\ +2\left(2.390018+3.376187+3.096162\right)\end{array}\right]\\ =66.96047\end{array}$

Hence, the value of integral is $I=66.96047$.

16 segment Simpson’s 1/3 rule.

The whole interval from $x_0=0\text{ to }{x}_n\text{=30}$ is divided into 16 intervals for $n=16$ and $h=1.875$

All the values are tabulated below.

$n$	$x_n$	$F\left(x_n\right)$	$\theta \left(x_n\right)$	$y_n=F\left(x_n\right)\cos \left[\left({\theta }_n\right)\right]$
0	0	0	0.8	0
1	1.875	2.841797	1.004053	1.525726
2	3.75	5.367188	1.152734	2.179025
3	5.625	7.576172	1.253955	2.360482
4	7.5	9.46875	1.315625	2.390018
5	9.375	11.04492	1.345654	2.465721
6	11.25	12.30469	1.351953	2.671355
7	13.125	13.24805	1.342432	2.999159
8	15	13.875	1.325	3.376187
9	16.875	14.18555	1.307568	3.691061
10	18.75	14.17969	1.298047	3.819728
11	20.625	13.85742	1.304346	3.648784
12	22.5	13.21875	1.334375	3.096162
13	24.375	12.26367	1.396045	2.132203
14	26.25	10.99219	1.497266	0.807535
15	28.125	9.404297	1.645947	–0.70608
16	30	7.5	1.85	–2.06693

Apply Simpson’s 1/3 rule to calculate work done.

According to Simpson’s 1/3 rule integral $I$ is written as,

$I=\frac{h}{3}\left[f\left(a\right)+f\left(b\right)+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

$I=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

In the above expression $a$ and $b$ are the lower and upper limit of the integral respectively and $h$ is the step size that is $x_n-x_{n-1}$. Here step size is $h=$ 1.875

Work done is given by,

$W={\displaystyle \underset{0}{\overset{30}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$

Here, $a=0$ and $b=30$, $h=$ 1.875

Substitute values of $h,$ $y_0,y_1,y_2,\mathrm{...},y_{16}$ using above table in equation (4).

Calculate $I$ for $n=16$ so equation (3) can be written as,

$\begin{array}{c}I=\frac{h}{3}\left[\begin{array}{l}{y}_0+y_{16}+4\left(y_1+y_3+y_5+y_7+y_9+y_{11}+y_{13}+y_{15}\right)\\ +2\left(y_2+y_4+y_6+y_8+y_{10}+y_{12}+y_{14}\right)\end{array}\right]\\ =\frac{1.875}{3}\left[\begin{array}{l}0+\left(-2.06693\right)+4\left(\begin{array}{l}1.525726+2.360482+2.465721+2.999159\\ +3.691061+3.648784\\ +2.132203-0.70608\end{array}\right)\\ +2\left(\begin{array}{l}2.179025+2.390018+2.671355\\ +3.376187+3.819728+3.096162+0.807535\end{array}\right)\end{array}\right]\\ =66.9258212\end{array}$

Hence, the work done is $I=66.92582125$.

To determine

To calculate: The work done for the given equations of $F\left(x\right)$ and $\theta \left(x\right)$ using Romberg integration to ${\epsilon }_s=0.5\%$.

Answer to Problem 36P

Solution: The work done is obtained to be $W=66.960\text{9 erg}$.

Explanation of Solution

Given Information:

The given expressions are as follows,

$\begin{array}{l}F\left(x\right)=1.6x-0.045x^2\\ \theta \left(x\right)=0.8+0.125x-0.009x^2+0.0002x^3\end{array}$

Work done in integral form (Refer Sec. 24.4).

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)dx}$

If the direction between the force and displacement changes between initial and final position, then the work done is written as,

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$ …… (1)

Here, $\theta \left(x\right)$ is the angle between force and displacement.

Formula Used:

Single segment trapezoidal rule.

$I=\left(b-a\right)\frac{f\left(a\right)+f\left(b\right)}{2}$

Multiple application trapezoidal rule.

$I=\frac{h}{2}\left[f\left(x_0\right)+f\left(x_n\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}\right]$

An estimate of relative percentage error.

$\left|{\epsilon }_a\right|=\left|\frac{O{\left(h^{n+2}\right)}_k-O{\left(h^n\right)}_{k-1}}{O{\left(h^{n+2}\right)}_k}\right|100\%$

Calculation:

The integral is,

$W={\displaystyle \underset{0}{\overset{30}{\int }}\left(1.6x-0.045x^2\right)}\cos \left(0.8+0.125x-0.009x^2+0.0002x^3\right)dx$

And,

$f\left(x\right)=\left(1.6x-0.045x^2\right)\cos \left(0.8+0.125x-0.009x^2+0.0002x^3\right)$

Here, $\theta \left(x\right)$ is in radians and conversion to degrees is necessary before each computation.

For Romberg Iteration- 1, 2, 4 and 8 segment trapezoidal rule integral needs to be calculated which will be used for complexity calculation in higher order correction of integral estimates.

The Integral for single segment trapezoidal rule is,

$I=\left(b-a\right)\frac{f\left(a\right)+f\left(b\right)}{2}$

Substitute $a=0$, $b=30$, $f\left(0\right)=0$ and $f\left(30\right)=-2.0669$

$\begin{array}{c}{I}_1=30\left(\frac{0-2.0669}{2}\right)\\ =-31.0035\end{array}$

The Integral for multiple application trapezoidal rules is,

$I=\frac{h}{2}\left[f\left(x_0\right)+f\left(x_n\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}\right]$

Here, $h$ is the step size, $h=\frac{b-a}{n}$

For calculation of $I_2$, $n=2$

Hence, $\begin{array}{c}h=\frac{30-0}{2}\\ =15\end{array}$

So, $x_0=0$, $x_1=15$ and $x_2=30$

$\begin{array}{c}{I}_2=\frac{15}{2}\left[0-2.0669+2\left(3.3761\right)\right]\\ =35.1397\end{array}$

For calculation of $I_4$, $n=4$

Hence, $\begin{array}{c}h=\frac{30}{4}\\ =7.5\end{array}$

So, $x_0=0$, $x_1=7.5$, $x_2=15$, $x_3=22.5$, $x_4=30$

$\begin{array}{c}{I}_4=\frac{7.5}{2}\left[f\left(x_0\right)+f\left(x_4\right)+2\left\{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right\}\right]\\ =\frac{7.5}{2}\left[0-2.0669+2\left\{2.4000+3.3761+3.0934\right\}\right]\\ =58.7703\end{array}$

For calculation of $I_8$, $n=8$

$\begin{array}{c}h=\frac{30-0}{8}\\ =3.75\end{array}$

So, $x_0=0,x_1=3.75,x_2=7.5,x_3=11.25$

$x_4=15,x_5=18.75,x_6=22.5,x_7=26.25,x_8=30$

Hence,

$\begin{array}{c}{I}_8=\frac{3.75}{2}\left[f\left(x_0\right)+f\left(x_8\right)+2\left\{\begin{array}{l}f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)\\ +f\left(x_5\right)+f\left(x_6\right)+f\left(x_7\right)\end{array}\right\}\right]\\ =64.9021\end{array}$

The complexity notation for Romberg iteration is,

$\begin{array}{l}O\left(h^4\right)=\frac{4}{3}{I}_k-\frac{1}{3}{I}_{k-1}\\ O\left(h^6\right)=\frac{16}{15}{I}_k-\frac{1}{15}{I}_{k-1}\\ O\left(h^8\right)=\frac{64}{63}{I}_k-\frac{1}{63}{I}_{k-1}\end{array}$

An estimate of relative percentage error is,

$\left|{\epsilon }_a\right|=\left|\frac{O{\left(h^{n+2}\right)}_k-O{\left(h^n\right)}_{k-1}}{O{\left(h^{n+2}\right)}_k}\right|100\%$

Setting up the table for Romberg iteration,

The first table of $O\left(h^2\right)$ is filled with the four results of trapezoidal rule obtained above, the latter tables are derived with the help of the above formula.

$\begin{array}{ccccc}n& O\left(h^2\right)& O\left(h^4\right)& O\left(h^4\right)& O\left(h^8\right)\\ 1& -31.0035& 57.1874\left({\epsilon }_s=62.7429\%\right)& 67.2777\left({\epsilon }_s=0.9461\%\right)& 66.9609\left({\epsilon }_s=0.0074\right)\\ 2& 35.1397& 66.6471\left({\epsilon }_s=13.4026\%\right)& 66.9659\left({\epsilon }_s=0.0297\%\right)& \\ 4& 58.7703& 66.9460\left({\epsilon }_s=3.1492\%\right)& & \\ 8& 64.9021& & & \end{array}$

From the Romberg table of iteration, the work done is obtained to be $W=66.960\text{9 erg}$.

To determine

To calculate: The work done for the given equations of $F\left(x\right)$ and $\theta \left(x\right)$ using Gauss Quadrature.

Answer to Problem 36P

Solution: The work done using Gauss Quadrature is obtained to be $66.812\text{9 erg}$.

Explanation of Solution

Given Information:

The given expressions are as follows,

$\begin{array}{l}F\left(x\right)=1.6x-0.045x^2\\ \theta \left(x\right)=0.8+0.125x-0.009x^2+0.0002x^3\end{array}$

Work done in integral form (Refer Sec. 24.4)

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)dx}$

If the direction between the force and displacement changes between initial and final position, then the work done is written as,

$W={\displaystyle \underset{x_0}{\overset{x_n}{\int }}F\left(x\right)\cos \left[\theta \left(x\right)\right]dx}$ …… (1)

Here, $\theta \left(x\right)$ is the angle between force and displacement.

Formula Used:

Change of variables formula,

$x=\frac{\left(b+a\right)+\left(b-a\right)x_d}{2}$

Gauss Quadrature formula for integral calculation,

$W\cong c_{0}f\left(x_0\right)+c_{1}f\left(x_1\right)+c_{2}f\left(x_2\right)$

Calculation:

The integral is,

$W={\displaystyle \underset{0}{\overset{30}{\int }}\left(1.6x-0.045x^2\right)}\cos \left(0.8+0.125x-0.009x^2+0.0002x^3\right)dx$

And,

$f\left(x\right)=\left(1.6x-0.045x^2\right)\cos \left(0.8+0.125x-0.009x^2+0.0002x^3\right)$

Change of variables is required so as to transform the original limits of given original integral to $-1$ to $1$ for calculating Gauss Quadrature integral,

$x=\frac{\left(b+a\right)+\left(b-a\right)x_d}{2}$

Substitute $a=0$ and $b=30$,

$\begin{array}{c}x=\frac{\left(30+0\right)+\left(30-0\right)x_d}{2}\\ =15+15x_d\end{array}$

Differentiate $x$,

$dx=15d{x}_d$ {By Chain Rule}

So, the function is,

$f\left(x\right)=\left(1.6x-0.045x^2\right)\cos \left(0.8+0.125x-0.009x^2+0.0002x^3\right)$

Substitute the value of $x$ and $dx$

$f\left(x\right)=\left\{\begin{array}{l}1.6\left(15+15x_d\right)\\ -0.045{\left(15+15x_d\right)}^2\end{array}\right\}\cos \left\{\begin{array}{l}0.8+0.125\left(15+15x_d\right)-0.009{\left(15+15x_d\right)}^2\\ +0.0002{\left(15+15x_d\right)}^3\end{array}\right\}$

The integral becomes,

$W={\displaystyle \underset{0}{\overset{30}{\int }}15\left\{\begin{array}{l}1.6\left(15+15x_d\right)\\ -0.045{\left(15+15x_d\right)}^2\end{array}\right\}\cos \left\{\begin{array}{l}0.8+0.125\left(15+15x_d\right)-0.009{\left(15+15x_d\right)}^2\\ +0.0002{\left(15+15x_d\right)}^3\end{array}\right\}d{x}_d}$

Therefore, Integrals is suitable for Gauss Quadrature calculation.

The weighting factors for 2-point Gauss Quadrature evaluation is,

$c_0=1,c_1=1$

And,

$x_0=-0.5773,x_1=0.5773$

The Integral is given by,

$W\cong c_{0}f\left(x_0\right)+c_{1}f\left(x_1\right)$

Substitute the values from above.

$\begin{array}{c}W\cong f\left(-0.5773\right)+f\left(0.5773\right)\\ \cong 34.6441+38.3001\\ =72.9442\end{array}$

Three-Point Gauss Quadrature factor will be tried out to reduce the truncation error.

The weighting factors for 3-Point Gauss Quadrature evaluation is,

$c_0=0.5555,c_1=0.8888,c_2=0.5555$

And,

$x_0=-0.7745,x_1=0,x_2=0.7745$

The Integral is given by,

$W\cong c_{0}f\left(x_0\right)+c_{1}f\left(x_1\right)+c_{2}f\left(x_2\right)$

Substitute the value from above.

$\begin{array}{c}W\cong 0.5555f\left(-0.7745\right)+0.8888f\left(0\right)+0.5555f\left(0.7745\right)\\ =66.8129\end{array}$

The obtained value is acceptable as the relative percentage error is less than1%.

Hence, the value of $W$ is $66.812\text{9 erg}$.