Chapter 24, Problem 35P

Perform the same computation as in Sec. 24.4, but use the following equations:

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

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

Use 4-, 8-, and 16-segment trapezoidal rules to compute the integral.

To determine

To calculate: The work done for the given equations of $F\left(x\right)$ and $\theta \left(x\right)$ using 4-, 8-, and 16-segment trapezoidal rule.

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

Answer to Problem 35P

Solution:

The value of $W$ after application of 4-segment Trapezoidal rule is $W=58.71675$.

The value of $W$ after application of 8-segment Trapezoidal rule is $W=64.8995062$.

The value of $W$ after application of 16-segment Trapezoidal rule is $W=66.419208$.

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:

Multiple Segment Trapezoidal Rule.

$I_n=\left(b-a\right)\frac{f\left(x_0\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}+f\left(x_n\right)}{2\left(n\right)}$

Calculation:

Calculate the work done.

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

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

Apply 4-Segment Trapezoidal rule.

Calculate the value of $△x$, when $n=4,x_n=30$ and $x_0=0$,

$\begin{array}{c}△x=\frac{x_n-x_0}{n}\\ =\frac{30-0}{4}\\ =7.5\end{array}$

Divide the interval from $0\text{ to }30$, and $△x=7.5$,

So, the value of x after each iteration is $x=0,7.5,15,22.5\text{ and }30$.

Calculate $F\left(0\right)$ for $x=0$,

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

Calculate $\theta \left(0\right)$ for $x=0$,

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

Calculate $F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)$ for $x=0$,

$\begin{array}{c}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)=0\cdot \cos \left(0.8\right)\\ =0\end{array}$

Calculate $F\left(x\right)$ for $x=7.5$,

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

Calculate $\theta \left(7.5\right)$ for $x=7.5$,

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

Calculate $F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)$ for $x=7.5$,

$\begin{array}{c}F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)=9.46875\times \cos \left(1.315625\right)\\ =2.39002\end{array}$

Calculate $F\left(x\right)$, for $x=15$,

$\begin{array}{c}F\left(15\right)=1.6\left(15\right)-0.045{\left(15\right)}^2\\ =13.875\end{array}$

Calculate $\theta \left(15\right)$ for $x=15$,

$\begin{array}{c}\theta \left(15\right)=0.8+0.125\left(15\right)-0.009{\left(15\right)}^2+0.0002{\left(15\right)}^3\\ =1.325\end{array}$

Calculate $F\left(15\right)\cdot \cos \left(\theta \left(15\right)\right)$ for $x=15$,

$\begin{array}{c}F\left(15\right)\cdot \cos \left(\theta \left(15\right)\right)=13.875\times \cos \left(1.325\right)\\ =3.37619\end{array}$

Similarly, calculate for $x=22.5\text{ and }30$ thentabulate the values as shown below.

$\begin{array}{cccc}x& F\left(x\right)& \theta \left(x\right)& F\left(x\right)\cos \left(\theta \left(x\right)\right)\\ 0& 0& 0.8& 0\\ 7.5& 9.46875& 1.315625& 2.39002\\ 15& 13.875& 1.325& 3.37618\\ 22.5& 13.21875& 1.334375& 3.09616\\ 30& 7.5& 1.85& -2.06692\end{array}$

Calculate the solution using Trapezoidal rule,

$W=\frac{△x}{2}\left(\begin{array}{l}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)+2F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)+2F\left(15\right)\cdot \cos \left(\theta \left(15\right)\right)\\ +2F\left(22.5\right)\cdot \cos \left(\theta \left(22.5\right)\right)+F\left(30\right)\cdot \cos \left(\theta \left(30\right)\right)\end{array}\right)$

Substitute function values from above table for $△x=7.5$.

$\begin{array}{c}W=\frac{7.5}{2}\left(0+2\cdot 2.39002+2\cdot 3.37618+2\cdot 3.09616-2.06692\right)\\ =58.71675\end{array}$

Hence, the value of $W$ after application of 4-segment Trapezoidal rule is $W=58.71675$.

Apply 8-Segment Trapezoidal rule.

Calculate the value of $△x$, when $n=8,x_n=30$ and $x_0=0$,

$\begin{array}{c}△x=\frac{x_n-x_0}{n}\\ =\frac{30-0}{8}\\ =3.75\end{array}$

Divide the interval from $0\text{ to }30$, and $△x=3.75$,

So, the value of x after each iteration is $x=0,3.75,7.5,11.25,15,18.75,22.5,26.25\text{ and }30$.

Calculate $F\left(0\right)$ for $x=0$,

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

Calculate $\theta \left(0\right)$ for $x=0$,

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

Calculate $F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)$ for $x=0$,

$\begin{array}{c}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)=0\cdot \cos \left(0.8\right)\\ =0\end{array}$

Calculate $F\left(3.75\right)$ for $x=3.75$,

$\begin{array}{c}F\left(3.75\right)=1.6\left(3.75\right)-0.045{\left(3.75\right)}^2\\ =5.3671875\end{array}$

Calculate $\theta \left(3.75\right)$ for $x=3.75$,

$\begin{array}{c}\theta \left(3.75\right)=0.8+0.125\left(3.75\right)-0.009{\left(3.75\right)}^2+0.0002{\left(3.75\right)}^3\\ =1.152734375\end{array}$

Calculate $F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)$ for $x=3.75$,

$\begin{array}{c}F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)=5.3671875\times \cos \left(1.152734375\right)\\ =2.1790249\end{array}$

Calculate $F\left(x\right)$ for $x=7.5$,

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

Calculate $\theta \left(7.5\right)$ for $x=7.5$,

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

Calculate $F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)$ for $x=7.5$,

$\begin{array}{c}F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)=9.46875\times \cos \left(1.315625\right)\\ =2.39002\end{array}$

Similarly, calculate $x=11.25,15,18.75,22.5,26.25\text{ and }30$ then tabulate all solutions as shown below.

$\begin{array}{cccc}x& F\left(x\right)& \theta \left(x\right)& F\left(x\right)\cos \left(\theta \left(x\right)\right)\\ 0& 0& 0.8& 0\\ 3.75& 5.3671875& 1.152734& 2.17902\\ 7.5& 9.46875& 1.315625& 2.39002\\ 11.25& 12.3046875& 1.3519535& 2.67135\\ 15& 13.875& 1.325& 3.37618\\ 18.75& 14.1796875& 1.2980468& 3.81973\\ 22.5& 13.21875& 1.334375& 3.09616\\ 26.25& 10.9921875& 1.4972656& 0.807535\\ 30& 7.5& 1.85& -2.06692\end{array}$

Calculate the solution using Trapezoidal rule,

$W=\frac{△x}{2}\left(\begin{array}{l}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)+2F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)+2F\left(7.5\right)\cdot \cos \left(\theta \left(7.5\right)\right)\\ \cdots +2F\left(26.25\right)\cdot \cos \left(\theta \left(26.25\right)\right)+F\left(30\right)\cdot \cos \left(\theta \left(30\right)\right)\end{array}\right)$

Substitute function values from above table $△x=3.75$.

$\begin{array}{c}W=\frac{3.75}{2}\left(\begin{array}{l}0+2\cdot 2.17902+2\cdot 2.39002+2\cdot 2.67135+2\cdot 3.37618\\ +2\cdot 3.81973+2\cdot 3.09616+2\cdot 0.807535-2.06692\end{array}\right)\\ =64.8995062\end{array}$

Hence, the value of $W$ after application of 8-segment Trapezoidal rule is $W=64.8995062$

Apply 16-Segment Trapezoidal rule.

Calculate the value of $△x$, when $n=8,x_n=30$ and $x_0=0$,

$\begin{array}{c}△x=\frac{x_n-x_0}{n}\\ =\frac{30-0}{16}\\ =1.875\end{array}$

Divide the interval from $0\text{ to }30$, and $△x=1.875$,

So, the value of x after each iteration is,

$x=\left(\begin{array}{l}0,1.875,3.75,5.625,7.5,9.375,11.25,13.125,15,16.875\\ ,18.75,20.625,22.5,24.375,26.25,28.125\text{ and }30\end{array}\right)$.

Calculate $F\left(0\right)$ for $x=0$,

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

Calculate $\theta \left(0\right)$ for $x=0$,

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

Calculate $F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)$ for $x=0$,

$\begin{array}{c}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)=0\cdot \cos \left(0.8\right)\\ =0\end{array}$

Calculate $F\left(1.875\right)$ for $x=1.875$,

$\begin{array}{c}F\left(1.875\right)=1.6\left(1.875\right)-0.045{\left(1.875\right)}^2\\ =2.841796875\end{array}$

Calculate $\theta \left(1.875\right)$ for $x=1.875$,

$\begin{array}{c}\theta \left(1.875\right)=0.8+0.125\left(1.875\right)-0.009{\left(1.875\right)}^2+0.0002{\left(1.875\right)}^3\\ =1.004052734375\end{array}$

Calculate $F\left(1.875\right)\cdot \cos \left(\theta \left(1.875\right)\right)$ for $x=1.875$,

$\begin{array}{c}F\left(1.875\right)\cdot \cos \left(\theta \left(1.875\right)\right)=2.841796875\times \cos \left(1.004052734375\right)\\ =1.525725560\end{array}$

Calculate $F\left(3.75\right)$ for $x=3.75$,

$\begin{array}{c}F\left(3.75\right)=1.6\left(3.75\right)-0.045{\left(3.75\right)}^2\\ =5.3671875\end{array}$

Calculate $\theta \left(3.75\right)$ for $x=3.75$,

$\begin{array}{c}\theta \left(3.75\right)=0.8+0.125\left(3.75\right)-0.009{\left(3.75\right)}^2+0.0002{\left(3.75\right)}^3\\ =1.152734375\end{array}$

Calculate $F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)$ for $x=3.75$,

$\begin{array}{c}F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)=5.3671875\times \cos \left(1.152734375\right)\\ =2.1790249\end{array}$

Similarly, calculate $x=\left(\begin{array}{l}5.625,7.5,9.375,11.25,13.125,15,16.875,18.75\\ ,20.625,22.5,24.375,26.25,28.125\text{ and }30\end{array}\right)$ then tabulate all the solutions as shown below.

$\begin{array}{cccc}x& F\left(x\right)& \theta \left(x\right)& F\left(x\right)\cos \left(\theta \left(x\right)\right)\\ 0& 0& 0.8& 0\\ 1.875& 2.8417968& 1.0040527& 1.52572\\ 3.75& 5.3671875& 1.152734& 2.17902\\ 5.625& 7.5761718& 1.253955& 2.36048\\ 7.5& 9.46875& 1.315625& 2.39002\\ 9.375& 11.044921& 1.345654& 2.46572\\ 11.25& 12.3046875& 1.3519535& 2.67135\\ 13.125& 13.248046& 1.342431& 2.99916\\ 15& 13.875& 1.325& 3.37618\end{array}$ $\begin{array}{cccc}16.875& 14.185546& 1.307568& 3.69106\\ 18.75& 14.1796875& 1.2980468& 3.81973\\ 20.625& 13.857421& 1.3043457& 3.64878\\ 22.5& 13.21875& 1.334375& 3.09616\\ 24.375& 12.263671& 1.396044& 2.13220\\ 26.25& 10.9921875& 1.4972656& 0.807535\\ 28.125& 9.404296& 1.645947& -0.706077\\ 30& 7.5& 1.85& -2.06692\end{array}$

Calculate the solution using Trapezoidal rule,

$W=\frac{△x}{2}\left(\begin{array}{l}F\left(0\right)\cdot \cos \left(\theta \left(0\right)\right)+2F\left(1.875\right)\cdot \cos \left(\theta \left(1.875\right)\right)+2F\left(3.75\right)\cdot \cos \left(\theta \left(3.75\right)\right)\\ \cdots +2F\left(28.125\right)\cdot \cos \left(\theta \left(28.125\right)\right)+F\left(30\right)\cdot \cos \left(\theta \left(30\right)\right)\end{array}\right)$

Substitute function values from above table for $△x=1.875$.

$\begin{array}{c}W=\frac{1.875}{2}\left(\begin{array}{l}0+2\cdot 1.52572+2\cdot 2.17902+2\cdot 2.36048+2\cdot 2.39002+2\cdot 2.46572\\ +2\cdot 2.67135+2\cdot 2.99916+2\cdot 3.37618+2\cdot 3.69106+2\cdot 3.81973\\ +2\cdot 3.64878+2\cdot 3.09616+2\cdot 2.13220+2\cdot 0.807535\\ +2\cdot \left(-0.706077\right)-2.06692\end{array}\right)\\ =66.41920875\end{array}$

Hence, the value of $W$ after application of 16-segment Trapezoidal rule is $W=66.419208$.