Chapter 24, Problem 46P

The upward velocity of a rocket can be computed by the following formula:

$v=u\text{In}\left(\frac{m_0}{m_0-qt}\right)-gt$

Where $v=\text{upword velocity,}$ $u=\text{velocity}$ at which fuel is expelled relative to the rocket, $m_0=\text{initial}$ mass of the rocket at time $t=0$,

$q=$ fuel consumption rate, and $g=\text{downword}$ acceleration of gravity ( $\left(\text{assumed constant}=\text{9}{\text{.8 m/s}}^2\right)$. If $u=1800\text{m/s}$, $m_0=160,000$ kg, and $q=2500\text{kg/s}$, use six-segment trapezoidal and Simpson's $1/3$ rule, six-point Gauss quadrature, and $O\left(h^8\right)$ Romberg methods to determine how high the rocket will fly in 30 s. In addition, use numerical differentiation to generate a graph of acceleration as a function of time.

To determine

To calculate: The height of the rocket at $t=30\sec$ using Six-segment trapezoidal, Simpson’s 1/3 rule, six-point Gauss quadrature and $O\left(h^8\right)$ Romberg methods. Also, generate a graph of acceleration as a function of time.

Answer to Problem 46P

Solution:

All the results are summarized in the table below,

	$1$	$2$	$3$	$4$
$n$	${\in }_a\to$	$4.0634\%$	$0.04927\%$	$0.01159\%$
$1$	$12668.115$	$10896.965$	$10875.5283$	$10880.57300$
$2$	$11339.7525$	$10880.8875$	$10879.31183$
$4$	$10995.60375$	$10879.70575$
$8$	$10908.68025$

Acceleration graph is shown below.

Explanation of Solution

Given Information:

The formula for the upward velocity of rocket is given as,

$v=u\ln \left(\frac{m_0}{m_0-qt}\right)-gt$

Here, $v=$ upward velocity, $u=1800m/s$ Velocity at which fuel is expelled relative to the rocket

$m_0=160,000\text{ kg}$ Initial mass of the rocket at time $t=a=0$, $q=2500kg/s$ Fuel consumption rate, $g=9.8m/s^2$, Downward acceleration of gravity, $b=30\text{ s}$ is the final time.

Formula Used:

6-segment Trapezoidal rule.

$I=\frac{b-a}{2n}\left[f\left(a\right)+2\left\{{\displaystyle \sum _{i=1}^{n-1}f\left(a+ih\right)}\right\}+f\left(b\right)\right]$

6-segmentSimpson’s 1/3 rule.

$I=\frac{b-a}{3n}\left[f\left(x_0\right)+4{\displaystyle \sum _{i_{odd}=1}^{n-1}f\left(x_i\right)}+2{\displaystyle \sum _{i_{even}=2}^{n-2}f\left(x_i\right)}+f\left(x_n\right)\right]$

Change of variable formula.

$t=\frac{\left(b+a\right)+\left(b-a\right)t_d}{2}$, $dt=\frac{\left(b-a\right)d{t}_d}{2}$

Here, $a\text{ and }b$ are the lower and upper limits of the integral respectively.

Error Calculation.

${\in }_a=\left|\frac{\text{Actual value}-\text{Approx value}}{\text{Actual value}}\right|100\%$

Calculation:

Calculate the upward velocity of the rocket.

$v=u\ln \left(\frac{m_0}{m_0-qt}\right)-gt$…… (1)

Substitute thegiven values in equation(1),

$v={\displaystyle \underset{0}{\overset{30}{\int }}\left(1800\ln \left(\frac{160,000}{160,000-2500t}\right)-9.8t\right)}dt$…… (2)

Apply 6-segment Trapezoidal rule for $f\left(t\right)=v$.

$I=\frac{b-a}{2n}\left[f\left(a\right)+2\left\{{\displaystyle \sum _{i=1}^{n-1}f\left(a+ih\right)}\right\}+f\left(b\right)\right]$

Here, $h=\frac{b-a}{n}$

Substitute $n=6$, $a=0$ and $b=30$.

$I=\frac{30-0}{2\left(6\right)}\left[f\left(0\right)+2\left\{{\displaystyle \sum _{i=1}^{5}f\left(0+i\left(5\right)\right)}\right\}+f\left(30\right)\right]$

Expand the above terms.

$I=\frac{30-0}{2\left(6\right)}\left[f\left(0\right)+2\left\{\begin{array}{l}f\left(5\right)+f\left(10\right)+f\left(15\right)\\ +f\left(20\right)+f\left(25\right)\end{array}\right\}+f\left(30\right)\right]$…… (3)

Calculate $f\left(0\right)$

$f\left(t\right)=1800\ln \left(\frac{160,000}{160,000-2500t}\right)-9.8t$

Substitute $t=0$

$\begin{array}{c}f\left(0\right)=1800\ln \left(\frac{160,000}{160,000-\left(2500\cdot 0\right)}\right)-\left(9.8\cdot 0\right)\\ =0\end{array}$

Calculate $f\left(5\right)$

$f\left(t\right)=1800\ln \left(\frac{160,000}{160,000-2500t}\right)-9.8t$

Substitute $t=5$

$\begin{array}{c}f\left(5\right)=1800\ln \left(\frac{160,000}{160,000-\left(2500\cdot 5\right)}\right)-\left(9.8\cdot 5\right)\\ =97.4222\end{array}$

Calculate $f\left(10\right)$

$f\left(t\right)=1800\ln \left(\frac{160,000}{160,000-2500t}\right)-9.8t$

Substitute $t=10$

$\begin{array}{c}f\left(10\right)=1800\ln \left(\frac{160,000}{160,000-\left(2500\cdot 10\right)}\right)-\left(9.8\cdot 10\right)\\ =207.818\end{array}$

Similarly, calculate all the function values.

$\begin{array}{l}f\left(15\right)=333.713,f\left(20\right)=478.448\\ f\left(25\right)=646.579,f\left(30\right)=844.541\end{array}$

Substitute above values in equation(3),

$\begin{array}{c}I=\frac{30}{2\left(6\right)}\left[0+2\left(\begin{array}{l}97.4222+207.818+333.713\\ +478.448+646.579\end{array}\right)+844.541\right]\\ =10931.2535\end{array}$

The formula for 6-segment Simpson’s 1/3 rule,

$I=\frac{b-a}{3n}\left[f\left(x_0\right)+4{\displaystyle \sum _{i_{odd}=1}^{n-1}f\left(x_i\right)}+2{\displaystyle \sum _{i_{even}=2}^{n-2}f\left(x_i\right)}+f\left(x_n\right)\right]$

Substitute the function values from above,

$\begin{array}{c}I=\frac{30-0}{3\left(6\right)}\left[\begin{array}{l}f\left(0\right)+4\left\{f\left(5\right)+f\left(15\right)+f\left(25\right)\right\}\\ +2\left\{f\left(10\right)+f\left(20\right)\right\}+f\left(30\right)\end{array}\right]\\ =\frac{30}{3\left(6\right)}\left[0+4\left\{97.4222+333.713+646.579\right\}+2\left\{207.818+478.448\right\}+844.541\right]\\ =10879.883\end{array}$

The given velocity Integral is,

$v={\displaystyle \underset{0}{\overset{30}{\int }}\left(1800\ln \left(\frac{160,000}{160,000-2500t}\right)-9.8t\right)}dt$

Change of variable before integrating the function,

$t=\frac{\left(b+a\right)+\left(b-a\right)t_d}{2}$…… (4)

And

$dt=\frac{\left(b-a\right)d{t}_d}{2}$…… (5)

Here, $a\text{ and }b$ are the lower and upper limits of the integral respectively.

Substitute the values of $a=0\text{ and }b=30$ in the equations(4) and (5),

$\begin{array}{l}t=15+15t_d\\ dt=15d{t}_d\end{array}$

Substitute the values of $\text{t and }dt$ in equation(2),

$v={\displaystyle {\int }_0^{30}\left(1800\ln \left(\frac{160,000}{160,000-2500\left(15+15t_d\right)}\right)-9.8\left(15+15t_d\right)\right)}15d{t}_d$

Thus, the transformed function is,

$f\left(t_d\right)=\left(1800\ln \left(\frac{160,000}{160,000-2500\left(15+15t_d\right)}\right)-9.8\left(15+15t_d\right)\right)15$

Apply Six-Point Gauss Quadrature formula.

$\begin{array}{l}I=0.1713245f\left(-0.932469514\right)+0.3607616f\left(-0.661209386\right)\\ +0.4679139f\left(-0.238619186\right)+0.4679139f\left(0.238619186\right)\\ +0.3607616f\left(0.661209386\right)+0.1713245f\left(0.932469514\right)\end{array}$

Calculate $f\left(-0.932469514\right)$.

$f\left(t_d\right)=\left(1800\ln \left(\frac{160,000}{160,000-2500\left(15+15t_d\right)}\right)-9.8\left(15+15t_d\right)\right)15$

Substitute $t_d=-0.932469514$.

$\begin{array}{c}f\left(-0.932469514\right)=\left(\begin{array}{l}1800\ln \left(\frac{160000}{160000-2500\left(15+15\left(-0.932469514\right)\right)}\right)\\ -9.8\left(15+15\left(-0.932469514\right)\right)\end{array}\right)\left(15\right)\\ =281.855\end{array}$

Calculate $f\left(-0.661209386\right)$.

$f\left(t_d\right)=\left(1800\ln \left(\frac{160,000}{160,000-2500\left(15+15t_d\right)}\right)-9.8\left(15+15t_d\right)\right)15$

Substitute $t_d=-0.661209386$.

$\begin{array}{c}f\left(-0.661209386\right)=\left(\begin{array}{l}1800\ln \left(\frac{160000}{160000-2500\left(15+15\left(-0.661209386\right)\right)}\right)\\ -9.8\left(15+15\left(-0.661209386\right)\right)\end{array}\right)\left(15\right)\\ =1486.79\end{array}$

Similarly, calculate all function values.

$\begin{array}{l}f\left(-0.238619186\right)=3628.30\\ f\left(0.238619186\right)=6527.55\\ f\left(0.661209386\right)=9654.10\\ f\left(0.932469514\right)=12024.4\end{array}$

Substitute the values in Six-Point Gauss Quadrature formula.

$\begin{array}{c}I=0.1713245\left(281.855\right)+0.3607616\left(1486.79\right)+0.4679139\left(3628.30\right)\\ +0.4679139\left(6527.55\right)+0.3607616\left(9654.10\right)+0.1713245\left(12024.4\right)\\ =10879.63166\end{array}$

Apply Romberg Integration.

Choose $h=30$(distance between two consecutive ordinates), the values of $v$ are,

$t$	$0$	$30$
$v$	$x_0=0$	$x_1=844.541$

Substitute value from above and calculate $I\left(h\right)$,

$\begin{array}{c}I\left(h\right)=\frac{h}{2}\left(x_0+x_1\right)\\ =\frac{30}{2}\left(0+844.541\right)\\ =12668.115\end{array}$

Choose $h=15$, the values of $v$ are,

$t$	$0$	$15$	$30$
$v$	$x_0=0$	$x_1=333.713$	$x_2=844.541$

Substitute values from above and calculate $I\left(\frac{h}{2}\right)$,

$\begin{array}{c}I\left(\frac{h}{2}\right)=\frac{h}{2}\left(x_0+2x_1+x_2\right)\\ =\frac{15}{2}\left(0+2\left(333.713\right)+844.541\right)\\ =11339.7525\end{array}$

Choose $h=7.5$, the values of $v$ are,

$t$	$0$	$7.5$	$15$	$22.5$	$30$
$v$	$x_0=0$	$x_1=150.856$	$x_2=333.713$	$x_3=559.241$	$x_4=844.541$

Substitute values from above and calculate $I\left(\frac{h}{4}\right)$,

$\begin{array}{c}I\left(\frac{h}{4}\right)=\frac{h}{2}\left(x_0+2\left(x_1+x_2+x_3\right)+x_4\right)\\ =\frac{7.5}{2}\left(0+2\left(150.856+333.713+559.241\right)+844.541\right)\\ =10995.60375\end{array}$

Choose $h=3.75$, the values of $v$ are,

$t$	$0$	$3.75$	$7.5$	$11.25$
$v$	$x_0=0$	$x_1=71.9349$	$x_2=150.856$	$x_3=237.725$

$15$	$18.75$	$22.5$	$26.25$	$30$
$x_4=333.713$	$x_5=440.275$	$x_6=559.241$	$x_7=692.966$	$x_8=844.541$

Substitute values from above and calculate $I\left(\frac{h}{8}\right)$,

$\begin{array}{c}I\left(\frac{h}{8}\right)=\frac{h}{2}\left(x_0+2\left(x_1+x_2+x_3+x_4+x_5+x_6+x_7\right)+x_8\right)\\ =\frac{3.75}{2}\left(0+2\left(\begin{array}{l}71.9349+150.856+237.725+333.713\\ +440.275+559.241+692.966\end{array}\right)+844.541\right)\\ =10908.68025\end{array}$

Substitute values from above and calculate $I\left(h,\frac{h}{2}\right)$,

$\begin{array}{c}I\left(h,\frac{h}{2}\right)=\frac{1}{3}\left[4I\left(\frac{h}{2}\right)-I\left(h\right)\right]\\ =\frac{1}{3}\left[4\left(11339.7525\right)-12668.115\right]\\ =10896.965\end{array}$

Calculate the approximate error,

${\in }_a=\left|\frac{\text{Actual value}-\text{Approx value}}{\text{Actual value}}\right|100\%$

Substitute values from above.

$\begin{array}{c}{\in }_a=\left|\frac{10896.965-11339.7525}{10896.965}\right|100\%\\ {\in }_a=4.0634\%\end{array}$

Substitute values from above and calculate $I\left(\frac{h}{2},\frac{h}{4}\right)$,

$\begin{array}{c}I\left(\frac{h}{2},\frac{h}{4}\right)=\frac{1}{3}\left[4I\left(\frac{h}{4}\right)-I\left(\frac{h}{2}\right)\right]\\ =\frac{1}{3}\left[4\left(10995.60375\right)-11339.7525\right]\\ =10880.8875\end{array}$

Substitute values from above and calculate $I\left(\frac{h}{4},\frac{h}{8}\right)$,

$\begin{array}{c}I\left(\frac{h}{4},\frac{h}{8}\right)=\frac{1}{3}\left[4I\left(\frac{h}{8}\right)-I\left(\frac{h}{4}\right)\right]\\ =\frac{1}{3}\left[4\left(10908.68025\right)-10995.60375\right]\\ =10879.70575\end{array}$

Substitute values from above and calculate $I\left(h,\frac{h}{2},\frac{h}{4}\right)$,

$\begin{array}{c}I\left(h,\frac{h}{2},\frac{h}{4}\right)=\frac{1}{3}\left[4I\left(\frac{h}{2},\frac{h}{4}\right)-I\left(h,\frac{h}{2}\right)\right]\\ =\frac{1}{3}\left[4\left(10880.8875\right)-10896.965\right]\\ =10875.5283\end{array}$

Calculate the approximate error,

${\in }_a=\left|\frac{\text{Actual value}-\text{Approx value}}{\text{Actual value}}\right|100\%$

Substitute values from above,

$\begin{array}{c}{\in }_a=\left|\frac{10875.5283-10880.8875}{10875.5283}\right|100\%\\ {\in }_a=0.04927\%\end{array}$

Substitute values from above and calculate $I\left(\frac{h}{2},\frac{h}{4},\frac{h}{8}\right)$,

$\begin{array}{c}I\left(\frac{h}{2},\frac{h}{4},\frac{h}{8}\right)=\frac{1}{3}\left[4I\left(\frac{h}{4},\frac{h}{8}\right)-I\left(\frac{h}{2},\frac{h}{4}\right)\right]\\ =\frac{1}{3}\left[4\left(10879.70575\right)-10880.8875\right]\\ =10879.31183\end{array}$

Substitute values from above and calculate $I\left(h,\frac{h}{2},\frac{h}{4},\frac{h}{8}\right)$,

$\begin{array}{c}I\left(h,\frac{h}{2},\frac{h}{4},\frac{h}{8}\right)=\frac{1}{3}\left[4I\left(\frac{h}{2},\frac{h}{4},\frac{h}{8}\right)-I\left(h,\frac{h}{2},\frac{h}{4}\right)\right]\\ =\frac{1}{3}\left[4\left(10879.31183\right)-10875.5283\right]\\ =10880.5730\end{array}$

Calculate the approximate error as,

${\in }_a=\left|\frac{\text{Actual value}-\text{Approx value}}{\text{Actual value}}\right|100\%$

Substitute values from above.

$\begin{array}{c}{\in }_a=\left|\frac{10880.57300-10879.31183}{10880.57300}\right|100\%\\ {\in }_a=0.01159\%\end{array}$

All above calculated results are summarized in the table below,

	$1$	$2$	$3$	$4$
$n$	${\in }_a\to$	$4.0634\%$	$0.04927\%$	$0.01159\%$
$1$	$12668.115$	$10896.965$	$10875.5283$	$10880.57300$
$2$	$11339.7525$	$10880.8875$	$10879.31183$
$4$	$10995.60375$	$10879.70575$
$8$	$10908.68025$

UseNumerical differentiation to calculate acceleration as a function of time.

Apply finite-divided differences for $h=2$,

$\begin{array}{c}a\left(0\right)=\frac{dv\left(0\right)}{dt}\\ =\frac{-3\left(76.9693\right)+4\left(37.5477\right)-0}{2\left(2\right)}\\ =18.3053\end{array}$

Apply centred differences to calculate the intermediate values,

$\begin{array}{c}a\left(2\right)=\frac{dv\left(2\right)}{dt}\\ =\frac{76.9693-0}{2\left(2\right)}\\ =19.2423\end{array}$

Similarly, calculate all intermediate values of acceleration.

Apply backward difference for the last value at t = 30 s,

$\begin{array}{c}a\left(30\right)=\frac{dv\left(30\right)}{dt}\\ =\frac{-3\left(844.5406\right)+4\left(761.255\right)+683.5345}{2\left(2\right)}\\ =43.0036\end{array}$

All of the results are displayed in the following table and graph,

Select the time and acceleration column and go to insert chart and scattered chart. The graph of acceleration verses time is plotted as follows: