91Ó°ÊÓ

The equation of the general linear model is defined as:

\({\bf{y}} = X\beta + \in \)

Where, \({\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is the parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is a residual vector.

(a)

The given equation is\(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\)and the given data from \(t = 0\) to is 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 380.4, 471.1, 571.7, 686.8 and 809.2.

Write the Design matrix, observational vector, and the parameter vector for the given equation and data set by using the information given in step 1.

Design matrix:

\(X = \left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)\)

Observational vector:

\({\bf{y}} = \left( {\begin{aligned}0\\{8.8}\\{29.9}\\{62.0}\\{104.7}\\{159.1}\\{222.0}\\{294.5}\\{380.4}\\{471.1}\\{571.7}\\{686.8}\\{809.2}\end{aligned}} \right)\)

And, the parameter vectorfor the given equation is,

\({\bf{\beta }} = \left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\\{{\beta _2}}\\{{\beta _3}}\end{aligned}} \right)\)

These are the best fit for the given data set and equation.

The normal equation is given by,

\({X^T}X\beta = {X^T}{\bf{y}}\)

The general least-squares equation is given by \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\), and to find the associated least-squares curve, the values of \({\beta _0},{\beta _1},{\beta _2},{\beta _3}\) are required, so find the values of \({\beta _0},{\beta _1},{\beta _2},{\beta _3}\) by using normal equation.

By using the obtained information from step 2, the normal equation will be:

\(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}{\bf{y}}\)

That implies;

\(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\\{{\beta _2}}\\{{\beta _3}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)^T}\left( {\begin{aligned}0\\{8.8}\\{29.9}\\{62.0}\\{104.7}\\{159.1}\\{222.0}\\{294.5}\\{380.4}\\{471.1}\\{571.7}\\{686.8}\\{809.2}\end{aligned}} \right)\)

Use the following steps to find the associated values for the obtained data in MATLAB.

1. Enter the data \(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\\{{\beta _2}}\\{{\beta _3}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)^T}\left( {\begin{aligned}0\\{8.8}\\{29.9}\\{62.0}\\{104.7}\\{159.1}\\{222.0}\\{294.5}\\{380.4}\\{471.1}\\{571.7}\\{686.8}\\{809.2}\end{aligned}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&0&0&0\\1&1&1&1\\1&2&4&8\\1&3&9&{27}\\1&4&{16}&{64}\\1&5&{25}&{125}\\1&6&{36}&{216}\\1&7&{49}&{343}\\1&8&{64}&{512}\\1&9&{81}&{729}\\1&{10}&{100}&{1000}\\1&{11}&{121}&{1331}\\1&{12}&{144}&{1728}\end{aligned}} \right)^T}\left( {\begin{aligned}0\\{8.8}\\{29.9}\\{62.0}\\{104.7}\\{159.1}\\{222.0}\\{294.5}\\{380.4}\\{471.1}\\{571.7}\\{686.8}\\{809.2}\end{aligned}} \right)\).
2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\\{{\beta _2}}\\{{\beta _3}}\end{aligned}} \right)\) is: \(\left( {\begin{aligned}{ - 0.8558}\\{4.7025}\\{5.5554}\\{ - 0.0274}\end{aligned}} \right)\)

Now, substitute the obtained values into \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\).

\(y = - 0.8558 + 4.7025t + 5.5554{t^2} - 0.0274{t^3}\)

Hence, the required least cubic curve is \(y = - 0.8558 + 4.7025t + 5.5554{t^2} - 0.0274{t^3}\).

(b)

Velocity is the derivative of the position function, so find the derivative of the obtained least-squares cubic curve \(y = - 0.8558 + 4.7025t + 5.5554{t^2} - 0.0274{t^3}\) with respect to \(t\).

\(\begin{aligned}y'\left( t \right) &= \frac{d}{{dt}}\left( { - 0.8558 + 4.7025t + 5.5554{t^2} - 0.0274{t^3}} \right)\\v\left( t \right) &= 4.7025 + 11.1108t - 0.822{t^2}\end{aligned}\)

To find the velocity at \(t = 4.5\), substitute \(t = 4.5\) into .

\(\begin{aligned}v\left( {4.5} \right) &= 4.7025 + 11.1108\left( {4.5} \right) - 0.822{\left( {4.5} \right)^2}\\ &= 53\end{aligned}\)

So, the velocity at \(t = 4.5\) is \(53{\rm{ ft/sec}}\).