Question

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from $t=0$ to $t=12$. The positions (in feet) were: 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.

a. Find the least-squares cubic curve $y={\beta }_0+{\beta }_{1}t+{\beta }_2{t}^2+{\beta }_3{t}^3$ for these data.

b. Use the result of part (a) to estimate the velocity of the plane when $t=4.5$ seconds.

Short answer

(a) The least-squares cubic curve is $y=-0.8558+4.7025t+5.5554t^2-0.0274t^3$.

(b) The velocity at $t=4.5$ is $53\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}$.

Step-by-step solution

The General Linear Model 

The equation of the general linear model is defined as:

$\mathbf{y}=X\beta +\in$

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

Find design matrix, observation vector, parameter vector for given data 

(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:

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

Observational vector:

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

And, the parameter vectorfor the given equation is,

$\beta =\left(\begin{array}{r}{\beta }_0\\ {\beta }_1\\ {\beta }_2\\ {\beta }_3\end{array}\right)$

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

Normal equation

The normal equation is given by,

$X^{T}X\beta =X^T\mathbf{y}$

Find least-squares cubic curve

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:

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

That implies;

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

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

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

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

Now, substitute the obtained values into $y={\beta }_0+{\beta }_{1}t+{\beta }_2{t}^2+{\beta }_3{t}^3$.

$y=-0.8558+4.7025t+5.5554t^2-0.0274t^3$

Hence, the required least cubic curve is $y=-0.8558+4.7025t+5.5554t^2-0.0274t^3$.

Find the velocity

(b)

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

$\begin{array}{rl}{y}^{\prime }\left(t\right)& =\frac{d}{dt}\left(-0.8558+4.7025t+5.5554t^2-0.0274t^3\right)\\ v\left(t\right)& =4.7025+11.1108t-0.822t^2\end{array}$

To find the velocity at $t=4.5$, substitute $t=4.5$ into .

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

So, the velocity at $t=4.5$ is $53\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}$.