17.5 Solving GLSP for Airplane Descent Path

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: 17.4 Model of Airplane Descent Path to Landing

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Solving GLSP for Airplane Descent Path

Recall 17.1 The General Linear-Systems Problem we found for our model of a Boeing 787's descent path to landing.

\underset{A}{\underset{⏟}{[\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 400 & 160000 & 64000000 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \end{matrix}]}} \underset{\vec{x}}{\underset{⏟}{[\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}]}} = \underset{\vec{b}}{\underset{⏟}{[\begin{matrix} 40000 \\ 0 \\ 0 \\ 0 \end{matrix}]}}

Recall that we modeled the descent path with the cubic polynomial

p (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}

Solve for the vector $\vec{x}$ .

\begin{proof}[Solution.]
To start, we need to transform matrix $A$ into 17.2 Row Echelon Form by multiplying on the left by elementary matrices. Notice that rows 1 and 3 are nice looking from the standpoint of elimination, while rows 2 and 4 look like they require more work for elimination. Let's switch row 3 to row 2, row 4 to row 3, and switch row 2 to row 4 using a permutation matrix (combination of 9.10 Transposition Matrix).

\begin{aligned} \underset{E_{1}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}]}} \cdot \underset{A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 1 & 400 & 1600000 & 64000000 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{1}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}]}} \cdot \underset{\vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ 0 \end{array}]}} \\ ⟹ \underset{E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \\ 1 & 400 & 1600000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ 0 \end{array}]}} \end{aligned}

We now have a structure that will be much easier to transform into row echelon form. We can zero out the bottom left corner by multiplying on the left by a 9.8 Shear Matrix. Recall that we get $c$ by taking the negative of what we are trying to cancel out and dividing by its pivot.

\begin{aligned} \underset{E_{2}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 1 & 0 & 0 & 0 \end{array}]}} \cdot \underset{E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \\ 1 & 400 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{2}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 1 & 0 & 0 & 0 \end{array}]}} \cdot \underset{E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ 0 \end{array}]}} \\ ⟹ \underset{E_{2} \cdot E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \\ 0 & 400 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{2} \cdot E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \end{aligned}

Now, we need to cancel out the values below the diagonal in column 2. Once again, we multiply by a shear matrix.

\begin{aligned} \underset{E_{3}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & - 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]}} \cdot \underset{E_{2} \cdot E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 800 & 480000 \\ 0 & 400 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{3}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & - 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]}} \cdot \underset{E_{2} \cdot E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \\ ⟹ \underset{E_{4}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 400 & 0 & 1 \end{array}]}} \cdot \underset{E_{3} \cdot E_{2} \cdot E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 800 & 480000 \\ 0 & 400 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{4}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 400 & 0 & 1 \end{array}]}} \cdot \underset{E_{3} \cdot E_{2} \cdot E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \\ ⟹ \underset{E_{4} \cdot E_{3} \cdot E_{2} \cdot E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 800 & 480000 \\ 0 & 0 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{4} \cdot E_{3} \cdot E_{2} \cdot E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \end{aligned}

Now, we need to eliminate the sub diagonal values in column 3. We see that we just need to eliminate 160000.

\begin{aligned} \underset{E_{5}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 200 & 1 \end{array}]}} \cdot \underset{E_{4} \cdot E_{3} \cdot E_{2} \cdot E_{1} \cdot A}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 800 & 480000 \\ 0 & 0 & 160000 & 64000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{E_{5}}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 200 & 1 \end{array}]}} \cdot \underset{E_{4} \cdot E_{3} \cdot E_{2} \cdot E_{1} \cdot \vec{b}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \\ ⟹ \underset{U}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 800 & 480000 \\ 0 & 0 & 0 & - 32000000 \end{array}]}} \cdot \underset{\vec{x}}{\underset{⏟}{[\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array}]}} & = \underset{\vec{y}}{\underset{⏟}{[\begin{array}{c} 40000 \\ 0 \\ 0 \\ - 40000 \end{array}]}} \end{aligned}

We have successfully transformed our original modeling matrix $A$ into row echelon form $U$ . Using 12.2 Backwards Substitution, we produce the following values of $\vec{x}$ :

\underset{\vec{x}}{\underset{⏟}{[\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}]}} = [\begin{matrix} 40000 \\ 0 \\ - 0.75 \\ 0.00125 \end{matrix}] = [\begin{matrix} 40000 \\ 0 \\ - \frac{3}{4} \\ \frac{1}{800} \end{matrix}]

We can conclude that the descent path is modeled by the cubic polynomial

p (x) = 40000 - 0.75 x^{2} + 0.00125 x^{3}

\end{proof}