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Model of Airplane Descent Path to Landing

Using the graph of a Boeing 787's descent path to landing, create a mathematical model of the path.
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Set up the GLSP

Airplane landings typically follow a smooth curve. A linear line would be too sharp. A quadratic path would cause us to crash. However, a cubic polynomial seems to provide a smooth curve, which we will use to model the descent.

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

Our goal is to find the parameters $a_{0}, a_{1}, a_{2}, a_{3}$ that fit the path of our Boeing 787. To determine the unknown coefficients, we impose the following conditions.

\begin{array}{ccc} Condition # & Verbal Description & Equation \\ 1. & The cruising altitude is 40000 ft & p (0) = 40000 \\ at the start of the descent. \\ 2. & The landing point has an altitude & p (400) = 0 \\ of 0. \\ 3. & The tangent line to the descent & p^{'} (0) = 0 \\ path is horizontal at the start. \\ 4. & The tangent line to the descent & p^{'} (400) = 0 \\ path is horizontal at the landing. \end{array}

Let's take a look at what each condition means.

Condition 1

We start our landing process at 40000 ft in the air and we assume this is 0 miles. In other words, we begin with our first equation $p (0) = 40000$ . By substituting in $x = 0$ and $p (0) = 400000$ , we see

\begin{aligned} 1 \cdot a_{0} + 0 \cdot a_{1} + 0 \cdot a_{2} + 0 \cdot a_{3} & = 40000 \end{aligned}

This is a linear equation based on the fact that we know the input and know the output. We can rewrite this as an 5.1 Inner Product Between Vectors.

[\begin{matrix} 1 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}] = 40000

Condition 2

We also know that our altitude will be 0 when we land. From our graph, we see that the altitude reaches 0 when $x$ is 400 miles. Plugging $x = 400$ and $p (400) = 0$ into our cubic polynomial, we see

1 \cdot a_{0} + 400 \cdot a_{1} + 400^{2} \cdot a_{2} + 400^{3} \cdot a_{3} = 0

Once again, we can write this linear equation as an inner product.

[\begin{matrix} 1 & 400 & 400^{2} & 400^{3} \end{matrix}] [\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}] = 0

Recall that matrices and vectors are data structures used to help us organize data. We can arrange our equations from conditions 1 and 2 into row form to produce 17.1 The General Linear-Systems Problem.

{[\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 400 & 400^{2} & 400^{3} \end{matrix}]}_{2 \times 4} {[\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}]}_{4 \times 1} = {[\begin{matrix} 40000 \\ 0 \end{matrix}]}_{2 \times 1}

When we have a short and wide matrix, we generally have too many (infinite!) possible solutions. We need to enforce more constraints to produce a meaningful solution.

Condition 3

Before landing, we want our plane to be traveling horizontally. Mathematically, we can say we want the tangent line of our path to have 0 slope at the moment it starts to descend. Written using mathematical notation, we see

p^{'} (0) = 0

We can also take the derivative of our function $p (x)$ .

\begin{aligned} p (x) & = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \\ ⟹ p^{'} (x) & = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} \end{aligned}

When we substitute in $x = 0$ and $p^{'} (0) = 0$ , we get at third equation.

0 \cdot a_{0} + 1 \cdot a_{1} + 2 \cdot 0 \cdot a_{2} + 3 \cdot 0^{2} \cdot a_{3} = 0

Note that we added $0 \cdot a_{0}$ so we can create a $4 \times 1$ 3.4 Row Vector of the scalars.

Once again, we can write this equation as an inner product.

[\begin{matrix} 0 & 1 & 0 & 0 \end{matrix}] [\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}] = 0

This is now a third condition we can put into a matrix.

Condition 4

We also want the tangent line at the end of the landing path to have a 0 slope. After all, we do not want to crash into the ground.

p^{'} (400) = 0

We can plug this value into our equation for $p^{'} (x)$ .

\begin{aligned} 0 \cdot a_{0} + 400 \cdot a_{1} + 2 \cdot 400 \cdot a_{2} + 3 \cdot 400^{2} \cdot a_{3} & = 0 \\ 0 \cdot a_{0} + 1 \cdot a_{1} + 800 \cdot a_{2} + 480000 \cdot a_{3} & = 0 \end{aligned}

And written as an inner product we see

[\begin{matrix} 0 & 1 & 800 & 480000 \end{matrix}] [\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{matrix}] = 0

Combining Conditions

Let's combine all four of these equations into a single system.

\begin{matrix} 1 \cdot a_{0} & + & 0 \cdot a_{1} & + & 0 \cdot a_{2} & + & 0 \cdot a_{3} & = 40000 \\ 1 \cdot a_{0} & + & 400 \cdot a_{1} & + & 160000 \cdot a_{2} & + & 64000000 \cdot a_{3} & = 0 \\ 0 \cdot a_{0} & + & 1 \cdot a_{1} & + & 0 \cdot a_{2} & + & 0 \cdot a_{3} & = 0 \\ 0 \cdot a_{0} & + & 1 \cdot a_{1} & + & 800 \cdot a_{2} & + & 480000 \cdot a_{3} & = 0 \end{matrix}

Each equation can be stored as a product. We can organize these products using matrices to form the system.

\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}]}}

We have successfully set up a general linear-systems problem from a real world context. See 17.5 Solving GLSP for Airplane Descent Path for the final solution to this modeling problem.