Problem 45 - Calculus

Question

A plane starts its descent from height \(y=h\) at \(x=-L\) to land at (0,0) . Choose \(a, b, c, d\) so its landing path \(y=a x^{3}+b x^{2}+c x+d\) is smooth. With \(d x / d t=V=\) constant, find \(d y / d t\) and \(d^{2} y / d t^{2}\) at \(x=0\) and \(x=-L\) (To keep \(d^{2} y / d t^{2}\) small, a coast-to-coast plane starts down \(L>100\) miles from the airport.)

Step-by-Step Solution

Answer

At x=0, \( \frac{dy}{dt} = 0 \) and \( \frac{d^2y}{dt^2} = 0 \). At x=-L, \( \frac{dy}{dt} = -\frac{3hV}{L} \) and \( \frac{d^2y}{dt^2} = -\frac{6hV^2}{L^2} \).

Step 1: Identify Boundary Conditions

The plane's path starts at the point \(x = -L, y = h\) and ends at \(x = 0, y = 0\). At \(x = 0\), the descent should be smooth, so both the slope (first derivative) and curvature (second derivative) should be zero.

Step 2: Set Up the Polynomial

The landing path is given by the polynomial \(y = ax^3 + bx^2 + cx + d\). We know that \(y(-L) = h\) and \(y(0) = 0\). Also, \(rac{dy}{dx}(0) = 0\) and \(rac{d^2y}{dx^2}(0) = 0\) to ensure smoothness.

Step 3: Apply Boundary Conditions for Height

Plug \(x = 0\) and \(y = 0\) into the polynomial to get \(d = 0\). Next, plug \(x = -L\) and \(y = h\), resulting in an equation involving \(a, b, c\) and \(h\): \(h = a(-L)^3 + b(-L)^2 + c(-L)\).

Step 4: Apply Derivative Conditions for Smoothness

Calculate the first derivative \(rac{dy}{dx} = 3ax^2 + 2bx + c\). At \(x = 0\), \(rac{dy}{dx} = 0\) implies that \(c = 0\). The second derivative \(rac{d^2y}{dx^2} = 6ax + 2b\) should also be zero at \(x = 0\), giving \(b = 0\).

Step 5: Simplify Polynomial with Conditions

Substitute \(b = 0\) and \(c = 0\) in the original polynomial, leading to \(y = ax^3 + d\). We already know \(d = 0\), so the equation simplifies further to \(y = ax^3\). From the condition \(h = a(-L)^3\), we derive \(a = -\frac{h}{L^3}\).

Step 6: Find \\( \frac{dy}{dt} \\) and \\( \frac{d^2y}{dt^2} \\)

With constant \( \frac{dx}{dt} = V\), apply the chain rule: \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (3ax^2) V\). At \(x = 0\), \( \frac{dy}{dt} = 0\). At \(x = -L\), calculate \( \frac{dy}{dt} = 3aL^2V = -\frac{3hV}{L}\). Similarly, find \( \frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} (\frac{dx}{dt})^2 = (6ax)V^2\). At \(x = 0\), it is \(0\). At \(x = -L\), calculate \( \frac{d^2y}{dt^2} = -\frac{6hV^2}{L^2}\).

Key Concepts

Boundary Conditions

Boundary conditions are crucial in many engineering and physics problems, and in this case, they help to define the parameters of the landing path of a plane. Here, we'll discuss the initial and final conditions that need to be met for a successful descent and landing.

The aircraft begins its descent at the point \(x = -L\), \(y = h\), indicating an initial height \(h\) above the runway. This point is the start of the descent path. The landing occurs at \(x = 0\), \(y = 0\), where the plane must be exactly at ground level or runway height.

In addition to these position-based boundary conditions, it's equally relevant to impose conditions on the smoothness of the landing. At \(x = 0\), both \(\frac{dy}{dx}\) (the slope) and \(\frac{d^2y}{dx^2}\) (the curvature) of the path must be zero. This ensures that the plane's descent is smooth and parallel to the ground when it touches down. These conditions guide the determination of the coefficients \(a, b, c, d\) in the polynomial equation for the landing path.

Polynomial Derivative

The concept of derivatives is fundamental when dealing with polynomial functions, especially in determining the shape and behavior of a path, such as an aircraft's landing trajectory. Derivatives tell us about the rate of change, helping to ensure the path's smoothness and safety.

For the polynomial landing path \(y = ax^3 + bx^2 + cx + d\), we are interested in both the first and second derivatives. The first derivative \(\frac{dy}{dx}\) gives us the slope of the tangent to the curve at any point \(x\). It is crucial for determining how steeply the plane descends. The first derivative is \(\frac{dy}{dx} = 3ax^2 + 2bx + c\). At \(x = 0\), a smooth landing requires \(\frac{dy}{dx} = 0\), informing us that \(c = 0\).

The second derivative \(\frac{d^2y}{dx^2}\) tells us about the curvature or the rate of change of the slope of the path. The expression \(\frac{d^2y}{dx^2} = 6ax + 2b\) must also equal zero at \(x = 0\) to ensure smoothness, resulting in \(b = 0\). These derivatives are integral to solving for the polynomial coefficients that define a landing path that is both practical and safe.

Chain Rule in Calculus

The chain rule is a fundamental tool in calculus that allows us to differentiate composite functions. It plays a vital role when the rate of change with respect to time, rather than distance, needs to be calculated. In this scenario, understanding how different rates of change interact is key to modeling flight paths.

When working with the descent path of a plane, you might be given \(\frac{dx}{dt} = V\), where \(V\) is a constant representing the speed of the plane along the x-axis. To find the rate of change of the descent path with respect to time, \(\frac{dy}{dt}\), the chain rule is applied:

• First, calculate \(\frac{dy}{dx}\).
• Then, use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

At any position \(x\) along the descent path, this gives the vertical speed. Specifically, it evaluates as zero at \(x = 0\), ensuring that the plane lands without vertical velocity. The chain rule thus bridges spatial and temporal analysis, providing insight into the dynamic landing process.

Smooth Landing Path

Creating a smooth landing path is critical for ensuring both passenger comfort and the structural integrity of the aircraft during landing. This requirement means the path must meet specific delineated conditions, resulting in smoother flight dynamics and transitions.

The concept of a smooth landing path involves minimizing abrupt changes in the descent path's slope and curvature. For a path given by \(y = ax^3\), the parameters must ensure continuity and gentleness as the plane approaches the runway. Achieving this involves setting the first and second derivatives of the path at the landing point (\(x = 0\)) to zero.

The chosen polynomial here results not just in smooth transitions but also an extendable path that can be modified depending on variables like descent starting point \(L\) and initial height \(h\). This flexibility helps in maintaining small values for \(\frac{d^2y}{dt^2}\), keeping rapid accelerations and decelerations minimal. These efforts ensure a safe and stable approach, adhering to engineering design principles and operational safety standards.