Problem 28 - University Calculus: Early Transcendentals

Question

In submarine location problems, it is often necessary to find a submarine's closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that the submarine travels on the parabolic path \(y=x^{2}\) and that the buoy is located at the point \((2,-1 / 2)\) a. Show that the value of \(x\) that minimizes the distance between the submarine and the buoy is a solution of the equation \(x=1 /\left(x^{2}+1\right)\) b. Solve the equation \(x=1 /\left(x^{2}+1\right)\) with Newton's method. (GRAPH CAN'T COPY)

Step-by-Step Solution

Answer

Solve \(x = \frac{1}{x^2 + 1}\) using Newton's method from an initial guess to find minimal distance.

Step 1: Understanding the Problem Formulation

The problem involves two tasks:1. Show that the value of \(x\) minimizing the distance between the submarine's path \(y = x^2\) and the buoy at \((2, -\frac{1}{2})\) is a solution of the equation \(x = \frac{1}{x^2 + 1}\). 2. Solve \(x = \frac{1}{x^2 + 1}\) using Newton's method.

Step 2: Finding the Distance Formula

The distance \(D\) between a point \((x, x^2)\) on the parabola and the point \((2, -\frac{1}{2})\) is given by the distance formula:\[D = \sqrt{(x - 2)^2 + (x^2 + \frac{1}{2})^2}.\]

Step 3: Expressing Distance in Terms of x

We express \(D^2\) rather than \(D\) to simplify calculations (since minimizing \(D\) is equivalent to minimizing \(D^2\)):\[D^2 = (x - 2)^2 + (x^2 + \frac{1}{2})^2.\]

Step 4: Differentiating for Minimum Distance

We minimize \(D^2\) by finding the derivative with respect to \(x\):\[\frac{d}{dx}((x - 2)^2 + (x^2 + \frac{1}{2})^2) = 2(x - 2) + 4x(x^2 + \frac{1}{2}).\]

Step 5: Setting Derivative Equal to Zero for Critical Points

Set the derivative equal to zero and solve:\[2(x - 2) + 4x(x^2 + \frac{1}{2}) = 0.\]Simplifies to:\[2x - 4 + 4x^3 + 2x = 0.\]

Step 6: Simplifying the Equation

Simplifying gives:\[4x^3 + 4x - 4 = 0\]Which can further simplify to:\[x^3 + x = 1.\]

Step 7: Transforming the Equation

Rewrite the equation \(x^3 + x = 1\) as \(x = \frac{1}{x^2 + 1}\), showing that \(x\) satisfying this will minimize the distance. Thus, we have derived the desired equation.

Step 8: Newton's Method Setup

We use Newton's method to solve \(x = \frac{1}{x^2 + 1}\). Define \(f(x) = x - \frac{1}{x^2 + 1}\) to transform the equation into finding the root \(f(x) = 0\).

Step 9: Computing Derivative for Newton's Method

Compute the derivative \(f'(x)\):\[f'(x) = 1 + \frac{2x}{(x^2 + 1)^2}.\]

Step 10: Applying Newton's Method

Apply the iterative formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.\]Choose a reasonable initial guess (for example, \(x_0 = 0\)) and iterate until convergence.

Key Concepts

Newton's Method

Newton's Method is a cornerstone in solving equations numerically. It helps us find approximate solutions to complex equations when algebra alone isn't sufficient. The method utilizes the concept of tangents to iteratively hone in on a solution.
To start, assume you have an initial guess, say at point \(x_0\). Newton's Method uses derivatives to essentially "nudge" your guess closer to the actual root of the equation by moving along the tangent. For our problem, where the equation is \(x - \frac{1}{x^2 + 1} = 0\), we define a function \(f(x)\) that captures this expression.

• First, compute the function's derivative, \(f'(x)\), which represents the slope of the tangent at any point \(x\).
• Then, use the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) to update your guess.

By iterating this process and refining the value repeatedly, you rapidly approach a solution. The beauty of Newton's Method is its efficiency and speed, especially useful in calculations needing quick refinements.

Parabolic Path

In mathematical terms, a parabolic path is represented by a quadratic equation, such as \(y = x^2\). It results in a U-shaped curve known as a parabola.
Submarines, like in this exercise, might follow such parabolic trajectories due to certain navigation patterns or maneuvers. This curve's properties lend themselves well to prediction and optimization tasks.
For the given problem, the parabola lies entirely in the plane, where each point \(x\) on the path corresponds to \(y = x^2\). The buoy is positioned at \((2, -\frac{1}{2})\), and our goal is to determine the point on the parabolic path that is nearest to this buoy. Understanding how the parabola frames the context of movement is crucial to developing solutions to minimize distance as outlined in the exercise.

Critical Points

Critical points are vital in calculus when seeking minimum or maximum values of functions. They occur where the derivative of a function equals zero, indicating potential points of interest such as a hilltop or a valley.
To find the critical points in our problem, we worked with the expression for \(D^2\), the squared distance between the submarine's path \( (x, x^2) \) and the buoy at \((2, -\frac{1}{2})\). By differentiating \(D^2\) and setting the derivative equal to zero, we isolate where the slope of the tangent to the distance function becomes flat.

• This solves the equation \(2(x-2) + 4x(x^2 + \frac{1}{2}) = 0\).
• On simplifying, it leads to \(4x^3 + 4x - 4 = 0\) which transforms into \(x^3 + x = 1\).

Understanding this concept allows us to identify candidate points where the distance might be minimized or maximized.

Distance Minimization

Distance minimization involves finding the point where two entities, in our case a submarine and buoy, are as close as possible.
The exercise specifies minimizing the distance between a submarine on a parabolic path and a fixed buoy position. This requires defining and working with a distance function. Here, the distance \(D\) from a point \((x, x^2)\) on the parabola to \((2, -\frac{1}{2})\) is calculated with the formula \(D = \sqrt{(x-2)^2 + (x^2 + \frac{1}{2})^2}\).
Squaring the distance simplifies the calculations, as minimizing \(D\) equates to minimizing \(D^2\). Thus, our task involves manipulating \(D^2\) by calculations such as derivatives to derive an equation that describes critical conditions where this minimum occurs. In this context, solving for \(x = \frac{1}{x^2 + 1}\) provides insight into where the minimum occurs on the submerged path close to the buoy.