Problem 31 - Calculus Early Transcendentals

Question

Use Newton's Method to approximate the coordinates of the point on the parabola \(y=x^{2}\) that is closest to the point \((1,0) .\)

Step-by-Step Solution

Answer

Use Newton's method iteratively to find the \(x\) value that minimizes \((x-1)^2 + x^4\); then solve for \(y=x^2\).

Step 1: Understand the Problem

We need to find the point on the parabola \(y = x^2\) that is closest to the given point \((1,0)\). This involves finding an \(x\)-coordinate for which the distance from the point \((x, x^2)\) to \((1,0)\) is minimized.

Step 2: Set Up the Distance Formula

The distance \(d\) from a point \((x, x^2)\) on the parabola to the point \((1, 0)\) can be expressed using the distance formula: \[d = \sqrt{(x - 1)^2 + (x^2 - 0)^2} = \sqrt{(x - 1)^2 + x^4}.\] To simplify the problem, we minimize the square of the distance, \(d^2\), which is \[f(x) = (x - 1)^2 + x^4.\]

Step 3: Find the Derivative

To use Newton's method, we need the derivative of \(f(x)\). Calculate the derivative: \[f'(x) = 2(x - 1) + 4x^3 = 2x - 2 + 4x^3.\]

Step 4: Apply Newton's Method

Newton's Method uses the iteration formula: \[x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)}.\] We need the second derivative, \[f''(x) = 2 + 12x^2.\]

Step 5: Initial Guess and Iteration

Choose an initial guess for \(x\), say \(x_0 = 0.5\). Apply Newton's Method:\[x_{n+1} = x_n - \frac{2x_n - 2 + 4x_n^3}{2 + 12x_n^2}.\]Calculate subsequent values for \(x\) until the values converge to a stable \(x\) coordinate. Typically three to four iterations are sufficient.

Step 6: Calculate Closest Point Coordinates

Once \(x\) converges to a stable value, \(x^*\), calculate the corresponding \(y\) value using \(y = (x^*)^2\). Ensure this \((x^* , (x^*)^2)\) is the closest point.

Key Concepts

Minimization Problem

When faced with a minimization problem such as finding the nearest point on a parabola to a specified point, we are essentially looking for a point that yields the smallest possible value of a certain function. Here, our function of interest is the distance between the point \( (x, x^2) \) on the parabola \( y = x^2 \) and the fixed point \( (1, 0) \).
The process involves finding the value of \( x \) that minimizes this distance.
To make the problem simpler, we often choose to minimize the square of the distance instead of the distance itself, as squaring removes the square root and simplifies calculations.

• This is because the square root operation in the distance formula adds computational complexity.
• Minimizing the square distance will also minimize the actual distance, as they reach their minimum values at the same point.

Parabola Equation

The parabola equation \( y = x^2 \) describes a symmetrical curve where each \( y \) value is the square of its corresponding \( x \) value.
This is a common quadratic function and represents a parabola opening upwards.
It's important to note that in the context of optimization problems like the one we are dealing with, the parabola guides us in understanding the behavior of the curve and the changes in points on it.

• The vertex of this parabola is at the origin \( (0, 0) \), making it symmetrical about the \( y \)-axis.
• Understanding this symmetry can often help in hypothesizing where the closest point might roughly lie.
• Parabolas are well-explored graphs, making them generally ideal for applying derivatives and methods such as Newton's.

Derivative Calculation

Derivatives are crucial in optimization problems like minimizing distance because they tell us how the function behaves as its input changes.
In this case, we found the derivative of the function representing squared distance: \( f(x) = (x - 1)^2 + x^4 \).
By differentiating, we obtained \( f'(x) = 2x - 2 + 4x^3 \), which helps us determine where the slope of the function is zero—potentially indicating a minimum point. To effectively apply Newton's Method, we also computed the second derivative, \((f''(x) = 2 + 12x^2)\), which provides information about the curvature.
Newton's Method iteratively refines guesses for the root of the derivative using these calculations.

• The first derivative indicates where the function has a zero slope, which could be a minimum or maximum.
• The second derivative helps determine the nature (concavity) of the curve around those critical points, confirming whether it's indeed a minimum.
• For Newton's Method, these derivatives enable us to efficiently zero in on the precise solution by repeatedly adjusting our guess.