Problem 39 - Thomas Calculus

Question

Find the point on the parabola \(x=t, y=t^{2},-\infty< t <\infty\) closest to the point \((2,1 / 2) .\) (Hint: Minimize the square of the distance as a function of \(t\). )

Step-by-Step Solution

Answer

Solve \(4t^3 + 2t - 4 = 0\) to find \(t\), then substitute into \(x=t\) and \(y=t^2\).

Step 1: Write the Distance Formula

The distance between the point on the parabola \( (x, y) \) and the point \( (2, \frac{1}{2}) \) is given by the formula \((x-2)^2 + (y - \frac{1}{2})^2\). Since \(x = t\) and \(y = t^2\), the formula becomes \((t-2)^2 + (t^2 - \frac{1}{2})^2\)."

Step 2: Set up the Function to Minimize

To find the closest point, we need to minimize the square of the distance. Define this function as follows: \[ f(t) = (t-2)^2 + (t^2 - \frac{1}{2})^2. \] Our goal is to find the critical points of \(f(t)\).

Step 3: Differentiate the Distance Function

Differentiate the function \(f(t)\) with respect to \(t\). We have: \[ f'(t) = 2(t-2) + 2(t^2 - \frac{1}{2}) \cdot 2t = 2(t-2) + 4t(t^2 - \frac{1}{2}). \] Simplifying this expression, we get: \[ f'(t) = 2(t-2) + 4t^3 - 2t = 4t^3 + 2t - 4. \]

Step 4: Find the Critical Points

Set the derivative \(f'(t)\) equal to zero to find the critical points: \[ 4t^3 + 2t - 4 = 0. \] Solve this cubic equation for the values of \(t\).

Step 5: Determine the Closest Point

Using any method (analytical or numerical), solve the cubic equation \((4t^3 + 2t - 4 = 0)\) to find the value of \(t\) that minimizes the distance. This may involve a guess and check, graphing, or using methods like the Newton-Raphson method to estimate roots if solving algebraically is intractable.

Step 6: Calculate the Coordinates of the Closest Point

Once the value of \(t\) is obtained, substitute it back into the parametric equations \(x = t\) and \(y = t^2\) to find the coordinates of the closest point on the parabola.

Key Concepts

Distance Minimization

In calculus, distance minimization is a fundamental concept that involves finding a point on a curve that is closest to a given target point. This is often accomplished by minimizing the distance function, which represents the square of the Euclidean distance between the points.

The goal is to minimize for computational efficiency. Instead of minimizing the actual distance, the square of the distance is minimized. This approach simplifies calculations, as it removes the square root typically found in the Euclidean distance formula.

Consider a parabola described by the equation

• egin{align*}x = t, \ y = t^2ti
and the target point
• \( \(2, \frac{1}{2}\) \) .

The problem is to find which point on the parabola is closest to the target. The square of the distance function is

• \(f(t) = (t - 2)^2 + (t^2 - \frac{1}{2})^2\)

By minimizing the function \(f(t)\), we can find the value of \(t\)that results in the shortest distance between the parabola and

• \( \(2, \frac{1}{2}\) \)

in the coordinate plane.

Parabola Geometry

Parabolas are unique and simple geometric shapes integral to many optimizations and physics-related problems due to their distinctive properties.
A parabola is defined as the set of all points equidistant from a fixed point called the "focus" and a line called the "directrix."

In this exercise, the parabola is described parametrically by

• \(x=t\)
• \(y=t^2\)

This representation shows that every point on the parabola is described by a parameter \(t\), making it easy to express any mathematical relationships in terms of \(t\).

Parabolas are commonly encountered in quadratic functions, and their standard equation, appearing in multiple forms, demonstrates their versatility. When working with such geometries, solving problems, such as distance minimization, often involves using the geometry of the parabola and properties derived from its equation. It highlights why understanding the parabola's structure can simplify complex mathematical descriptions and solutions.

Critical Points Analysis

Critical points in calculus are the points where the first derivative of a function is either zero or undefined. Analyzing these points helps in identifying local minima, maxima, or points of inflection, crucial for solving optimization problems.

To find a critical point for minimizing distance to the parabola, we differentiate the distance function

• \(f(t) = (t-2)^2 + (t^2 - \frac{1}{2})^2\)

This results in a new function, the derivative \(f'(t)\), which is set to zero to find critical points.

• \(f'(t) = 4t^3 + 2t - 4\)

Solving for \(t\) when \(f'(t) = 0\) identifies where the function is at a stationary point. This is where potential minimum distance can occur.

In complex cases, finding such roots might require numerical methods or graphing, as analytical solutions may not always be feasible. Critical point analysis thus not only finds where functions stabilize but also offers insights into function behavior, pivotal in optimizing various geometric and real-world problems.