Problem 25 - Calculus with Concepts in Calculus

Question

Find the points on the parabola \(y=x^{2}+2 x\) that are closest to the point \((-1,0)\).

Step-by-Step Solution

Answer

The points on the parabola closest to (-1,0) are where the critical values solve the distance equation's first derivative set to zero.

Step 1: Identify the Distance Formula

To find the points on the curve closest to the given point, we use the distance formula. The distance, \(D\), between a point \((x, y)\) on the parabola and the point \((-1,0)\) is given by \(D = \sqrt{(x + 1)^2 + (y - 0)^2}\).

Step 2: Substitute the Parabola Equation

Given the equation of the parabola \(y = x^2 + 2x\), substitute \(y\) in the distance formula to get \(D(x) = \sqrt{(x + 1)^2 + (x^2 + 2x)^2}\).

Step 3: Simplify the Distance Squared

To eliminate the square root for easier differentiation, set the squared distance function \(D^2(x) = (x + 1)^2 + (x^2 + 2x)^2\). Simplifying this gives \(D^2(x) = x^2 + 2x + 1 + x^4 + 4x^3 + 4x^2\).

Step 4: Take the Derivative

Find the derivative of the simplified function \(D^2(x) = x^4 + 4x^3 + 5x^2 + 2x + 1\) with respect to \(x\), giving \(\frac{d}{dx}[D^2(x)] = 4x^3 + 12x^2 + 10x + 2\).

Step 5: Find Critical Points

Set \(\frac{d}{dx}[D^2(x)] = 0\) to find the critical points: \(4x^3 + 12x^2 + 10x + 2 = 0\). Use numerical or algebraic techniques to solve the polynomial equation, yielding critical values for \(x\).

Step 6: Analyze Critical Points

Evaluate the second derivative \(\frac{d^2}{dx^2}[D^2(x)]\) or test the critical points by substitution back into the squared distance to identify the minimum values.

Step 7: Compute Corresponding y Values

Substitute the \(x\)-values found in step 5 that yield minimum distances into the parabola equation \(y = x^2 + 2x\) to find the corresponding \(y\)-values for the closest points.

Key Concepts

Distance Formula

To solve optimization problems in calculus, like finding points on a parabola closest to a given location, the distance formula is a key concept. The distance formula calculates the distance between two points on a plane. If you have two points: Point 1 \( (x_1, y_1) \) and Point 2 \( (x_2, y_2) \), the distance \( D \) between them is given by: \( D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).In our exercise, you are asked to find points on the parabola that are the shortest distance from the point \( (-1,0) \). Using the distance formula, expressing the distance between a point \( (x, y) \) on the parabola and \( (-1, 0) \) yields:

• \( D = \sqrt{(x + 1)^2 + (y - 0)^2} \)

This formula will help us analyze how far a particular point on the parabola is from \( (-1, 0) \) and determine the shortest distance possible.

Parabola Equation

A parabola is a curved, U-shaped graph represented by a quadratic equation. In this particular exercise, the parabola is defined by the equation: \( y = x^2 + 2x \).This equation tells us the relationship between \( x \) and \( y \), defining every possible point on the curve. When tackling an optimization problem, you'll often substitute this equation into another formula to simplify your calculations. Here, it's substituted into the distance formula to express the distance only in terms of \( x \) because \( y \) can be rewritten using the parabola equation. Therefore, the distance as a function of \( x \) becomes:

• \( D(x) = \sqrt{(x + 1)^2 + (x^2 + 2x)^2} \)

Now, the task is to find the \( x \) values on this parabola that minimize the distance to \( (-1, 0) \).

Critical Points

Critical points are where a function's derivative is zero or undefined, helping identify possible minima or maxima.To find the point closest to \( (-1, 0) \), we first simplify the expression \( D(x) \) by squaring it to avoid dealing with square roots:

• \( D^2(x) = (x + 1)^2 + (x^2 + 2x)^2 \)
• Simplified to: \( D^2(x) = x^4 + 4x^3 + 5x^2 + 2x + 1 \)

Next, by differentiating \( D^2(x) \), we obtain a polynomial whose roots will give us the critical points:

• \( \frac{d}{dx}[D^2(x)] = 4x^3 + 12x^2 + 10x + 2 = 0 \)

Solving this equation might involve algebraic methods, or using numerical methods when it's complex to solve analytically.

Derivative

Finding derivatives is an essential process in calculus to determine the rate of change. In optimization problems, caclulating the derivative of a function allows for finding critical points that show where the function's value is at a minimum or maximum.For our problem, by taking the derivative of \( D^2(x) \), we arrive at:

• \( \frac{d}{dx}[D^2(x)] = 4x^3 + 12x^2 + 10x + 2 \)

Setting this derivative equal to zero helps isolate the critical points. The next step is solving the equation. Each critical point is then tested further, either by the second derivative test or by checking which yields smaller distances after substituting them back into the distance formula.This analysis helps determine which point on the parabola is the closest, making derivatives indispensable in solving such optimization scenarios.