Problem 349 - Calculus Volume 1

Question

For the following exercises, consider the points on the given graphs. Use a calculator to graph the functions. Where is the parabola \(y=x^{2}\) closest to point \((2,0) ?\)

Step-by-Step Solution

Answer

The parabola is closest to the point (2,0) at (1,1).

Step 1: Understanding the Problem

To find where the parabola \(y=x^{2}\) is closest to the point \((2,0)\), we need to determine the point on the parabola that minimizes the distance to \((2,0)\). This involves finding a point \((x, x^2)\) on the parabola such that the distance to \((2,0)\) is minimized.

Step 2: Define the Distance Formula

The distance \(D\) between any point \((x, x^2)\) on the parabola and the point \((2,0)\) can be expressed as follows: \[ D = \sqrt{(x - 2)^2 + (x^2 - 0)^2} \] This simplifies to: \[ D = \sqrt{(x - 2)^2 + x^4} \] We will minimize this expression.

Step 3: Simplify for Optimization

To make the optimization easier, we minimize the square of the distance instead of the distance itself, since minimizing \(f(x)\) and \(f^2(x)\) leads to the same optimal point. The squared distance \(D^2\) is: \[ D^2 = (x - 2)^2 + x^4 = x^4 + x^2 - 4x + 4 \]

Step 4: Find the Derivative

To find the minimum, take the derivative of \(D^2(x)\) with respect to \(x\): \[ \frac{d}{dx}(x^4 + x^2 - 4x + 4) = 4x^3 + 2x - 4 \] Set this derivative to zero to find the critical points.

Step 5: Solve the Derivative Equation

The derivative equation \(4x^3 + 2x - 4 = 0\) needs to be solved for \(x\). Let's simplify and solve it. Divide the whole equation by 2: \[ 2x^3 + x - 2 = 0 \] Finding the roots of this polynomial involves trying values or using a numerical solver.

Step 6: Verify and Choose the Closest Point

Checking roots numerically or graphically, we find that \(x \approx 1\) is a solution. Substituting \(x = 1\) back into the parabola equation, the point is \((1,1)\). Calculate the distance from \((1,1)\) to \((2,0)\) and confirm it's the minimum distance found.

Key Concepts

Distance Formula

The distance formula is a crucial tool in mathematics that helps measure the distance between two points in a coordinate plane. In this exercise, we want to find the distance from any point \((x, y)\) on the parabola \(y = x^2\) to a fixed point \((2,0)\). The distance \(D\) between these two points is given by the formula:

• \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In this exercise, \((x_1, y_1)\) is a point on the parabola, meaning it has coordinates \((x, x^2)\), and \((x_2, y_2)\) is the point \((2,0)\). Therefore, the distance becomes:

• \[ D = \sqrt{(x - 2)^2 + (x^2 - 0)^2} \]

This formula allows you to find how close or far the parabola is from \((2,0)\) at any given \(x\)-coordinate.

Derivative

In calculus, the derivative is a way to understand how functions change. It gives a precise way to talk about the rate of change or slope of a function at any given point. To minimize the distance to the point \((2,0)\), we need to find the derivative of the squared distance function,\(D^2(x) = x^4 + x^2 - 4x + 4\).This function represents the square of the distance formula because minimizing the square is mathematically simpler and finds the same point as minimizing the distance.The derivative \(\frac{d}{dx}D^2(x)\) is calculated as:

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

By setting this derivative to zero, we determine where the function is flat, potentially indicating a minimum point.

Parabola

A parabola is a curved symmetrical shape found as the graph of a quadratic function like \(y = x^2\). It is a U-shaped curve that opens upwards if the quadratic term's coefficient is positive, as in this case. Understanding the parabola's shape and position helps in visualizing and solving problems like the one at hand, where we need to determine which point on this curve is closest to a specific point.We focus only on the points of the parabola form \((x, x^2)\) since that is the equation \(y = x^2\) given to us in the problem. These points are crucial for setting up the distance equations and carrying out the optimization process later using calculus.

Critical Points

Critical points are where the derivative of a function is either zero or undefined. They help locate the minimum or maximum values within a given interval, making them particularly useful in optimization problems like this one. When finding the minimum distance between the parabola \(y = x^2\) and the point \((2,0)\), we solve the equation derived from the derivative \(4x^3 + 2x - 4 = 0\) for \(x\).Solving for these leads us to potential points where distance might be minimal. Once the critical points are identified, examine them and determine which provide the minimal distance by substituting back into the distance equation and calculating the actual distance.