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Magazine Chapter 4: Problem 18
Find the point(s) on the parabola \(y=\frac{1}{8} x^{2}\) closest to the point (0,6).
Short Answer
Expert verified
The point(s) on the parabola \(y=\frac{1}{8} x^{2}\) closest to the point (0,6) are \(\pm (\frac{4}{\sqrt[3]{2}},1)\).
Step by step solution
01
Express the distance
The first step is to express the distance between a generic point (x,y) on the curve \(y=\frac{1}{8} x^{2}\) and the given point (0,6) by the distance formula. As we're going to square this distance and minimize it, we can avoid the square root. So \(d^2 = (x-0)^2 + (\frac{1}{8} x^{2}-6)^2\).
02
Simplify the equation
Simplify the above equation to obtain a formula for \(d^2\). This results in \(d^2 = x^2 + (\frac{1}{64} x^4 - \frac{3}{4}x^2 + 36)\). Further simplifying we get \(d^2 = \frac{1}{64}x^4 - \frac{1}{2}x^2 + x^2 + 36 = \frac{1}{64}x^4 + \frac{1}{2}x^2 + 36\).
03
Differentiate the equation
Differentiate the equation \(d^2 = \frac{1}{64}x^4 + \frac{1}{2}x^2 + 36\) to find the points where \(d^2\) reaches a minimum. The derivative \(d'(x) = \frac{1}{16}x^3 + x\).
04
Find the minimum points
Set \(d'(x) = \frac{1}{16}x^3 + x = 0\), to find the x coordinate(s) where \(d^2\) reaches a minimum. Factor the equation to \(x( \frac{1}{16}x^2 + 1) =0\). The solutions for the x coordinates are \(x = 0\) and \(x = \pm \frac{4}{\sqrt[3]{2}}\)
05
Find the y coordinates
Substitute the x-values into the equation of the curve \(y=\frac{1}{8} x^{2}\) to find the corresponding y values. For \(x = 0\), we have \(y = 0\), and for \(x = \pm \frac{4}{\sqrt[3]{2}}\), we have \(y = \pm 1\).
06
Verify the results
By comparing \(d^2\) for different x values, one can verify that \(x = \pm \frac{4}{\sqrt[3]{2}}\) indeed minimize the distance to the point (0,6).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
Parabolas are a familiar sight in mathematics and appear frequently in calculus. They are defined by a quadratic equation, often taking the form \(y = ax^2+bx+c\). In this specific exercise, we deal with the parabola represented by \(y = \frac{1}{8}x^2\). This equation is simplified, having its vertex at the origin (0,0), and it opens upwards.
- Vertex at (0,0)
- Upward opening
Distance Formula
The distance formula is used to determine the distance between two points in a coordinate plane. For two points, \((x_1, y_1)\) and \((x_2, y_2)\), the formula is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This exercise simplifies the problem by working with squared distances. By minimizing the expression for \(d^2\), we avoid having to deal with a square root, making the computation process easier.
In the context of this problem, the two points are \((x, \frac{1}{8}x^2)\) on the parabola and (0,6), respectively. Thus, the distance squared becomes \(d^2 = x^2 + (\frac{1}{8}x^2 - 6)^2\). Simplifying this equation lets us later apply calculus techniques for optimization.
In the context of this problem, the two points are \((x, \frac{1}{8}x^2)\) on the parabola and (0,6), respectively. Thus, the distance squared becomes \(d^2 = x^2 + (\frac{1}{8}x^2 - 6)^2\). Simplifying this equation lets us later apply calculus techniques for optimization.
Derivative
The derivative is a core concept of calculus, representing the rate at which a function changes at any given point. In optimization problems, derivatives help identify where functions reach their minimum or maximum values.
In this problem, once the expression for \(d^2\) is simplified to \(d^2 = \frac{1}{64}x^4 + \frac{1}{2}x^2 + 36\), we take its derivative \(d'(x) = \frac{1}{16}x^3 + x\). Calculating the derivative helps us find critical points, where the function does not change, indicating potential minimum distances between two points.
In this problem, once the expression for \(d^2\) is simplified to \(d^2 = \frac{1}{64}x^4 + \frac{1}{2}x^2 + 36\), we take its derivative \(d'(x) = \frac{1}{16}x^3 + x\). Calculating the derivative helps us find critical points, where the function does not change, indicating potential minimum distances between two points.
Minimization Problem
Minimization is a central topic in optimization, where the goal is to find the smallest possible value of a function. In this exercise, we are tasked with finding the points on the parabola that are closest to another point.
- Set derivative to zero
- Solve for critical points
- Verify minimum by comparing values
