/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Find the points on the parabola ... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 23

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

Short Answer

Expert verified
The closest points are found by solving the equation \(4x^3 + 12x^2 + 10x + 2 = 0\).

Step by step solution

01

Write the Distance Formula

To determine the points on the parabola closest to the point (-1, 0), we begin by using the distance formula between two points (x, y) and (-1, 0). The distance formula is given by \(d = \sqrt{(x + 1)^2 + y^2}\).
02

Substitute the Parabola Equation

Since the point (x, y) is on the parabola \(y = x^2 + 2x\), substitute \(y = x^2 + 2x\) into the distance formula to get \(d = \sqrt{(x + 1)^2 + (x^2 + 2x)^2}\).
03

Minimize the Distance Squared

Instead of minimizing \(d\), minimize its square to avoid dealing with the square root. Thus, consider \(d^2 = (x + 1)^2 + (x^2 + 2x)^2\).
04

Simplify the Expression

Simplify the expression by expanding \((x + 1)^2\) and \((x^2 + 2x)^2\):- \((x + 1)^2 = x^2 + 2x + 1\)- \((x^2 + 2x)^2 = x^4 + 4x^3 + 4x^2\)Thus, \(d^2 = x^4 + 4x^3 + 5x^2 + 2x + 1\).
05

Find the Derivative

Differentiate \(d^2\) with respect to \(x\):\[\frac{d(d^2)}{dx} = 4x^3 + 12x^2 + 10x + 2\].
06

Solve for Critical Points

Set the derivative equal to zero to find critical points:\[4x^3 + 12x^2 + 10x + 2 = 0\]. This polynomial can be solved using methods like the quadratic formula or numerical techniques.
07

Evaluate Critical Points for Minimum Distance

Once you find the values of \(x\) that satisfy the equation, substitute each back into the parabola equation \(y = x^2 + 2x\) to find the corresponding \(y\)-values. Calculate the distance \(d\) for each to determine which gives the minimum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola Properties
A parabola is a symmetrical, U-shaped curve that is widely used in mathematics, especially in the study of quadratic equations. In this problem, the parabola given is described by the equation \(y = x^2 + 2x\). This specific parabola opens upwards due to the positive coefficient of the \(x^2\) term. Understanding the properties of this parabola helps us find the closest point on it to the point \((-1, 0)\).
  • Vertex: The vertex of a parabola in the form \(y = ax^2 + bx + c\) is found using the formula \(x = -\frac{b}{2a}\). For our parabola, \(a = 1\) and \(b = 2\), so the vertex is at \(x = -1\).
  • Axis of Symmetry: This line passes through the vertex and divides the parabola into two symmetric halves. For this parabola, the axis of symmetry is \(x = -1\).
  • Direction: The parabola opens upwards because the coefficient of \(x^2\) is positive.
Knowing these properties helps to understand the parabola's shape and can guide us in finding the minimum distance point analytically.
Critical Points
Critical points are where the derivative of a function is zero or undefined, often leading to local minimum or maximum values. In distance optimization problems, these points help us determine where the smallest (or largest) distance occurs.
  • We look for the critical points of the function \(d^2\), which represents our squared distance formula: \(d^2 = x^4 + 4x^3 + 5x^2 + 2x + 1\).
  • To find these critical points, we take the derivative of \(d^2\) and set it equal to zero: \(\frac{d(d^2)}{dx} = 4x^3 + 12x^2 + 10x + 2 = 0\).
Solving this equation gives us potential \(x\) values where the minimum distance may occur. These \(x\) values are then checked to find the corresponding \(y\) values on the parabola to pinpoint the exact critical points.
Derivatives
A derivative essentially measures how a function changes as its input changes. In optimization, derivatives are crucial because they indicate points of increase, decrease, or constant values within a function.
  • For this problem, the derivative of the squared distance function \(d^2\) gives us insights into how the distance changes as \(x\) changes.
  • The differential \(\frac{d(d^2)}{dx} = 4x^3 + 12x^2 + 10x + 2\), when set to zero, indicates places where the slope is flat (critical points), meaning potential minimum or maximum distances.
  • Using calculus, the critical points are found by solving the equation: \(4x^3 + 12x^2 + 10x + 2 = 0\).
Knowing how to work with derivatives helps find these critical points and make informed decisions about the closest point on the parabola.
Distance Formula
The distance formula is a mathematical tool used to calculate the distance between two points in a coordinate plane. Distances are commonly used in various optimization problems.
  • The formula used here is \(d = \sqrt{(x + 1)^2 + y^2}\), where \((x, y)\) is a point on the parabola, and \((-1, 0)\) is the specific point of interest.
  • To simplify the problem, we work with the squared distance \(d^2 = (x + 1)^2 + (x^2 + 2x)^2\). Minimizing \(d^2\) avoids dealing with square roots, which makes the calculations simpler.
  • This formula and its transformation are tools for determining the shortest path or closest point, which in many practical problems, is linked to efficiency and optimization.
Understanding these concepts allows us to apply the formula effectively to find the optimum points on a curve.

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Most popular questions from this chapter

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=4 x y+2 x^{2} y-x y^{2} $$

Show that the surfaces \(z=\sqrt{x^{2}+y^{2}}\) and \(10 z=\) \(25+x^{2}+y^{2}\) have the same tangent plane at \((3,4,5)\).

Find an equation of the plane tangent to the given surface at the given point. $$ \sin (x y)=2-z^{2} ;\left(\pi, \frac{1}{2},-1\right) $$

Find the extreme values of \(f\) in the region described by the given inequalities. In each case assume that the extreme values exist. $$ f(x, y)=16-x^{2}-4 y^{2} ; x^{4}+2 y^{4} \leq 1 $$

It seems reasonable that an increase in taxation on a commodity would decrease the production of that commodity. The following argument supports that claim. Assume that all required derivatives exist. For any \(x \geq 0\), let \(P_{0}(x)\) be the profit before taxes on \(x\) units produced. Let \(P(x, t)\) denote the profit after taxes on \(x\) units produced with \(\operatorname{tax} t\) on each unit. Assume that at any tax rate \(t\) the company will maximize its profits by producing \(f(t)\) units so that $$ \begin{aligned} &\frac{\partial P}{\partial x}(f(t), t)=0 \\ &\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)<0 \end{aligned} $$ (The conditions in (9) and (10) are just those required for the Second Derivative Test.) a. Show that \(P(x, t)=P_{0}(x)-t x\). b. Using (a), show that $$ \frac{\partial P}{\partial x}(x, t)=P_{0}^{\prime}(x)-t \quad \text { and } \quad \frac{\partial^{2} P}{\partial x^{2}}(x, t)=P_{0}^{\prime \prime}(x) $$ c. From \((9)\) and \((\mathrm{b})\), show that \(P_{0}^{\prime}(f(t))-t=0\). d. By differentiating both sides of the equation in (c) and by using (b) and (10), show that $$ f^{\prime}(t)=\frac{1}{P_{0}^{\prime \prime}(f(t))}=\frac{1}{\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)}<0 $$ (Thus the production tends to decrease as the tax rate increases.)
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