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Chapter 4: Problem 8

A point moves in the \((x, y)\) plane on the line \(2 x+3 y-4=0\). Where will it be when the sum of the squares of its distances from (1,0) and (-1,0) is smallest?

Short Answer

Expert verified
The point is \left( \frac{8}{13}, \frac{4}{39} \right).

Step by step solution

01

Represent Distance Formula

Use the distance formula to express distances from points \( (1,0) \) and \( (-1,0) \) to a generic point \( (x,y) \). The distance from \( (x,y) \) to \( (1,0) \) is given by: \[ D_1 = \sqrt{(x-1)^2 + y^2} \]. The distance from \( (x,y) \) to \( (-1,0) \) is given by: \[ D_2 = \sqrt{(x+1)^2 + y^2} \].
02

Sum of Squares of Distances

Express the sum of the squares of these distances: \[ S = D_1^2 + D_2^2 \]. Thus: \[ S = (x-1)^2 + y^2 + (x+1)^2 + y^2 \].
03

Simplify the Expression

Simplify the expression for \ S \: \[ S = (x-1)^2 + y^2 + (x+1)^2 + y^2 = x^2 - 2x + 1 + y^2 + x^2 + 2x + 1 + y^2 \] Combining like terms: \[ S = 2x^2 + 2y^2 + 2 \].
04

Constraint of the Line Equation

The point \( (x,y) \) lies on the line \(2x + 3y - 4 = 0\). Solving for \ y \: \[ y = -\frac{2}{3}x + \frac{4}{3} \].
05

Substitute the Constraint

Substitute \( y = -\frac{2}{3} x + \frac{4}{3} \) into the expression for \ S \: \[ S = 2x^2 + 2 \left( -\frac{2}{3} x + \frac{4}{3} \right)^2 + 2 \].
06

Expand and Simplify

Expand and simplify the substituted expression: \[ S = 2x^2 + 2 \left( \frac{4}{9} x^2 - \frac{16}{9} x + \frac{16}{9} \right) + 2 = 2x^2 + \frac{8}{9} x^2 - \frac{32}{9} x + \frac{32}{9} + 2 = \frac{26}{9} x^2 - \frac{32}{9} x + \frac{50}{9} \].
07

Complete the Square

Complete the square to find the minimum value: \[ S = \frac{26}{9} \left( x^2 - \frac{16}{13}x + \frac{50}{26} \right) = \frac{26}{9} \left( x - \frac{8}{13} \right)^2 + \frac{50}{9} - \frac{26}{169} \times 64 = \frac{26}{9} \left( x - \frac{8}{13} \right)^2 + \frac{450}{117} - \frac{416}{117} = \frac{26}{9} \left( x - \frac{8}{13} \right)^2 + \frac{34}{117} = \frac{26}{9} \left( x - \frac{8}{13} \right)^2 + \frac{34}{117} = \frac{26}{9} \left( x - \frac{8}{13} \right)^2 + \frac{17}{9} \] which is minimized at \ x = \frac{8}{13} \.
08

Find the Corresponding y-Coordinate

Substitute \ x = \frac{8}{13} \ back into the line equation to find the corresponding \ y \ value: \[ y = \-frac{2}{3} \frac{8}{13} + \frac{4}{3} = \frac{4}{39} \].

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Key Concepts

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

Distance Formula
The distance formula is a crucial concept in geometry. It helps in determining the distance between two points in a coordinate plane. The formula is derived from the Pythagorean Theorem and is given by: \( D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This formula calculates the linear distance between two points \((x_1, y_1)\) and \((x_2, y_2)\). In the exercise, we use this formula to express distances from a point \((x,y)\) on the line 2x+3y-4=0 to points (1,0) and (-1,0). Specifically, the distances are: \[ D_1 = \sqrt{(x-1)^2 + y^2} \text{ and } D_2 = \sqrt{(x+1)^2 + y^2} \] Understanding how to apply the distance formula is fundamental in solving optimization problems in geometry.
Sum of Squares
The sum of squares is another key concept in this exercise. It involves adding the squares of different quantities, which can be useful in various optimization problems. In our case, we aim to minimize the sum of squares of distances from a point on a line to two other fixed points. The sum of squares of distances can be expressed as: \[ S = D_1^2 + D_2^2 \] Substituting the distance formulas, we get: \[ S = (x-1)^2 + y^2 + (x+1)^2 + y^2 \] Simplifying this expression leads us to: \[ S = 2x^2 + 2y^2 + 2 \] By working with sum of squares, we can ensure that we always end up with positive values since squares are inherently non-negative.
Completing the Square
Completing the square is a method used to simplify quadratic expressions. It involves rewriting a quadratic in the form of a squared binomial plus or minus a constant. In our exercise, after substituting the constraint equation into the distance sum formula, we obtain: \[ S = \frac{26}{9}x^2 - \frac{32}{9}x + \frac{50}{9} \] Completing the square for this expression helps to identify the minimum value: \[ S = \frac{26}{9}(x - \frac{8}{13})^2 + \frac{17}{9} \] This form shows that the expression reaches its minimum value when the squared term equals zero, i.e., when \(x = \frac{8}{13}\). Completing the square is a powerful tool in optimization because it transforms difficult quadratics into more manageable forms.
Constraint Equation
The constraint equation is crucial in problems where variables are limited by certain conditions. In our exercise, the point \((x,y)\) must lie on the line given by: \[ 2x + 3y - 4 = 0 \] To work with this constraint, we solve for one variable in terms of the other. For instance, solving for \( y \) we get: \[ y = -\frac{2}{3}x + \frac{4}{3} \] Substituting this relationship back into the expression for the sum of squares allows us to reduce the problem to a single variable. Constraints are essential in optimization because they define the feasible region where the optimal solution must lie.

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Most popular questions from this chapter

Show that \(\frac{d}{d x} \int_{\cos x}^{\sin x} \sqrt{1-t^{2}} d t=1.\)

The operating equation for a synchrotron in the relativistic range is $$ q B=\omega m\left[1-(\omega R)^{2} / c^{2}\right]^{-1 / 2} $$ where \(q\) and \(m\) are the charge and rest mass of the particle being accelerated, \(B\) is the magnetic field strength, \(R\) is the orbit radius, \(\omega\) is the angular frequency, and \(c\) is the speed of light. If \(\omega\) and \(B\) are varied (all other quantities constant), show that the relation between \(d \omega\) and \(d B\) can be written as $$ \frac{d B}{B^{3}}=\left(\frac{q}{m}\right)^{2} \frac{d \omega}{\omega^{3}} $$ or as \(\quad \frac{d B}{B}=\frac{d \omega}{\omega}\left[1-(\omega R / c)^{2}\right]^{-1}.\)

About how much (in percent) does an error of \(1 \%\) in \(x\) and \(y\) affect \(x^{3} y^{2} ?\)

A function \(f(x, y, z)\) is called homogeneous of degree \(n\) if \(f(t x, t y, t z)=t^{n} f(x, y, z)\) For example, \(z^{2} \ln (x / y)\) is homogeneous of degree 2 since $$ (t z)^{2} \ln \frac{t x}{t y}=t^{2}\left(z^{2} \ln \frac{x}{y}\right) $$ Euler's theorem on homogeneous functions says that if \(f\) is homogeneous of degree \(n,\) then $$ x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}+z \frac{\partial f}{\partial z}=n f $$ Prove this theorem. Hints: Differentiate \(f(t x, t y, t z)=t^{n} f(x, y, z)\) with respect to \(t,\) and then let \(t=1 .\) It is convenient to call \(\partial f / \partial(t x)=f_{1}(\) that is, the partial derivative of \(f\) with respect to its first variable), \(f_{2}=\partial f / \partial(t y),\) and so on. Or, you can at first call \(t x=u, t y=v, t z=w\). (Both the definition and the theorem can be extended to any number of variables.)

If \(z=x^{2}+2 y^{2}, x=r \cos \theta, y=r \sin \theta,\) find the following partial derivatives. Repeat if \(z=r^{2} \tan ^{2} \theta\). $$\left(\frac{\partial z}{\partial x}\right)_{\theta}$$
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