Problem 28 - Thomas Calculus

Question

Find the point on the line \(\frac{x}{a}+\frac{y}{b}=1\) that is closest to the origin.

Step-by-Step Solution

Answer

The closest point is \( \left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right) \).

Step 1: Express y in terms of x

Starting with the given line equation \( \frac{x}{a} + \frac{y}{b} = 1 \), we can express \( y \) in terms of \( x \) by solving for \( y \): \( \frac{y}{b} = 1 - \frac{x}{a} \). Multiply through by \( b \) to obtain \( y = b - \frac{b}{a}x \).

Step 2: Define the distance formula to the origin

The distance \( D \) from a point \( (x, y) \) to the origin \( (0,0) \) can be given by the formula \( D = \sqrt{x^2 + y^2} \). We aim to minimize this distance, so we'll work with \( D^2 = x^2 + y^2 \) to simplify calculus later.

Step 3: Substitute y in terms of x into the distance formula

Substitute \( y = b - \frac{b}{a}x \) into the expression for \( D^2 \): \( D^2 = x^2 + (b - \frac{b}{a}x)^2 \). Simplify it to get \( D^2 = x^2 + b^2 - 2\frac{b^2}{a}x + \frac{b^2}{a^2}x^2 \).

Step 4: Differentiate with respect to x

Differentiate \( D^2 = x^2 + b^2 - 2\frac{b^2}{a}x + \frac{b^2}{a^2}x^2 \) with respect to \( x \). The derivative is \( 2x - 2\frac{b^2}{a} + 2\frac{b^2}{a^2}x \). Simplifying, this becomes \( 2x + 2\frac{b^2}{a^2}x - 2\frac{b^2}{a} \).

Step 5: Set the derivative equal to zero and solve for x

To find the minimum distance, set the derivative \( 2x + 2\frac{b^2}{a^2}x - 2\frac{b^2}{a} \) equal to zero: \( 2x(1 + \frac{b^2}{a^2}) - 2\frac{b^2}{a} = 0 \). Solving, we find \( x = \frac{ab^2}{a^2 + b^2} \).

Step 6: Calculate y using x

Substitute \( x = \frac{ab^2}{a^2 + b^2} \) back into the equation for \( y \): \( y = b - \frac{b}{a}\left(\frac{ab^2}{a^2 + b^2}\right) \). Simplify to get \( y = \frac{a^2b}{a^2 + b^2} \).

Step 7: Present the closest point

The point on the line that is closest to the origin is \( \left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right) \).

Key Concepts

Distance Formula

The distance formula is a fundamental concept in geometry, used to calculate the distance between two points in a plane. In the problem given, we're looking to find the distance from a point on a line to the origin of the coordinate plane, which is (0,0). The formula we use is:

• For any point \( (x, y) \), the distance \( D \) to the origin (0,0) is given by \( D = \sqrt{x^2 + y^2} \).

This formula comes from the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, \( x \) and \( y \) represent the other two sides, and \( D \) represents the hypotenuse.
To simplify calculations, especially when using calculus, we often use the distance squared function \( D^2 = x^2 + y^2 \). This eliminates the square root, making the derivative easier to compute.

Derivative

The derivative is a key tool in calculus that helps us find the rate of change of a function's value with respect to its variables. In the context of optimization, we use derivatives to identify points where a function reaches its minimum or maximum value.
For our problem, we aim to find where the distance from the origin is minimized. We derived the expression \( D^2 = x^2 + y^2 \) and differentiated it with respect to \( x \), leading us to:

• The derivative of \( D^2 \) with respect to \( x \), is \( 2x + 2\frac{b^2}{a^2}x - 2\frac{b^2}{a} \).

The derivative helps us find where the slope of the \( D^2 \) function is zero, which indicates a minimum or maximum point.

Minimization

Minimization is the process of finding the smallest value that a function can take. In the case of this problem, we're interested in the minimum distance from the line \( \frac{x}{a}+\frac{y}{b}=1 \) to the origin. Achieving this involves finding the point where the derivative of the distance function is zero and the concavity of the function confirms it as a minimum.
Setting the derivative \( 2x + 2\frac{b^2}{a^2}x - 2\frac{b^2}{a} \) to zero allows us to solve for \( x \). The solution \( x = \frac{ab^2}{a^2 + b^2} \) gives us the \( x \)-coordinate of the point on the line closest to the origin.
Substituting back to find \( y \), we determine the point at which the minimum distance occurs is \( \left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right) \).

Line Equation

A line equation in a plane describes all points (

• For a standard form of the line: \( \frac{x}{a} + \frac{y}{b}=1 \).

This equation can be manipulated in various ways, like solving for \( y \) in terms of \( x \) to express it in a slope-intercept form \( y = b - \frac{b}{a}x \). This is often useful in optimization problems to have \( y \) explicitly written as a function of \( x \).
This simplification allows substitution directly into the distance formula, reducing calculations to only one variable and simplifying the process of finding minimum or maximum points. Understanding how to manipulate line equations is crucial in solving optimization problems in calculus, as it simplifies complex relations into workable formulas.