Using Calculus to Find the Minimum Distance

Oct 13, 2025 by thesweetdesigns.com 21 views

Introduction

In mathematics, particularly in geometry and algebra, finding the closest point on a line to a given point is a fundamental problem. This problem has numerous applications in various fields, including physics, engineering, and computer science. In this article, we will focus on finding the point on the line $\frac{x}{a} + \frac{y}{b} = 9$ that is closest to the origin.

Understanding the Problem

The problem requires finding the point on the line $\frac{x}{a} + \frac{y}{b} = 9$ that is closest to the origin. The origin is the point (0, 0) in a coordinate system. To find the closest point, we need to minimize the distance between the point on the line and the origin.

Mathematical Formulation

Let's denote the point on the line as (x, y). The equation of the line is given by $\frac{x}{a} + \frac{y}{b} = 9$ . We want to find the point (x, y) that minimizes the distance to the origin.

The distance between two points (x1, y1) and (x2, y2) is given by the formula:

d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}

In this case, the distance between the point (x, y) and the origin (0, 0) is:

d = \sqrt{x^2 + y^2}

We want to minimize this distance.

Using Calculus to Find the Minimum Distance

To find the minimum distance, we can use calculus. We will use the concept of partial derivatives to find the critical points of the distance function.

Let's define the distance function as:

d(x, y) = \sqrt{x^2 + y^2}

We want to find the critical points of this function, which are the points where the partial derivatives of the function are equal to zero.

The partial derivatives of the distance function are:

\frac{\partial d}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}

\frac{\partial d}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}

We set these partial derivatives equal to zero and solve for x and y:

\frac{x}{\sqrt{x^2 + y^2}} = 0

\frac{y}{\sqrt{x^2 + y^2}} = 0

Solving these equations, we get:

x = 0

y = 0

However, this is not the point on the line that we are looking for. We need to find the point on the line that minimizes the distance to the origin.

Using the Equation of the Line

We can use the equation of the line to find the point on the line that minimizes the distance to the origin.

The equation of the line is:

\frac{x}{a} + \frac{y}{b} = 9

We can rewrite this equation as:

y = \frac{9ab^2}{a^2 + b^2}

Substituting this expression for y into the distance function, we get:

d(x, y) = \sqrt{x^2 + \left(\frac{9ab^2}{a^2 + b^2}\right)^2}

We want to minimize this distance.

Minimizing the Distance

To minimize the distance, we can use the concept of Lagrange multipliers. We introduce a new variable λ and form the Lagrangian function:

L(x, y, λ) = \sqrt{x^2 + \left(\frac{9ab^2}{a^2 + b^2}\right)^2} - λ\left(\frac{x}{a} + \frac{y}{b} - 9\right)

We take the partial derivatives of the Lagrangian function with respect to x, y, and λ, and set them equal to zero:

\frac{\partial L}{\partial x} = \frac{x}{\sqrt{x^2 + \left(\frac{9ab^2}{a^2 + b^2}\right)^2}} - \frac{λ}{a} = 0

\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2 + \left(\frac{9ab^2}{a^2 + b^2}\right)^2}} - \frac{λ}{b} = 0

\frac{\partial L}{\partial λ} = \frac{x}{a} + \frac{y}{b} - 9 = 0

Solving these equations, we get:

x = \frac{9a^2b}{a^2 + b^2}

y = \frac{9ab^2}{a^2 + b^2}

This is the point on the line that minimizes the distance to the origin.

Conclusion

In this article, we have found the point on the line $\frac{x}{a} + \frac{y}{b} = 9$ that is closest to the origin. The point is given by:

\left(\frac{9a^2b}{a^2 + b^2}, \frac{9ab^2}{a^2 + b^2}\right)

This result has numerous applications in various fields, including physics, engineering, and computer science.

References

[1] "Calculus" by Michael Spivak
[2] "Linear Algebra and Its Applications" by Gilbert Strang
[3] "Geometry: A Comprehensive Introduction" by Jeffrey R. Chasnov
Q&A: Finding the Closest Point on a Line to the Origin =====================================================

Q: What is the problem of finding the closest point on a line to the origin?

A: The problem of finding the closest point on a line to the origin is a fundamental problem in mathematics, particularly in geometry and algebra. It involves finding the point on a given line that is closest to the origin (0, 0) in a coordinate system.

Q: What is the equation of the line in this problem?

A: The equation of the line is given by $\frac{x}{a} + \frac{y}{b} = 9$ . This is a linear equation in two variables, x and y.

Q: How do we find the point on the line that minimizes the distance to the origin?

A: To find the point on the line that minimizes the distance to the origin, we can use calculus. We introduce a new variable λ and form the Lagrangian function. We then take the partial derivatives of the Lagrangian function with respect to x, y, and λ, and set them equal to zero. Solving these equations, we get the point on the line that minimizes the distance to the origin.

Q: What is the point on the line that minimizes the distance to the origin?

A: The point on the line that minimizes the distance to the origin is given by:

\left(\frac{9a^2b}{a^2 + b^2}, \frac{9ab^2}{a^2 + b^2}\right)

Q: What are the applications of this result?

A: This result has numerous applications in various fields, including physics, engineering, and computer science. For example, it can be used to find the closest point on a line to a given point in a coordinate system, which is useful in computer graphics and game development.

Q: How do we use the equation of the line to find the point on the line that minimizes the distance to the origin?

A: We can use the equation of the line to find the point on the line that minimizes the distance to the origin by substituting the expression for y into the distance function. We then minimize the distance function to find the point on the line that minimizes the distance to the origin.

Q: What is the concept of Lagrange multipliers?

A: The concept of Lagrange multipliers is a technique used in calculus to find the maximum or minimum of a function subject to a constraint. In this problem, we use Lagrange multipliers to find the point on the line that minimizes the distance to the origin.

Q: How do we take the partial derivatives of the Lagrangian function?

A: We take the partial derivatives of the Lagrangian function with respect to x, y, and λ, and set them equal to zero. Solving these equations, we get the point on the line that minimizes the distance to the origin.

Q: What are the steps to find the point on the line that minimizes the distance to the origin?

A: The steps to find the point on the line that minimizes the distance to the origin are:

Write down the equation of the line.
Substitute the expression for y into the distance function.
Minimize the distance function to find the point on the line that minimizes the distance to the origin.
Use Lagrange multipliers to find the point on the line that minimizes the distance to the origin.

Q: What are the benefits of finding the closest point on a line to the origin?

A: The benefits of finding the closest point on a line to the origin include:

Finding the closest point on a line to a given point in a coordinate system.
Solving problems in physics, engineering, and computer science.
Understanding the concept of Lagrange multipliers.

Q: What are the limitations of finding the closest point on a line to the origin?

A: The limitations of finding the closest point on a line to the origin include:

The equation of the line must be linear.
The distance function must be minimized.
The point on the line that minimizes the distance to the origin must be found using Lagrange multipliers.