Problem 3 - A First Course in Mathematical Modeling

Question

Use the method of Lagrange multipliers to solve Problems. Find the minimum distance from the surface \(x^{2}+y^{2}-z^{2}=1\) to the origin.

Step-by-Step Solution

Answer

The minimum distance is 1.

Step 1: Define the Objective Function

To find the minimum distance from a point on the surface to the origin, we need to minimize the distance function. The distance of a point \(x, y, z\) from the origin is given by \((x^2 + y^2 + z^2)^{1/2}\). However, to simplify the calculus, we minimize the square of the distance:\[ g(x, y, z) = x^2 + y^2 + z^2 \]

Step 2: Define the Constraint

The constraint we are given is the surface equation:\[ h(x, y, z) = x^2 + y^2 - z^2 - 1 = 0 \]

Step 3: Set Up the Lagrange Function

The Lagrange function combines the objective function and the constraint using a Lagrange multiplier, \( \lambda\):\[ \mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda (x^2 + y^2 - z^2 - 1) \]

Step 4: Find the Partial Derivatives

Calculate the partial derivatives of \mathcal{L}\ with respect to \(x, y, z, ext{ and } \lambda\):\[\text{\(\dfrac{\partial \mathcal{L}}{\partial x}\): } 2x + 2x\lambda = 0\]\[\text{\(\dfrac{\partial \mathcal{L}}{\partial y}\): } 2y + 2y\lambda = 0\]\[\text{\(\dfrac{\partial \mathcal{L}}{\partial z}\): } 2z - 2z\lambda = 0\]\[\text{\(\dfrac{\partial \mathcal{L}}{\partial \lambda}\): } x^2 + y^2 - z^2 - 1 = 0\]

Step 5: Solve the Equations System

From the partial derivatives, solve for the variables and \lambda\:- From \(\dfrac{\partial \mathcal{L}}{\partial x} = 0\): \ \Rightarrow x (1 + \lambda) = 0 \Rightarrow x = 0 \text{ unless } \lambda = -1.- From \(\dfrac{\partial \mathcal{L}}{\partial y} = 0\): \ \Rightarrow y (1 + \lambda) = 0 \Rightarrow y = 0 \text{ unless } \lambda = -1.- From \(\dfrac{\partial \mathcal{L}}{\partial z} = 0\): \ \Rightarrow z (1 - \lambda) = 0 \Rightarrow z = 0 \text{ unless } \lambda = 1.- Use the constraint \(x^2 + y^2 - z^2 = 1\).By checking combinations, when \(x = y = 0\): \z^2 = 1 \Rightarrow z = \pm 1.

Step 6: Verify and Compute Minimum Distance

Since \lambda\ = -1 \text{ and } \lambda\ = 1\ cannot simultaneously satisfy all equations except with trivial zero solution not on contour, choices: \x = y = 0, z = \pm 1\ are valid. Compute distance for these points:\[ \text{Distance: } \sqrt{x^2 + y^2 + z^2} = \sqrt{0 + 0 + 1} = 1 \]Thus, the minimum distance is 1.

Key Concepts

Calculus

Calculus plays a crucial role in many advanced mathematical concepts, including optimization and finding minimum and maximum values of functions. Here, we're focusing on Lagrange multipliers, a technique involving calculus that helps solve problems with constraints. For our problem, it involves finding the minimum distance from a surface to a point.
Generally, calculus helps us understand how functions change and the principles behind these changes. It does so by:

• Derivatives: These show the rate of change of a function, giving insights into its behavior and slope.
• Integrals: Contrary to derivatives, integrals calculate area under curves, helping us understand accumulation.

In the specific context of this exercise, we efficiently use derivatives to find out the necessary conditions at which a function reaches its extremum, given constraints. Understanding derivatives in particular is vital as they help compute the gradient, a vector indicating the direction of steepest ascent of a function.

Constraint Optimization

Constraint optimization is a powerful mathematical tool used when a solution must satisfy certain limitations or conditions. Here, we used Lagrange multipliers to achieve this, merging calculus techniques with sophisticated algebraic insights.
When dealing with constraint optimization, we're often interested in maximizing or minimizing a function, knowing that the solution lies on or within a surface defined by another function. For our problem:

• Objective Function: The distance from a point to the origin squared. Not just minimizing this function directly helps because it simplifies our derivatives and equations.
• Constraint: The surface equation, which is the condition that our points must satisfy.

By constructing the Lagrange function, we bring together these parts into a single equation, which cleverly incorporates both the objective function and the constraint using a new variable, the Lagrange multiplier, \( \lambda \). This addition helps convert the difficult problem of minimizing a function subject to a constraint into solving a set of partial derivative equations.

Distance Minimization

Distance minimization is all about finding the closest point on a curve or surface to a given point, like the origin. This is often critical in physics and engineering, where determining shortest paths or minimum energy configurations makes processes efficient.
In this particular problem:

• We started with the notion that the shortest distance from the origin to any point \((x, y, z)\) is given by \( \sqrt{x^2 + y^2 + z^2} \).
• To simplify calculus, instead of minimizing the square root expression, we minimized its square, \( x^2 + y^2 + z^2 \), effectively giving the same point.

Finally, after both defining the problem properly and solving the partial derivatives using the constraint, we found that the minimum distance from the origin to the surface is 1. This is because the point \((0, 0, \pm1)\) on the surface is closest to the origin, satisfying our constraint \( x^2 + y^2 - z^2 = 1 \). This reveals the beauty of distance minimization techniques: they provide a straightforward solution to a seemingly complex geometric problem.