Question

Find the shortest distance from the origin to cach of the following quadric surfaces.\(2 z^{2}+6 x y=3\)

Short answer

\( d = \sqrt{3/2} \)

Step-by-step solution

Identify the objective

The goal is to find the shortest distance from the origin (0,0,0) to the quadric surface defined by the equation \(2z^2 + 6xy = 3\).

Set up the distance formula

The distance from any point \((x, y, z)\) to the origin \((0, 0, 0)\) is given by the formula: \(d = \sqrt{x^2 + y^2 + z^2}\).

Formulate the optimization problem

To find the minimum distance, we need to minimize the function \(d = \sqrt{x^2 + y^2 + z^2}\) subject to the constraint \(2z^2 + 6xy - 3 = 0\).

Use the method of Lagrange multipliers

The method of Lagrange multipliers introduces a new variable (the Lagrange multiplier \(\lambda\)) and considers the Lagrangian: \(L(x, y, z, \lambda) = x^2 + y^2 + z^2 - \lambda(2z^2 + 6xy - 3)\).

Compute partial derivatives

Compute the partial derivatives of the Lagrangian with respect to each variable and set them equal to zero: \(abla L = (\frac{\partial L}{\partial x}, \frac{\partial L}{\partial y}, \frac{\partial L}{\partial z}, \frac{\partial L}{\partial \lambda}) = 0\).

Solve for extrema

Solving the system of equations derived from the partial derivatives will provide the critical points. These include the following equations: \[ \begin{cases} \frac{\partial L}{\partial x} = 2x - 6 y \lambda = 0 \ \frac{\partial L}{\partial y} = 2y - 6 x \lambda = 0 \ \frac{\partial L}{\partial z} = 2z - 4 z \lambda = 0 \ \frac{\partial L}{\partial \lambda} = - (2z^2 + 6xy - 3) = 0 \end{cases} \]

Find critical points

From \(2x - 6 y \lambda = 0\), we get \(x = 3 y \lambda / 2\). From \(2y - 6 x \lambda = 0\), we get \(y = 3 x \lambda / 2\). Substituting these back into the equations and solving the system of equations, we find \( x = y = 0, z^2 = 3/2\).

Verify and find minimum distance

The minimum distance should satisfy the constraint. The critical point that satisfies the constraint \(x = y = 0\) and \(z = \pm \sqrt{3/2}\) gives: \(d = \sqrt{0^2 + 0^2 + (\sqrt{3/2})^2}\) \(d = \sqrt{3/2}\).

Key concepts

Lagrange multipliers

Lagrange multipliers are a powerful tool in the field of optimization, especially when dealing with problems that have constraints. They allow us to find the extrema of a function subject to constraints. The basic idea is to convert the constrained problem into an unconstrained one by introducing an extra variable, called the Lagrange multiplier, denoted as \( \lambda \). This variable helps incorporate the constraint into the function we seek to optimize. The method works by setting up a new function called the Lagrangian, which combines the original function and the constraint multiplied by \( \lambda \).
For instance, in our problem, we need to find the shortest distance to the quadric surface \(2z^2 + 6xy = 3\). We define a Lagrangian \(L(x, y, z, \lambda) = x^2 + y^2 + z^2 - \lambda(2z^2 + 6xy - 3) \). This equation implicitly includes the constraint, making it possible to find the solution by solving for the critical points where the partial derivatives of L with respect to each variable (x, y, z) and \( \lambda \) are zero.

Optimization problems in mathematics

Optimization problems in mathematics deal with finding the maximum or minimum values of a function. These types of problems are prevalent in various fields, including economics, engineering, and physics. Optimization often involves functions that depend on multiple variables and come with certain constraints.
For our problem, the goal is to minimize the distance from the origin to the given quadric surface, subject to the equation \(2z^2 + 6xy = 3\). To do this, we formulated an optimization problem where we aim to minimize the function \(d = \sqrt{x^2 + y^2 + z^2}\) under the given constraint. By solving this, we determine the critical points and find the point minimizing the distance.
Using Lagrange multipliers simplifies solving such constrained optimization problems by reducing them into a system of equations we can solve algebraically.

Distance formula in three dimensions

The distance formula in three dimensions is used to calculate the distance between two points in 3D space. For any two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), the distance \(d\) between them is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
When the problem involves finding the distance from the origin \((0, 0, 0)\) to a point \((x, y, z)\), the formula simplifies to:
\[ d = \sqrt{x^2 + y^2 + z^2} \]
In our case, this formula is crucial for determining the shortest distance between the origin and the quadric surface \(2z^2 + 6xy = 3\). We aim to minimize this distance under the given constraint, leading us to the distance minimizing point (x, y, z).

Quadric surfaces in mathematics

Quadric surfaces are a type of surface in three-dimensional space defined by a second-degree polynomial equation in three variables (x, y, and z). These surfaces encompass many familiar shapes, including ellipsoids, hyperboloids, and paraboloids.
In our exercise, the quadric surface is given by \(2z^2 + 6xy = 3\). Understanding the nature of this surface helps in formulating and solving the optimization problem. Each type of quadric surface has unique properties that influence how we approach problems involving distances and optimizations.
Quadric surfaces often appear in physics and engineering problems, such as in the study of optical systems, stress analysis in materials, and in the analysis of potential energy fields.