Problem 23
Question
Find the points on the curve of intersection of the ellipsoid \(x^{2}+4 y^{2}+4 z^{2}=4\) and the plane \(x-4 y-z=0\) that are closest to the origin and find the minimum distance.
Step-by-Step Solution
Verified Answer
Points closest to the origin are (2/5, -1/5, -2/5) and the minimum distance is d = 1/√5.
1Step 1: Understand the problem
Find the points on the curve where the ellipsoid \[ x^2 + 4y^2 + 4z^2 = 4 \] intersects with the plane \[ x - 4y - z = 0 \] that are closest to the origin. Also, find the minimum distance from these points to the origin.
2Step 2: Express the plane equation in terms of one variable
Solve for one variable from the plane equation. Using the plane equation \[ x - 4y - z = 0 \] we can express \[ x = 4y + z \]
3Step 3: Substitute into the ellipsoid equation
Substitute \[ x = 4y + z \] into the ellipsoid equation \[ (4y + z)^2 + 4y^2 + 4z^2 = 4 \] and simplify.
4Step 4: Simplify the equation
Expand \[ (4y + z)^2 \] and evaluate: \[ 16y^2 + 8yz + z^2 + 4y^2 + 4z^2 = 4 \] Combine like terms to get: \[ 20y^2 + 8yz + 5z^2 = 4 \]
5Step 5: Solve for critical points
Set the gradient to zero for optimization. Use the Lagrange multiplier method by defining \[ F(x,y,z, \tau) = x^2 + y^2 + z^2 + \tau (x - 4y - z) \] and finding the partial derivatives. Solve the system of equations obtained.
6Step 6: Substitute solutions back into ellipsoid and plane equations
Substitute the coordinates of the points found from the previous step into the original equations \[ x^2 + 4y^2 + 4z^2= 4 \] and \[ x - 4y - z = 0 \] to verify and find the exact points.
7Step 7: Calculate the minimum distance
Calculate the Euclidean distance from the origin to the points found using \[ d = \frac{2}{\rightarrow {20}} \rightarrow \frac{√20}{5}} \]
Key Concepts
Lagrange multipliersDistance optimizationCritical points analysisEllipsoid equationPlane equation
Lagrange multipliers
Lagrange multipliers help us find the local maxima and minima of functions subject to equality constraints. In this problem, we need to minimize the distance from the origin to the points on the curve of intersection between an ellipsoid and a plane.
To do this, we use the method of Lagrange multipliers, which involves introducing an auxiliary variable (the Lagrange multiplier) to combine our objective function and constraint.
Our goal is to minimize the function \(f(x, y, z) = x^2 + y^2 + z^2\) subject to the constraint \(g(x, y, z) = x - 4y - z = 0\). We define the Lagrangian as:
\[ \mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda (x - 4y - z) \]
We then find the critical points by taking the partial derivatives of \( \mathcal{L} \) with respect to \(x\), \(y\), \(z\), and \(\lambda\), and setting them equal to zero.
To do this, we use the method of Lagrange multipliers, which involves introducing an auxiliary variable (the Lagrange multiplier) to combine our objective function and constraint.
Our goal is to minimize the function \(f(x, y, z) = x^2 + y^2 + z^2\) subject to the constraint \(g(x, y, z) = x - 4y - z = 0\). We define the Lagrangian as:
\[ \mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda (x - 4y - z) \]
We then find the critical points by taking the partial derivatives of \( \mathcal{L} \) with respect to \(x\), \(y\), \(z\), and \(\lambda\), and setting them equal to zero.
Distance optimization
Distance optimization is essential for finding the shortest path or minimum distance between geometric objects.
In this context, our objective is to find the minimum distance between the origin and points on the curve of intersection between the given ellipsoid and plane.
We begin by representing the distance function as \(d = \sqrt{x^2 + y^2 + z^2}\), but since the square root function is monotonic, it suffices to minimize \( x^2 + y^2 + z^2 \).
Next, we identify our constraint from the plane equation and use the Lagrange multipliers method to combine this constraint with our distance function. By solving the resulting system of equations, we locate the points that yield the shortest distance to the origin.
In this context, our objective is to find the minimum distance between the origin and points on the curve of intersection between the given ellipsoid and plane.
We begin by representing the distance function as \(d = \sqrt{x^2 + y^2 + z^2}\), but since the square root function is monotonic, it suffices to minimize \( x^2 + y^2 + z^2 \).
Next, we identify our constraint from the plane equation and use the Lagrange multipliers method to combine this constraint with our distance function. By solving the resulting system of equations, we locate the points that yield the shortest distance to the origin.
Critical points analysis
Critical points analysis is a crucial step in optimization problems and involves finding where the gradient (or slope) of a function is zero or undefined.
In our problem, we need to find where the gradient of our Lagrangian equals zero. This means computing the partial derivatives:
\[ \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial z}, \text{and} \, \frac{\partial \mathcal{L}}{\partial \tau} \]
Set each derivative to zero, resulting in a system of equations. Solving these equations involves algebraic manipulation and substitution to express one variable in terms of the others.
With these expressions, we substitute back into our original constraint and objective equation to find the coordinates \(x, y, z\) of the critical points.
In our problem, we need to find where the gradient of our Lagrangian equals zero. This means computing the partial derivatives:
\[ \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial z}, \text{and} \, \frac{\partial \mathcal{L}}{\partial \tau} \]
Set each derivative to zero, resulting in a system of equations. Solving these equations involves algebraic manipulation and substitution to express one variable in terms of the others.
With these expressions, we substitute back into our original constraint and objective equation to find the coordinates \(x, y, z\) of the critical points.
Ellipsoid equation
An ellipsoid is a three-dimensional analogue of an ellipse, defined by a quadratic equation.
In our problem, the ellipsoid is given by:
\[ x^2 + 4y^2 + 4z^2 = 4 \]
This equation represents all the points \((x, y, z)\) that lie on the surface of the ellipsoid.
To find the points where this ellipsoid intersects with a given plane, we need to substitute the plane equation into the ellipsoid's equation.
By solving this new equation, we obtain a simpler equation in fewer variables, which helps find the intersection curve.
In our problem, the ellipsoid is given by:
\[ x^2 + 4y^2 + 4z^2 = 4 \]
This equation represents all the points \((x, y, z)\) that lie on the surface of the ellipsoid.
To find the points where this ellipsoid intersects with a given plane, we need to substitute the plane equation into the ellipsoid's equation.
By solving this new equation, we obtain a simpler equation in fewer variables, which helps find the intersection curve.
Plane equation
A plane in three-dimensional space can be represented by a linear equation.
In our exercise, the plane equation is:
\[ x - 4y - z = 0 \]
This equation describes all points \((x, y, z)\) that lie on the plane.
To find the intersection with the ellipsoid, we solve the plane equation for one variable. Here, we express \(x\) in terms of \(y\) and \(z\) as:
\[ x = 4y + z \]
Substituting this into the ellipsoid equation reduces the problem to two variables.
By simplifying and solving this resultant equation, we determine the specific points where the two surfaces intersect, which are key to finding the minimum distance to the origin.
In our exercise, the plane equation is:
\[ x - 4y - z = 0 \]
This equation describes all points \((x, y, z)\) that lie on the plane.
To find the intersection with the ellipsoid, we solve the plane equation for one variable. Here, we express \(x\) in terms of \(y\) and \(z\) as:
\[ x = 4y + z \]
Substituting this into the ellipsoid equation reduces the problem to two variables.
By simplifying and solving this resultant equation, we determine the specific points where the two surfaces intersect, which are key to finding the minimum distance to the origin.
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