Problem 41
Question
The plane \(x+y+2 z=2\) intersects the paraboloid \(z=x^{2}+y^{2}\) in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.
Step-by-Step Solution
Verified Answer
The nearest and farthest points on the ellipse from the origin are found by solving the Lagrange system.
1Step 1: Set Up the Equations
First, we have the equation of the plane, \(x + y + 2z = 2\), and the paraboloid, \(z = x^2 + y^2\). To find the intersection, substitute \(z = x^2 + y^2\) into the plane equation: \[x + y + 2(x^2 + y^2) = 2\]
2Step 2: Formulate in Terms of Lagrange Multipliers
To find the points on the ellipse that are nearest to and farthest from the origin, use the distance squared function \(f(x, y, z) = x^2 + y^2 + z^2\). We'll apply the Lagrange multiplier method with the constraint \(g(x, y, z) = x + y + 2(x^2 + y^2) - 2 = 0\).
3Step 3: Find the Gradients
Compute the gradients: \[abla f = (2x, 2y, 2z), abla g = (1 + 4x, 1 + 4y, 2) \]The equation \(abla f = \lambda abla g\) gives the system: \(2x = \lambda (1 + 4x)\), \(2y = \lambda (1 + 4y)\), \(2z = 2\lambda\).
4Step 4: Solve the System of Equations
Solve the system for \(x, y, z, \lambda\). First, from \(2z = 2\lambda\), we have \(z = \lambda\). Using \(z = x^2 + y^2\), substitute \(\lambda = x^2 + y^2\) back into equations \(2x = \lambda (1 + 4x)\) and \(2y = \lambda (1 + 4y)\) to find possible \(x, y\) values.
5Step 5: Calculate Intersection Points
Using the constraints and solutions from Step 4, calculate the specific values of \(x, y, z\) that satisfy both the intersection and distance conditions. Substitute \(x, y,\) and the corresponding \(z = x^2 + y^2\) back into the plane equation to verify they satisfy the constraint.
6Step 6: Evaluate Distance Squared
Plug the potential solutions from the intersection into the distance squared function \(x^2 + y^2 + z^2\). Compare the values calculated to find the minimum and maximum distances.
Key Concepts
Lagrange MultipliersPlane and Paraboloid IntersectionDistance from Origin
Lagrange Multipliers
The method of Lagrange multipliers is a powerful technique in calculus used to find the local maxima and minima of a function subject to equality constraints. In this context, we want to find the points on the ellipse, formed by the intersection of a plane and a paraboloid, that are nearest to and farthest from the origin.
To do this, we consider the function that represents the square of the distance from a point i.e., \(f(x, y, z) = x^2 + y^2 + z^2\). This is our objective function.
However, we cannot optimize \(f(x, y, z)\) directly because of the intersection condition given by the constraint \(g(x, y, z) = x + y + 2(x^2 + y^2) - 2 = 0\). This is where Lagrange multipliers come into play.
We set up the Lagrangian function: \( \ L(x, y, z, \lambda) = x^2 + y^2 + z^2 - \lambda (x + y + 2(x^2 + y^2) - 2) \ \), where \(\lambda\) is the Lagrange multiplier.
The goal is to compute the gradients of both \(f\) and \(g\), set them equal with a multiplier \(\lambda\), and solve for the variables. This ultimately provides the points on the ellipse that satisfy the constraint while locating extreme distances from the origin.
To do this, we consider the function that represents the square of the distance from a point i.e., \(f(x, y, z) = x^2 + y^2 + z^2\). This is our objective function.
However, we cannot optimize \(f(x, y, z)\) directly because of the intersection condition given by the constraint \(g(x, y, z) = x + y + 2(x^2 + y^2) - 2 = 0\). This is where Lagrange multipliers come into play.
We set up the Lagrangian function: \( \ L(x, y, z, \lambda) = x^2 + y^2 + z^2 - \lambda (x + y + 2(x^2 + y^2) - 2) \ \), where \(\lambda\) is the Lagrange multiplier.
The goal is to compute the gradients of both \(f\) and \(g\), set them equal with a multiplier \(\lambda\), and solve for the variables. This ultimately provides the points on the ellipse that satisfy the constraint while locating extreme distances from the origin.
Plane and Paraboloid Intersection
To grasp the concept of a plane intersecting a paraboloid, let's break down these surfaces first.
The equation of our plane is \(x + y + 2z = 2\). Geometrically, this is a flat surface where any point on it satisfies the equation.
This plane cuts through a 3D space. The paraboloid, described by \(z = x^2 + y^2\), is like a 3D bowl where every point has height \(z\), determined by its distance from the origin in the \(xy\)-plane.
When these two surfaces intersect, the resulting curve is an ellipse. To find the equation of the intersection, we substitute \(z = x^2 + y^2\) into the plane equation:
\[ x + y + 2(x^2 + y^2) = 2\] This represents the ellipse in the \(xy\)-plane. When visualizing or solving any problem involving intersections like this, ensure that you substitute appropriately to consolidate the variables and obtain a clear equation that represents the intersection itself.
The equation of our plane is \(x + y + 2z = 2\). Geometrically, this is a flat surface where any point on it satisfies the equation.
This plane cuts through a 3D space. The paraboloid, described by \(z = x^2 + y^2\), is like a 3D bowl where every point has height \(z\), determined by its distance from the origin in the \(xy\)-plane.
When these two surfaces intersect, the resulting curve is an ellipse. To find the equation of the intersection, we substitute \(z = x^2 + y^2\) into the plane equation:
\[ x + y + 2(x^2 + y^2) = 2\] This represents the ellipse in the \(xy\)-plane. When visualizing or solving any problem involving intersections like this, ensure that you substitute appropriately to consolidate the variables and obtain a clear equation that represents the intersection itself.
Distance from Origin
Finding the distance from the origin is a classic problem in many geometric scenarios, often tackled using some form of distance formula.
In a three-dimensional space, the distance of a point \((x, y, z)\) from the origin \((0, 0, 0)\) is given by the formula:
\[\sqrt{x^2 + y^2 + z^2}\]In this exercise, however, we're more interested in the square of the distance, \(x^2 + y^2 + z^2\), than the distance itself.
This approach simplifies the computation since working with squares avoids dealing with square roots, which can complicate derivatives.
Using this squaring method allows us to apply Lagrange multipliers more effectively, which helps in finding the critical points on the ellipse where these distances are either minimized or maximized.
The solutions to the exercise show both the closest and farthest points on the ellipse from the origin by comparing this squared distance value.
In a three-dimensional space, the distance of a point \((x, y, z)\) from the origin \((0, 0, 0)\) is given by the formula:
\[\sqrt{x^2 + y^2 + z^2}\]In this exercise, however, we're more interested in the square of the distance, \(x^2 + y^2 + z^2\), than the distance itself.
This approach simplifies the computation since working with squares avoids dealing with square roots, which can complicate derivatives.
Using this squaring method allows us to apply Lagrange multipliers more effectively, which helps in finding the critical points on the ellipse where these distances are either minimized or maximized.
The solutions to the exercise show both the closest and farthest points on the ellipse from the origin by comparing this squared distance value.
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