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Magazine Chapter 10: Problem 5
Find the minimum distance from the point (1,1,11) to the paraboloid given by the equation \(z=x^{2}+y^{2}\). Minimum distance = ___________.
Short Answer
Expert verified
The minimum distance between the point (1, 1, 11) and the paraboloid \(z = x^2 + y^2\) is approximately \(10.598\).
Step by step solution
01
Write down the given information
We are given a point \(P(1, 1, 11)\) and a paraboloid defined by the equation \(z = x^2 + y^2\).
02
Write down the distance formula in 3D
The distance formula in 3D, between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), is given by
\[
d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}
\]
03
Plug in the given point and the paraboloid equation into the distance formula
Since we are finding the distance from point \(P(1, 1, 11)\) to the paraboloid \(z = x^2 + y^2\), we can set \((x_1, y_1, z_1) = (1, 1, 11)\) and \((x_2, y_2, z_2) = (x, y, x^2 + y^2)\) in the distance formula:
\[
d = \sqrt{(1 - x)^2 + (1 - y)^2 + (11 - (x^2 + y^2))^2}
\]
04
Minimize the square of the distance function
To simplify the problem, we can minimize the square of the distance function, because minimizing the square of a function will also minimize the function itself:
\[
d^2 = (1 - x)^2 + (1 - y)^2 + (11 - (x^2 + y^2))^2
\]
05
Find the partial derivatives of the distance squared function
Calculate the partial derivatives of the distance squared function with respect to x and y:
\[
\frac{\partial d^2}{\partial x} = -2(1-x) + 4x(11-(x^2+y^2))
\]
\[
\frac{\partial d^2}{\partial y} = -2(1-y) + 4y(11-(x^2+y^2))
\]
06
Set the partial derivatives to zero and solve the system of equations
Set the partial derivatives to zero to find the critical points of the distance squared function:
\[
\begin{cases}
-2(1-x) + 4x(11-(x^2+y^2)) = 0 \\
-2(1-y) + 4y(11-(x^2+y^2)) = 0
\end{cases}
\]
Solving this system of equations, we get \(x = \frac{1}{3}\) and \(y = \frac{1}{3}\).
07
Find the minimum distance by plugging the critical points into the distance formula
Plug the critical points (\(x = \frac{1}{3}\) and \(y = \frac{1}{3}\)) back into the distance formula:
\[
d = \sqrt{(1 - \frac{1}{3})^2 + (1 - \frac{1}{3})^2 + (11 - ((\frac{1}{3})^2 + (\frac{1}{3})^2))^2}
\]
Evaluating this expression, we get the minimum distance \(d \approx 10.598 \).
08
Final Answer
The minimum distance between the point (1, 1, 11) and the paraboloid \(z = x^2 + y^2\) is approximately \(10.598\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance Formula in 3D
When tackling geometric problems in three dimensions, an essential tool is the distance formula in 3D. This formula allows us to compute the distance between two points in a three-dimensional space. Specifically, given two points, \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\), the distance \(d\) between them is defined by the following equation:
\[d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}\]
This formula is a direct extension of Pythagoras' theorem into three dimensions. When measuring the distance to a surface, such as a paraboloid, you consider a point on the surface and the given point, computing the distance between them by plugging their coordinates into the distance formula.
To optimize this calculation in the context of finding a minimum distance, we may only need to consider certain aspects or simplify the problem, such as by minimizing the square of the distance instead for easier computation, which leads us into optimization techniques in multivariable calculus.
\[d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}\]
This formula is a direct extension of Pythagoras' theorem into three dimensions. When measuring the distance to a surface, such as a paraboloid, you consider a point on the surface and the given point, computing the distance between them by plugging their coordinates into the distance formula.
To optimize this calculation in the context of finding a minimum distance, we may only need to consider certain aspects or simplify the problem, such as by minimizing the square of the distance instead for easier computation, which leads us into optimization techniques in multivariable calculus.
Partial Derivatives
In multivariable calculus, we often deal with functions of several variables, and partial derivatives are the tools we use to find how a function changes with respect to each variable independently. The partial derivative of a function \(f(x, y)\) with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and represents the rate of change of \(f\) with respect to \(x\) while keeping \(y\) constant.
In practical applications, such as finding the minimum distance to a surface, partial derivatives help us understand the behavior of the distance function with respect to each individual coordinate. By taking the partial derivative of the squared distance function with respect to \(x\) and \(y\), we set the stage for finding the critical points where the distance might be at a minimum or maximum.
In practical applications, such as finding the minimum distance to a surface, partial derivatives help us understand the behavior of the distance function with respect to each individual coordinate. By taking the partial derivative of the squared distance function with respect to \(x\) and \(y\), we set the stage for finding the critical points where the distance might be at a minimum or maximum.
Optimization in Multivariable Calculus
The process of finding either the maximum or minimum values of a function falls under the umbrella of optimization in multivariable calculus. When the function involves multiple variables, the solution often involves finding points where all partial derivatives of the function are zero; these points are called critical points. Once the partial derivatives of the function are set to zero, we can solve the resulting system of equations to find these critical points.
In the context of finding the minimum distance between a point and a paraboloid, the distance function involves variables \(x\) and \(y\) of the point on the paraboloid. By setting the partial derivatives with respect to \(x\) and \(y\) equal to zero, we locate the critical points which are potential candidates for the minimum distance. Additional checks, like the second derivative test, can confirm the nature of these critical points, ensuring they correspond to a minimum rather than a maximum.
In the context of finding the minimum distance between a point and a paraboloid, the distance function involves variables \(x\) and \(y\) of the point on the paraboloid. By setting the partial derivatives with respect to \(x\) and \(y\) equal to zero, we locate the critical points which are potential candidates for the minimum distance. Additional checks, like the second derivative test, can confirm the nature of these critical points, ensuring they correspond to a minimum rather than a maximum.
