Problem 31 - Calculus an Applied Approach

Question

Find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.) $$ \begin{array}{l}{\text { Plane: } x+y+z=1,(2,1,1)} \\ {\text { Minimize } d^{2}=(x-2)^{2}+(y-1)^{2}+(z-1)^{2}}\end{array} $$

Step-by-Step Solution

Answer

The minimum distance from the point (2, 1, 1) to the plane \(x + y + z = 1\) is \(\sqrt{2}\).

Step 1: Simplify the function

Using the equation of the plane \(x + y + z = 1\), we can express z as \(z = 1 - x - y\). We substitute this into the equation for the squared distance, to get a function of only two variables, \(x\) and \(y\). The resulting function is: \(d^{2} = (x-2)^{2} + (y-1)^{2} + ((1-x-y)-1)^{2}\).

Step 2: Find the minimum

To find the minimum of this function, compute the partial derivatives in respect to \(x\) and \(y\), set them to zero, and solve for \(x\) and \(y\). The partial derivatives are: \(\frac{\partial d^2}{\partial x} = 2x - 4 - 2(1 - x - y) = 2x + 2y - 2\) and \(\frac{\partial d^2}{\partial y} = 2y - 2 - 2(1 - x - y) = 2x + 2y - 2\). By setting these to zero, we get two equations: \(2x + 2y - 2 = 0\) and \(2x + 2y - 2 = 0\). Solving for \(x\) and \(y\) gives \(x = 1, y = 0\).

Step 3: Find the minimum distance

The minimum distance is the square root of the minimum squared distance. By substituting \(x = 1\) and \(y = 0\) into the function \(d^{2}\), we get \(d^{2} = (1-2)^{2} + (0-1)^{2} + ((1-1-0)-1)^{2} = 1 + 1 + 0 = 2\). Therefore, the minimum distance is \(\sqrt{2}\).

Key Concepts

Calculus Optimization

Calculus optimization is a critical process in mathematics where one seeks to find the maximum or minimum value of a function. The goal is to identify the points at which a function reaches its extreme values, which can be essential in various practical applications such as engineering and economics.

Optimization problems often involve functions with multiple variables and constraints. The objective is to find the optimal values of these variables within a given domain. To achieve this, one must understand how to set up the problem, identify constraints, and apply appropriate calculus methods like finding derivatives, which give insights into the behavior of the function around its critical points.

For example, when seeking to minimize the distance between a point and a surface, you’d start by defining the distance squared between the surfaces as a function of the relevant variables. Then, by computing the derivatives, you can find those variables that make this function as small as possible, thus solving the optimization problem.

Partial Derivatives

Partial derivatives play a pivotal role in multivariable calculus, particularly when dealing with functions of more than one variable. They measure the rate of change of a function with respect to one of the variables, while keeping the other variables constant. These are crucial for finding the extremum points of multivariable functions.

When computing the partial derivative of a function with respect to a given variable, we treat all other variables as constants and differentiate as we would in single-variable calculus. The notation \( \frac{\partial f}{\partial x} \) denotes the partial derivative of a function \( f \) with respect to the variable \( x \).

In the context of optimization problems, setting the partial derivatives equal to zero gives us a system of equations. When solved simultaneously, these equations can lead to finding the critical points where minimum or maximum values are located, assuming they exist and are within the domain under consideration.

Distance Formula

The distance formula is a fundamental tool in various branches of mathematics and physics. It derives from the Pythagorean theorem and provides a way to calculate the distance between two points in any dimensional space. For instance, the distance \( d \) between the two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in a 2D space is given by \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

In the context of optimization problems, rather than directly minimizing the distance, it is often more convenient to minimize the square of the distance—this simplifies calculations and avoids the complexity of square roots. The extremum points obtained this way will be the same as those obtained by minimizing the actual distance since the squaring function is strictly increasing for positive arguments.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. This branch of mathematics is particularly concerned with functions mapping vectors to real numbers, and it deals with topics such as partial derivatives, gradient, divergence, and multiple integrals. In optimization problems in multivariable calculus, the goal is to find points at which multivariable functions reach their highest or lowest values given certain constraints.

In the specific context of minimizing the distance from a point to a surface, the function representing the distance squared from a given point to the surface will typically involve multiple variables corresponding to the coordinates in space. Multivariable calculus provides the tools necessary to solve for the minimum by taking partial derivatives with respect to each of these variables and setting them to zero to find critical points which will lead to solutions for the optimization problem.