Problem 33 - Calculus for Biology and Medicine

Question

The distance between the origin \((0,0,0)\) and the point \((x, y, z)\) is $$ \sqrt{x^{2}+y^{2}+z^{2}} $$ Find the minimum distance between the origin and the plane \(x+y+z=1 .\) (Hint: Minimize the squared distance between the origin and the plane.)

Step-by-Step Solution

Answer

The minimum distance is \(\sqrt{\frac{1}{3}}\).

Step 1: Represent the Problem

We need to find the minimum distance from the origin \((0, 0, 0)\) to the plane given by the equation \(x + y + z = 1\). The distance from any point \((x, y, z)\) to the origin is given by the formula: \(\sqrt{x^2 + y^2 + z^2}\). Instead of minimizing this distance directly, we will minimize its square, \(x^2 + y^2 + z^2\), for simplicity.

Step 2: Express the Constraint

The point \((x, y, z)\) must lie on the plane, which provides the constraint \(x + y + z = 1\).

Step 3: Use the Method of Lagrange Multipliers

To minimize \(x^2 + y^2 + z^2\) subject to the constraint \(x + y + z = 1\), we use the method of Lagrange Multipliers. Define the Lagrangian: \(\mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda (1 - (x + y + z))\). Here, \(\lambda\) is the Lagrange multiplier.

Step 4: Compute Partial Derivatives

Calculate the partial derivatives of \(\mathcal{L}\) with respect to \(x, y, z,\) and \(\lambda\), then set each derivative to zero:- \(\frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0\)- \(\frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda = 0\)- \(\frac{\partial \mathcal{L}}{\partial z} = 2z - \lambda = 0\)- \(\frac{\partial \mathcal{L}}{\partial \lambda} = 1 - (x + y + z) = 0\)

Step 5: Solve the System of Equations

From \(2x = \lambda\), \(2y = \lambda\), and \(2z = \lambda\), we have \(2x = 2y = 2z = \lambda\). Thus, \(x = y = z\). Substitute \(x = y = z\) into the constraint \(x + y + z = 1\) to get:\(3x = 1\), thus, \(x = \frac{1}{3}\), \(y = \frac{1}{3}\), and \(z = \frac{1}{3}\).

Step 6: Calculate the Minimum Squared Distance

Now find the squared distance using the point \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\) that lies on the plane:\[x^2 + y^2 + z^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3}.\]

Step 7: Take the Square Root for the Minimum Distance

The actual minimum distance from the origin to the plane is the square root of the squared distance:\[\sqrt{\frac{1}{3}}.\]

Key Concepts

Distance from a Point to a Plane

When you want to find the distance from a point to a plane, it's like finding out how far away something is from a flat surface. You can imagine standing at a specific point and wondering how far you are from a really big wall that goes on forever.
The standard way to calculate this distance involves using a formula. For a point \((x_0, y_0, z_0)\) and a plane represented by the equation \(Ax + By + Cz + D = 0\), the distance \(d\) from the point to the plane is given by:

• \[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]

In our original exercise, although the formula isn't directly used, understanding this concept builds the foundation for more complex problems.
The minimized \( \sqrt{x^2 + y^2 + z^2} \) actually represents the line closest from the origin to the plane.

Minimization Problem

Minimization is an important concept in calculus and optimization, where the goal is to find the minimum value of a function under certain constraints.
In the original exercise, we were tasked with finding the smallest possible distance between the origin and a plane. Instead of working directly with the square root expression for distance, we minimized the squared distance. This is a common technique in math because squaring a number preserves order (larger numbers remain larger) and gets rid of square roots which can be cumbersome.

• Instead of minimizing \(\sqrt{x^2 + y^2 + z^2}\), we attempt to minimize \(x^2 + y^2 + z^2\).
• Squaring helps simplify our calculations while preserving the problem’s essence.

This approach makes solving the problem with calculus techniques, like Lagrange multipliers, more straightforward.

Partial Derivatives

Partial derivatives are like a special extra tool from calculus used when you have functions of several variables. They show how a function changes if you change just one of those variables, while keeping the others fixed. In our task of minimizing distance, we used partial derivatives to explore how the distance changes along different axes.
Specifically, we introduced Lagrange multipliers, a method which involves introducing an auxiliary variable,\(\lambda\), to account for constraints. The Lagrangian function, formed here, represents both the primary objective (minimization of \(x^2 + y^2 + z^2\)) and the constraint \(x + y + z = 1\).

• The partial derivative with regard to each variable (in this case, \(x\), \(y\), \(z\), and \(\lambda\)) helps us find where the function has potential minima or maxima, by setting them to zero.
• For example, \(\frac{\partial \mathcal{L}}{\partial x}\) gives us \(2x - \lambda = 0\), helping us understand how changes in \(x\) affect the Lagrangian.

Using these equations, we determine relations between the variables that allow us to solve the system efficiently.