Problem 31 - College Algebra and 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.) Plane: \(x+y+z=1,(2,1,1)\) Minimize \(d^{2}=(x-2)^{2}+(y-1)^{2}+(z-1)^{2}\)

Step-by-Step Solution

Answer

The minimum distance from the point (2, 1, 1) to the plane \(x+y+z=1\) is \(\frac{√3}{2}\)

Step 1: Formulate the distance equation

To get the minimum distance from a point to a plane, start by specifying the equation for the distance from the point to an arbitrary point on the plane, and then square it in order to make the calculus easier. The square of the distance from the given point (2, 1, 1) to an arbitrary point (x, y, z) on the plane is \((x-2)^{2}+(y-1)^{2}+(z-1)^{2}\)

Step 2: Substitute z from the plane equation

From the equation of the plane \(x+y+z=1\), you can express \(z\) in terms of \(x\) and \(y\). It becomes \(z=1-x-y\). Substitute this into the distance-squared equation: \(d^{2}=(x-2)^{2}+(y-1)^{2}+[(1-x-y)-1]^{2}\)

Step 3: Simplify the distance equation and differentiate

Simplify \(d^{2}\) to get a function of \(x\) and \(y\) only. This results in: \(d^{2}=2x^{2}+2y^{2}-4x-2y+2\). To find the minimum of \(d^{2}\), differentiate the equation with respect to \(x\) and \(y\). Equate the partial derivatives to 0. Hence, \(\frac{∂d^{2}}{∂x} = 4x - 4 = 0\) and \(\frac{∂d^{2}}{∂y} = 4y - 2 = 0\)

Step 4: Solve the equations

Solving these equations gives the values of \(x\) and \(y\) at the point of minimum distance. Hence, \(x = 1\) and \(y = \frac{1}{2}\). Substitute these values into the equation of the plane to get \(z = 1-x-y = \frac{1}{2}\)

Step 5: Calculate the minimum distance

The minimum distance is then given by substituting the values of \(x\), \(y\), and \(z\) into the formula for \(d\), where \(d = \sqrt{(x-2)^{2}+(y-1)^{2}+(z-1)^{2}}\). Substituting gives \(d = \sqrt{(1-2)^{2} + (\frac{1}{2}-1)^{2}+(\frac{1}{2}-1)^{2}}=\frac{√3}{2}\)

Key Concepts

Distance Squared Method

The method of minimizing the square of the distance is a clever approach in optimization problems involving distances.
By focusing on the square of the distance, we eliminate the square root, making calculations simpler and derivatives easier to handle.
This is particularly useful in calculus, as it transforms our distance function into a polynomial, which is smoother to differentiate and analyze. When exploring the distance from a point to a surface, like a plane, we first define the distance function. Instead of \(d\), the square of the distance \(d^2\) is often used:

• For a point (2, 1, 1) and a plane \(x + y + z = 1\), the square of the distance \(d^2\) to a point (x, y, z) on the plane is expressed as: \(d^2 = (x-2)^2 + (y-1)^2 + (z-1)^2\).

This simplification aids considerably when utilizing calculus techniques, particularly partial derivatives, to find the minimum value.

Partial Derivatives

Differentiation, specifically partial derivatives, plays a crucial role when optimizing multivariable functions.
With a function like our squared distance \(d^2\), we treat the variables separately and focus on changing one variable at a time while keeping others constant.
This is what makes partial derivatives so useful.To find the critical points that potentially minimize \(d^2\):

• Compute \(\frac{∂d^2}{∂x}\) and \(\frac{∂d^2}{∂y}\), and equate them to zero.

For instance, in our problem:

• \(\frac{∂d^2}{∂x} = 4x - 4\) is set to zero, leading to \(x = 1\).
• \(\frac{∂d^2}{∂y} = 4y - 2\) is also set to zero, leading to \(y = \frac{1}{2}\).

This approach lets us isolate and solve variables step-by-step, leading to a more manageable path to the solution.

Plane Equation

Understanding the equation of a plane is essential in problems involving distances in three dimensions.
A plane equation generally takes the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the coefficients, providing the orientation, while \(d\) adjusts its position.For our specific problem, the plane is described by \(x + y + z = 1\).
The equation neatly sums the coordinate contributions to equal a constant, determining the plane's layout in space.To solve for \(z\) as done in step 2 of our solution:

• Re-arrange to find \(z = 1 - x - y\), allowing substitution into the distance squared equation.

This conversion is crucial as it simplifies our function \(d^2\) to depend solely on two variables, \(x\) and \(y\), making differentiation straightforward.

Optimization in Calculus

Optimization in calculus focuses on finding the extremum values through derivative analysis, which is either a minimum or maximum.
The steps generally involve identifying critical points, testing boundaries, and using derivatives to verify the nature of these points.For function minimization, like our \(d^2\), the essential steps include:

• Finding where derivatives (like \(\frac{∂d^2}{∂x}\) and \(\frac{∂d^2}{∂y}\)) are zero, implying potential minima or maxima.
• Analyzing these points to confirm whether they represent a true minimum.

In the case of our problem, after determining \(x = 1\) and \(y = \frac{1}{2}\), we further calculate \(z\) and use these in the original distance formula to verify the minimum distance: \(d = \frac{\sqrt{3}}{2}\).Optimization problems are essentially about comparing possible solutions derived from derivative equations to identify the most efficient outcome.