Problem 9 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find the point of the plane \(z=2 x+3 y-12\) that is 1\. closest to the origin. 2\. closest to (4,5,6)

Step-by-Step Solution

Answer

Closest to origin: (0, 0, -12); Closest to (4,5,6): Solve for critical points.

Step 1 - Understand the Problem

We need to find the point on the plane defined by the equation \(z = 2x + 3y - 12\) that is closest to a given point. Let’s address each part separately: first find the point closest to the origin, and then find the point closest to \((4,5,6)\).

Step 2 - Write the Distance Formula to the Origin

The distance from a point \((x, y, z)\) on the plane to the origin \((0, 0, 0)\) is given by: \[ D = \sqrt{x^2 + y^2 + z^2} \]Substitute \(z = 2x + 3y - 12\) into this formula.

Step 3 - Minimize the Distance to the Origin

Instead of minimizing \(D\), minimize \(D^2 = x^2 + y^2 + (2x + 3y - 12)^2\) to avoid square roots. Expanding and simplifying gives: \[ D^2 = x^2 + y^2 + (2x + 3y - 12)^2 = x^2 + y^2 + 4x^2 + 9y^2 + 4xy + ...\] Use partial derivatives to find \(x\) and \(y\) that minimize \(D^2\).

Step 4 - Use Partial Derivatives for the Origin problem

Find the partial derivatives: \(\frac{\partial D^2}{\partial x} = 0\) and \(\frac{\partial D^2}{\partial y} = 0\). Solve these equations together to find the critical points that represent the \((x, y)\) coordinates of the closest point to the origin.

Step 5 - Compute the Solution for the Origin

Solve the system of equations from Step 4. Substitute back to find \(z\) using \(z = 2x + 3y - 12\) for the coordinates of the closest point. You get the point where these conditions are satisfied: \((x, y, z)\).

Step 6 - Write the Distance Formula to (4,5,6)

Find the distance from point \((x, y, z)\) on the plane to \((4, 5, 6)\) using: \[ D = \sqrt{(x-4)^2 + (y-5)^2 + (z-6)^2} \]Substitute \(z = 2x + 3y - 12\) into this distance formula.

Step 7 - Minimize the Distance to (4,5,6)

Minimize \(D^2 = (x-4)^2 + (y-5)^2 + (2x + 3y - 12 -6)^2\) to find a critical point. Expand and simplify the expression, then use partial derivatives to find \((x, y)\).

Step 8 - Use Partial Derivatives for the (4,5,6) problem

Find the partial derivatives: \(\frac{\partial D^2}{\partial x} = 0\) and \(\frac{\partial D^2}{\partial y} = 0\). Solve these equations together for the critical \(x\) and \(y\) values.

Step 9 - Compute the Solution for the Point (4,5,6)

Solve the system of equations from Step 8. Finally, substitute \(x\) and \(y\) back to find \(z\) using \(z = 2x + 3y - 12\) for the coordinates of the point closest to \((4, 5, 6)\).

Key Concepts

Distance Minimization

In multivariable calculus, the concept of distance minimization involves finding the smallest possible distance between a point on a surface and a given reference point. When dealing with distances in three-dimensional (3D) space, the Euclidean distance formula is commonly used. For a plane defined by an equation like \(z = 2x + 3y - 12\), we are tasked with calculating the closest point on this plane to another point, such as the origin \((0,0,0)\) or any arbitrary point like \((4,5,6)\). This distance is mathematically expressed as:

• For the origin: \[ D = \sqrt{x^2 + y^2 + z^2} \]
• For the point \((4,5,6)\): \[ D = \sqrt{(x-4)^2 + (y-5)^2 + (z-6)^2} \]

To simplify the process and avoid dealing with square roots, it is often beneficial to minimize the square of the distance \(D^2\) instead. This requires solving a system of equations obtained by finding derivatives and setting them to zero. This leads us to our next concept, partial derivatives.

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus, used to understand how a function changes as each of its independent variables is changed, while others are held constant. When minimizing distance squared \(D^2\), partial derivatives help us determine the rate of change of \(D^2\) with respect to each of the variables \(x\) and \(y\).
For our problem, we are particularly interested in:

• \(\frac{\partial D^2}{\partial x}\)
• \(\frac{\partial D^2}{\partial y}\)

Taking these derivatives and setting them equal to zero helps find the critical points of the function. These critical points potentially represent the locations on the plane \(z = 2x + 3y - 12\) that are closest to reference points, such as the origin or \((4,5,6)\). Partial derivatives thus enable a systematic approach to distance minimization.

Critical Points

Critical points are values of \(x\) and \(y\) where the first derivatives (partial derivatives in this context) are zero, leading potentially to local minima, maxima, or saddle points on the plane. Solving for critical points is instrumental in determining where the closest points lie on the plane with respect to another point.
In our exercise, once we compute the partial derivatives \(\frac{\partial D^2}{\partial x} = 0\) and \(\frac{\partial D^2}{\partial y} = 0\), we solve these equations simultaneously. These provide the \((x, y)\) coordinates which, when substituted back into the plane equation \(z = 2x + 3y - 12\), give the complete \((x, y, z)\) coordinates for the point closest to the reference point. These critical points define the positions on the plane that minimize the distance to points like the origin or \((4,5,6)\).

Planes in 3D Space

Planes in 3D space are flat, two-dimensional surfaces that extend infinitely along their directions. They can be defined by linear equations of the form \(z = ax + by + c\). In our problem, we have the plane \(z = 2x + 3y - 12\).
Key characteristics of planes include:

• The normal vector, perpendicular to the plane, which helps in determining distances and angles relative to other geometrical elements in space.
• Intersection with other geometrical entities, such as lines or points, which leads to critical points or solutions when solving for conditions such as distance minimization.

Understanding the equation and properties of planes allows us to apply various calculus techniques, like partial derivatives, to solve real-world problems involving 3D space scenarios, such as determining the closest point on a plane to another point in space.