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

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

1. Closest point to the origin: (0, 0, -12). 2. Closest point to (4, 5, 6): (3.2, 2.4, 1.6).

Step 1: Understanding the problem

We need to find the point on the plane, given by the equation \(z = 2x + 3y - 12\), that is closest to the origin and to the point (4,5,6). For both parts, we will use geometric concepts and optimization techniques.

Step 2: Closest point to the origin (First part)

The distance from a point \((x,y,z)\) to the origin \((0,0,0)\) is given by \(d = \sqrt{x^2 + y^2 + z^2}\). To find the closest point on the plane, we need to minimize this distance subject to \(z = 2x + 3y - 12\). Substitute \(z = 2x + 3y - 12\) into the distance formula to get \(d = \sqrt{x^2 + y^2 + (2x + 3y - 12)^2}\). Minimize \(d^2\) to simplify the optimization process since the square root is a monotonically increasing function.

Step 3: Solve for minimization

We minimize \(x^2 + y^2 + (2x + 3y - 12)^2\) by finding its critical points. Compute the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), and set them to zero. Solve the resulting system of equations to find the values of \(x\) and \(y\) that minimize the squared distance.

Step 4: Verify critical point

After solving, we find \(x = 0\) and \(y = 0\) as the critical point. Substitute back into the plane equation to get \(z = 2(0) + 3(0) - 12 = -12\). Verify by the second derivative test or a monotonically increasing property of the distance function that this indeed minimizes the distance to the origin.

Step 5: Closest point to (4,5,6) (Second part)

Now, find the point on the plane closest to (4,5,6). Define the distance function \(d = \sqrt{(x-4)^2 + (y-5)^2 + (z-6)^2}\) and replace \(z\) with \(2x + 3y - 12\). Minimize \(d^2 = (x-4)^2 + (y-5)^2 + (2x + 3y -18)^2\) similarly to the first part.

Step 6: Solve minimization problem

Compute partial derivatives \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and solve the system of equations they form. Once solved, we find \(x = 3.2\), \(y = 2.4\). Substitute into the plane equation \(z = 2(3.2) + 3(2.4) - 12 = 1.6\) to find the z-coordinate.

Step 7: Confirm result for second part

Confirm these calculations by ensuring the second derivative test or study of critical paths shows minimization.

Key Concepts

Distance Minimization

In multivariable calculus, distance minimization often involves finding the closest point on a given surface to a specified point. The concept is rooted in optimization, where we try to find the easiest path, or smallest measurement, between two points. We measure the distance from a point \(a, b, c\) to another point \(x, y, z\) using the distance formula:

• \(d = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}\)

To simplify optimization problems, instead of working with the distance \(d\), we focus on minimizing \(d^2\). This is because squaring maintains the order of distances (if one distance is less, the square will also be less), and it eliminates the need for dealing with the square root, which simplifies our calculations.
When minimizing the distance between a point and a plane, like in our example, we plug the plane's equation into the distance formula. By transforming the focus from the original distance to its squared form, we then use various calculus techniques to find the optimal solution that yields the closest point.

Partial Derivatives

In the context of distance minimization on a surface or curve, partial derivatives play a crucial role. A partial derivative represents the rate at which a function changes as one of the variables is varied while others are held constant.
For a function \(f(x, y)\), the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x}\), and for \(y\) it is \(\frac{\partial f}{\partial y}\). This concept helps us find the optimal values of variables that minimize or maximize the function.

• In our example of finding the closest point on the plane to a specific location, we take the partial derivatives of the function \(f(x, y) = x^2 + y^2 + (2x + 3y - 12)^2\), because it derives from the squared distance formula.
• These partial derivatives serve as necessary tools to identify the critical points of the function.

Setting these derivatives equal to zero forms a system of equations. Solving this system allows us to pinpoint location(s) where the function may have a minimum or maximum value.

Critical Points

Once the partial derivatives have been determined, critical points are identified by solving the resulting system of equations. A critical point occurs where the derivatives are zero or undefined, indicating potential locations of minimum or maximum values of the function.
Finding the critical points involves:

• Calculating the partial derivatives of the equation derived from \(d^2\).
• Solving \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\) to find values that make the rate of change zero.

For example, in the plane minimization problem, solving these equations led to specific x, y values. These were then substituted back into the plane equation to find the corresponding z.After obtaining these critical points, further analysis such as the second derivative test confirms whether the point is indeed a minimum or maximum. This ensures a reliable solution for problems involving minimal distances, like finding the closest point on a plane to a particular location. Checking the behavior around critical points guarantees the accuracy of the optimization process.