Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the minimum distance between the point \((1,2,0)\) and the quadric cone \(z^{2}=x^{2}+y^{2}\).

Step-by-Step Solution

Answer

The minimum distance is \(\sqrt{2}\).

Step 1: Define the Distance Function

The first step is to define the distance between a point on the cone and the point \((1, 2, 0)\). A general point on the cone in terms of \(x\), \(y\), and \(z\) where \(z^2 = x^2 + y^2\) is \((x,y,z)\). The distance \(d\) is given by:\[d = \sqrt{(x-1)^2 + (y-2)^2 + z^2}\]Given \(z^2 = x^2 + y^2\), substitute to express \(d\) solely in terms of \(x\) and \(y\):\[d = \sqrt{(x-1)^2 + (y-2)^2 + x^2 + y^2}\].

Step 2: Simplify the Distance Function

Next, simplify the expression for the distance:\[d = \sqrt{2x^2 + 2y^2 - 2x - 4y + 5}\].This distance function is what we will use to find the minimum distance.

Step 3: Minimize the Distance Function

To find the minimum distance, we should minimize \(d^2\) (the square of the distance) instead, to avoid dealing with the square root. Set:\[f(x, y) = 2x^2 + 2y^2 - 2x - 4y + 5\].Find the critical points by setting the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) to zero:- \(\frac{\partial f}{\partial x} = 4x - 2 = 0\) implies \(x = \frac{1}{2}\),- \(\frac{\partial f}{\partial y} = 4y - 4 = 0\) implies \(y = 1\).

Step 4: Verify Critical Points and Calculate Minimum Distance

Substitute \(x = \frac{1}{2}\) and \(y = 1\) back into the constraint \(z^2 = x^2 + y^2\) to find \(z\):\[z^2 = \left(\frac{1}{2}\right)^2 + 1^2 = \frac{1}{4} + 1 = \frac{5}{4}\].Calculate \(z\) to obtain \(z=\pm \frac{\sqrt{5}}{2}\).Finally, substitute the values \((x = \frac{1}{2}, y = 1, z = \frac{\sqrt{5}}{2})\) into the original distance formula:\[d = \sqrt{\left(\frac{1}{2} - 1\right)^2 + (1-2)^2 + \left(\frac{\sqrt{5}}{2}\right)^2}= \sqrt{\frac{1}{4} + 1 + \frac{5}{4}} = \sqrt{2}\].The minimum distance is thus \(\sqrt{2}\).

Key Concepts

Optimization Problems

Optimization problems in calculus involve finding maximum or minimum values of a function, subject to certain constraints. In our case, the goal is to find the minimum distance between a point and a given surface, specifically a quadric cone described by an equation. An optimization problem like this usually involves determining values for variables so that a particular measurement, in this case distance, is minimized. It begins with formula definition and proceeds with finding critical points, which serve as candidates for these extreme values. In essence, we apply calculus techniques to obtain points that make the function as small as possible, satisfying any necessary conditions or constraints imposed by the problem.

Solving optimization problems can require creativity and experience, as they frequently involve derivative calculations, critical point identification, and sometimes considering boundaries or additional mathematical conditions.

Partial Derivatives

Partial derivatives play a crucial role in solving optimization problems involving multiple variables. They allow us to examine how a function changes in response to slight alterations in one variable, keeping others constant.

In our exercise, the distance function has two variables, \(x\) and \(y\), which calls for partial derivatives with respect to each. Computing \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) results in equations that help identify critical points. These critical points are where the derivative equals zero or does not exist, indicating potential minimum or maximum values. Through careful evaluation of these points, we can distinguish whether they represent a true minimum, which in this instance indicates the minimum distance between the point and the quadric cone. Partial derivatives help uncover these tiny shifts at the heart of multidimensional calculus problems.

Constraints in Calculus

Constraints in calculus provide boundaries or necessary conditions that solutions to problems must respect. For this exercise, the primary constraint is the equation of the quadric cone \(z^2 = x^2 + y^2\). It explicitly ties together the variables so that the function representing distance accurately models the problem.

Incorporating constraints requires substituting them into the overall equation or system being studied. This replaces one of the variables, making it satisfy the constraint inherently, often simplifying the problem. Many optimization problems involve more than one constraint, potentially necessitating Lagrange multipliers or similar techniques for tackling them. However, for this problem, the single constraint needed to link \(z^2\) with \(x^2 + y^2\) ensures our final solution remains valid within the condition specified by the cone's equation.

Euclidean Distance

Euclidean distance is a staple concept in geometry and calculus, measuring the straight-line distance between two points in space. It extends into three dimensions by considering each coordinate direction: \(x\), \(y\), and \(z\). The common distance formula, \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\), is adapted to our specific conditions, replacing \(z\) using the quadric surface equation.

Calculating Euclidean distance is foundational to numerous applications, from physics to machine learning. It quantifies closeness in real terms and, when minimized, indicates the closest points between objects. This basic geometric understanding supports finding the optimal solution in optimization problems, ensuring accurate and meaningful results both in pure mathematics and practical applications.