Problem 48 - Calculus and its applications

Question

Find the point on the circle \(x^{2}+y^{2}=1\) that is closest to the point (2,1) .

Step-by-Step Solution

Answer

The point \((\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})\) on the circle is closest to (2,1).

Step 1: Set up optimization

Minimize \( D^2 = (x-2)^2 + (y-1)^2 \) subject to \( x^2+y^2=1 \). Use Lagrange multipliers or parametrize.

Step 2: Parametric approach

On the unit circle, closest point is in direction of \( (2,1) \): \( (x,y) = \frac{(2,1)}{\sqrt{5}} = \left(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right) \).

Key Concepts

Distance Formula

The Distance Formula is a mathematical equation used to determine the distance between two points in a plane. It is derived from the Pythagorean theorem. For two points with coordinates

• Point 1: \((x_1, y_1)\)
• Point 2: \((x_2, y_2)\)

The distance \(D\) between these points is given by the formula:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula is always useful when trying to quantify how far two locations are from each other on a two-dimensional plane. In the given exercise, we are interested in minimizing the distance between a point on a circle and a specific external point, i.e., \((2,1)\). To simplify calculations, it is often practical to consider the square of the distance \(D^2\), which retains the same minima as \(D\) without the computational complexity of a square root.

Minimization Problem

The concept of a minimization problem is central to optimization in calculus. It involves finding the values of variables that result in the lowest (or sometimes highest) value of a given function.In our exercise, the objective is to locate a point on the circle that minimizes the distance to an external point. By defining the function in terms of the squared distance:\[D^2 = (x - 2)^2 + (y - 1)^2\]we can minimize this function, as minimizing the square of the distance achieves the same result as minimizing the actual distance.When solving minimization problems, setting the derivative of the function to zero is a common technique, as it helps identify any critical points. These critical points may represent local minima or maxima. Our focus here is specifically on the global minimum concerning the given constraint (that the point lies on a circle). In practice, solving a minimization problem not only applies to distance but can also extend to various fields such as economics, physics, and engineering.

Constrained Optimization

Constrained optimization deals with problems where there are restrictions or conditions that the solution must satisfy. In our problem, the constraint is given by the equation of the circle \(x^2 + y^2 = 1\). This means that any point \((x, y)\) that we find as a solution must also lie on this circle.To manage this constraint, one common method is using Lagrange multipliers in calculus. However, since we are minimizing a simple function like squared distance in a geometric context, direct substitution or using the constraint equation to express one variable in terms of another can often provide effective results.For example, you might express \(y\) in terms of \(x\) from the circle's equation:\[y = \pm\sqrt{1-x^2}\]Then, substitute into the distance equation and solve for \(x\). Constraints ensure that the solutions are realistic and meaningful for the specific problem space, making constrained optimization a powerful approach in various science and engineering problems.

Equation of a Circle

The equation of a circle is a mathematical representation defining all points in a plane that are at a constant distance, known as the radius, from a fixed point called the center. The standard form of a circle's equation with center at \((h, k)\) and radius \(r\) is:\[(x - h)^2 + (y - k)^2 = r^2\]In the exercise presented, the equation of the circle is \(x^2 + y^2 = 1\), which is a simplification with the circle centered at the origin \((0,0)\) and a radius of 1. This specific scenario is often referred to as the unit circle.The equation provides a constraint on the potential solution space, limiting the \((x, y)\) points we consider to only those that satisfy the circle's equation. By understanding the properties of the circle's equation, one can accurately navigate geometric problems and solve questions related to areas, angles, and distances in a two-dimensional space. Recognizing this constraint is crucial when dealing with optimization problems that concern circular boundaries.