91Ó°ÊÓ

The distance formula is a tool used to determine the distance between two points in a plane. In a coordinate system, if you have two points, say

• Point A at \(x_1, y_1\)
• Point B at \(x_2, y_2\)

the distance \(d\) between them is calculated as \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula comes from the Pythagorean theorem, applied in a coordinate system to find the hypotenuse of a right triangle. Here, we applied it to find the distance between a point on the curve \(y=x^2\) and the point \(0,1\). By substituting the curve equation into the formula, we simplified it to find a function of \(x\), which allows us to easily differentiate it.

The derivative is a fundamental concept in calculus that describes how a function changes as its input changes. When we take the derivative of a function, we are finding the rate at which the function's y-values change with respect to changes in x. In optimization problems, derivatives help us identify points where the function reaches its highest or lowest values, known as extrema. For the function \(f(x) = x^2 + (x^2-1)^2\), we found its derivative to be \(f'(x) = 2x + 4x(x^2-1)\). Differentiation allows us to explore the behavior of the function, especially when we set the derivative equal to zero to find critical points. These are potential candidates for local minima or maxima.

Critical points are values of x where the first derivative of a function is zero or undefined. These points are crucial in finding the extrema, like minima or maxima, of the function. For solving optimization problems, such as finding the closest point on a curve to a given point, identifying critical points allows us to determine where a function might change its increasing or decreasing behavior. From the derivative \(f'(x) = 2x + 4x(x^2-1)\), solving \(f'(x) = 0\) gave us three critical points: \(x=0\) and \(x = \pm\sqrt{3}/2\). These are the x-values where the function could potentially achieve minimum distance from point \(0,1\).

Minimization refers to the process of finding the smallest value of a function, often for specific boundary conditions or constraints. In the context of calculus optimization, it involves using derivatives to identify critical points and then testing these points to determine which yields the lowest function value. In our problem, after finding the critical points, we evaluated the function at each point to see which provided the minimum distance. By substituting the critical points into the function \(f(x) = x^2 + (x^2-1)^2\), we determined that the points \(x= \pm\sqrt{3}/2\) resulted in the lowest value, indicating that these coordinates on the curve \(y=x^2\) are closest to the point \(0,1\). Thus, creating a successful minimization of the distance.