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When dealing with optimization problems in calculus, particularly those concerning distance, it's essential to understand the Distance Formula. This formula calculates the distance between two points in a plane and is represented as \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It's derived from the Pythagorean Theorem and is widely used in geometry and calculus problems.
In our example, the problem asks us to find the point on the curve \(y=\sqrt{1-x}\) closest to the origin, \((0, 0)\). Finding this shortest distance involves plugging the coordinates of any point on the curve into the Distance Formula. This connects seamlessly with the broader topic of mathematical optimization, where we aim to find the minimum value of the distance.

In calculus, a derivative represents how a function changes as its input changes. It provides the rate of change or the slope of the function at any given point. For optimization problems, derivatives are crucial because they help us find the minimum or maximum values of a function.
In the given exercise, we have a function \(D^2 = x^2 + 1 - x\), derived from the square of our distance formula. By differentiating this function with respect to \(x\), we obtain \(\frac{d}{dx}(x^2 + 1 - x) = 2x - 1\). Finding where this derivative equals zero allows us to pinpoint the values where the distance can be minimized. This process effectively helps us navigate the way functions behave and uncover their extremum points.

Critical points occur where the derivative of a function is zero or undefined. These points are significant because they can represent locations of minimums, maximums, or saddle points on the graph of a function.
In our problem, after differentiating \(D^2\) and setting the derivative to zero, we found \(2x - 1 = 0\). Solving for \(x\) gives us \(x = \frac{1}{2}\). This \(x\) value is a critical point of our distance function. To ensure this critical point is a minimum, we need to verify it lies within the domain of the function, and potentially check the second derivative if necessary.
Critical points serve as the cornerstone for determining where functions may reach their lowest or highest values, especially in the realm of calculus-based optimization.

Mathematical modeling involves translating a real-world scenario into mathematical language. It enables us to solve practical problems by crafting equations and functions that represent physical phenomena.
In the exercise, we created a mathematical model representing the distance from a point on the curve \(y=\sqrt{1-x}\) to the origin. We did this by adapting the Distance Formula into the equation \(D^2 = x^2 + 1 - x\), which serves as our model for solving the problem. Through this model, calculus techniques such as differentiation and identification of critical points were applied to find the point on the curve closest to the origin.
Mathematical modeling is a powerful tool in optimization, guiding us to formulate problems and apply mathematics to find optimal solutions.