91Ó°ÊÓ

In calculus optimization, one common problem involves minimizing the distance between a given point and a curve. It's crucial to understand that directly minimizing the distance can sometimes be difficult due to the complexity of square root calculations. In these cases, it can be beneficial to minimize the square of the distance instead. This method maintains the relative position of points but simplifies the process of finding the minimum value.

The distance minimization process involves determining the point on a curve where the distance to another given point is the smallest possible. Using the given parabola in our exercise, expressed as \( x = 2y^2 \), and the external point \((10, 0)\), we aim to minimize the squared distance rather than the distance itself. This simplifies differentiation because the square root in the original distance formula is eliminated when we minimize:

• \( d^2 = (x - 10)^2 + y^2 \)

By substituting \( x = 2y^2 \) into the squared distance formula, you achieve a simpler expression to work with. This approach is particularly useful in optimization problems where derivative calculations are necessary.

Derivative calculation is a cornerstone of optimization problems in calculus. When determining where a function reaches its minimum (or maximum), computing the derivative helps identify these points. In the context of minimizing distance, we derive the squared distance function to find the values of \( y \) that result in a minimum distance.

Our expanded expression for \( d^2 \) was:

• \( d^2 = 4y^4 - 39y^2 + 100 \)

To find the critical points, we differentiate this quadratic expression with respect to \( y \). The derivative \( \frac{d}{dy}(4y^4 - 39y^2 + 100) \) results in:

• \( 16y^3 - 78y \)

Setting the derivative equal to zero allows us to determine the potential values of \( y \) where the minima or maxima occur. Solving \( 16y^3 - 78y = 0 \) determines these values.

• Factoring gives: \( y(16y^2 - 78) = 0 \)

Identifying critical points is essential when working on optimization problems. These points occur where a function's derivative is zero or undefined and can indicate potential minima, maxima, or points of inflection.

In maintaining our focus on minimizing distance, the critical points are found by solving the equation derived from setting the derivative equal to zero. Here we've identified the possible values of \( y \) as:

• \( y = 0 \)
• \( 16y^2 - 78 = 0 \)

From this, further solving gives a solution for the critical values of \( y \). We found:

• \( y = \pm \frac{\sqrt{78}}{4} \)

With \( y \) values determined, substitute back into the parabola equation \( x = 2y^2 \) to find the corresponding \( x \) values, leading to points on the parabola nearest to \((10, 0)\). Checking each of these points confirms which one provides the absolute minimum distance. These calculations ultimately lead to determining the points responsible for the shortest distance efficiently.