Problem 13 - Calculus of a Single Variable

Question

Minimum Distance In Exercises \(13-16,\) find the point on the graph of the function that is closest to the given point. $$ f(x)=x^{2},\left(2, \frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The solution involves finding the minimum of the distance function by taking its derivative and setting it equal to zero. This gives the \(x\) coordinate of the nearest point on the curve. Then, the \(y\) coordinate can be found by substituting this \(x\) value into the equation of the function. Complete final solution depends upon solving \(D'(x)=0\) which isn't a simple calculation.

Step 1: Formulate The Distance Function

The distance between the point \((x,y)\) where \(y=f(x)=x^2\) on the graph and the given point \((2, \frac{1}{2})\) is given by \[D(x) = \sqrt{(x-2)^2+(x^2-\frac{1}{2})^2}\]

Step 2: Differentiate The Distance Function

Differentiate the distance function with respect to \(x\). The derivative is given by \[D'(x) = \frac{1}{\sqrt{(x-2)^2+(x^2-\frac{1}{2})^2}} \cdot [(x-2) + (x^2-\frac{1}{2}) \cdot 2x]\]

Step 3: Find The Value Of x that Minimizes The Distance

Set the derivative equal to zero and solve for \(x\), which will give the \(x\)- coordinate of the point on the curve that is closest to \((2, \frac{1}{2})\). Solve \[D'(x) = 0\] which simplifies to \[(x-2) + (x^2-\frac{1}{2}) \cdot 2x = 0\]

Step 4: Find The Corresponding y-coordinate

After finding the \(x\) coordinate, substitute this into the function \(f(x)=x^2\) to find the corresponding \(y\) coordinate. The point \((x,y)\) is the point on the curve \(f(x)=x^2\) that is closest to \((2, \frac{1}{2})\)

Key Concepts

Distance Function

When tackling calculus optimization problems involving spatial relationships, the distance function becomes a crucial tool. This function measures the distance between any two points in the Cartesian plane. Specifically, when we have a fixed point, say \(A\) with coordinates \( (2, \frac{1}{2}) \), and a variable point \(B\) on the graph of a function \(y = f(x)\), the distance function \(D(x)\) represents the distance between \(A\) and \(B\).

Mathematically, the formula for the distance between two points \( (x_1, y_1) \), and \( (x_2, y_2) \) is derived from the Pythagorean theorem and given as: \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. For our calculus problem, this becomes the function \(D(x)\) that you'll solve to find the point closest to \(A\).

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes with respect to a change in its input value. In optimization problems, we differentiate the function of interest to find the points at which it has a maximum or minimum value.

In the case of our distance function \(D(x)\), finding its derivative \(D'(x)\) is a crucial step. The derivative provides valuable information about the rate at which the distance changes as we move along the curve of \(f(x)\). By setting this derivative equal to zero, we can identify potential points where the function reaches its minimum distance—a fundamental step in optimization problems.

Minimizing Distance

Minimizing distance is the goal of our optimization problem—specifically, finding the point on the graph that is closest to a given point. After obtaining the distance function and its derivative, the task becomes to determine where this derivative equals zero, signaling a potential minimum.

We search for the value(s) of \(x\) that satisfy \(D'(x) = 0\). These values are candidates for where the distance between our fixed point and the curve is at its minimum. However, it's important to verify that these are indeed minima by checking the second derivative, or by using the first derivative test which examines the sign change of \(D'(x)\) around the critical points.

Quadratic Function

The quadratic function has the general form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and its graph is a parabola. In our optimization problem, \(f(x) = x^2\), which is a simple quadratic function.

A parabola has a single line of symmetry and either a maximum or minimum point called the vertex. The importance of the quadratic function in our context lies in its simplicity and the fact that its minimum or maximum can be straightforwardly determined using differentiation. The quadratic nature of \(f(x)\) in our problem ensures that there's only one minimum distance to the given point—which simplifies determining the optimal point on the curve that is nearest to \( (2, \frac{1}{2}) \).