Problem 71 - Thomas Calculus

Question

\begin{equation} \begin{array}{l}{\text { a. How close does the curve } y=\sqrt{x} \text { come to the point }} \\ {(3 / 2,0) ?(\text {Hint} : \text { If you minimize the square of the distance, }} \\ {\text { you can avoid square roots.) }} \\\ {\text { b. Graph the distance function } D(x) \text { and } y=\sqrt{x} \text { together and }} \\ {\text { reconcile what you see with your answer in part (a). }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The curve comes closest at point (1,1) with minimum distance of \(\sqrt{\frac{5}{4}}\).

Step 1: Define Distance Formula

The distance between a point \((x, \sqrt{x})\) on the curve and the point \((\frac{3}{2}, 0)\) is given by \( D(x) = \sqrt{(x - \frac{3}{2})^2 + (\sqrt{x} - 0)^2} \). To minimize \( D(x) \), we minimize \( D^2(x) \) to avoid square roots.

Step 2: Simplify Distance Squared

The function to minimize is \( D^2(x) = (x - \frac{3}{2})^2 + x \). Expand the terms: \((x - \frac{3}{2})^2 = x^2 - 3x + \frac{9}{4}\), so \( D^2(x) = x^2 - 3x + \frac{9}{4} + x \). Simplify this to get \( D^2(x) = x^2 - 2x + \frac{9}{4} \).

Step 3: Derivative and Critical Points

Find \( D^2(x) \)'s critical points by taking its derivative and setting it to zero: \( \frac{d}{dx}(x^2 - 2x + \frac{9}{4}) = 2x - 2 \). Setting this equal to zero gives \( 2x - 2 = 0 \) or \( x = 1 \).

Step 4: Verify Minimum

The second derivative, \( \frac{d^2}{dx^2}(x^2 - 2x + \frac{9}{4}) = 2 \), is positive, confirming a minimum at \( x = 1 \). The closest distance is thus at \((1, \sqrt{1}) = (1, 1)\).

Step 5: Calculate Minimum Distance

Compute the squared minimal distance from the point \((1, 1)\) to \((\frac{3}{2}, 0)\): \( D^2(1) = (1 - \frac{3}{2})^2 + (1 - 0)^2 = (\frac{-1}{2})^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4} \). So, the minimum distance \( D \) is \( \sqrt{\frac{5}{4}} \).

Step 6: Graph and Analyze

Graph \( D^2(x) \) and \( y = \sqrt{x} \) using tools or sketching. Notice the closest point \((1, 1)\) is where \( D^2(x) \) is minimized, matching our analysis in parts a and b.

Key Concepts

Critical Points

In calculus, critical points help us find where a function's slope is zero or undefined, which often reveals maximums, minimums, or points of inflection. These points are crucial in distance minimization problems because they can indicate where the shortest or longest distance might occur.

To find critical points, we take the derivative of the function we're analyzing—in this case, the squared distance function—to see where these rates of change become zero. For the distance minimization problem involving \[D^2(x) = x^2 - 2x + \frac{9}{4}\], we calculate its derivative \[\frac{d}{dx}(x^2 - 2x + \frac{9}{4}) = 2x - 2\].

Setting the derivative equal to zero \(2x - 2 = 0\) lets us find the critical point: \(x = 1\). This helps us determine the spot on the curve where the distance to the given point might be minimized. By working through these calculations, we confirm that the critical point corresponds to the minimum distance.

Derivative

The term 'derivative' refers to the rate at which a function changes at any given point. It's a fundamental concept in calculus that provides insight into the behavior of functions and allows us to solve optimization problems. In the context of distance minimization, derivatives are used to find where the function could reach its minimum or maximum value.

When minimizing squared distance, we derive the function\(D^2(x) = x^2 - 2x + \frac{9}{4}\) to obtain \(\frac{d}{dx}(x^2 - 2x + \frac{9}{4}) = 2x - 2\). This derivative tells us the slope of the tangent line to our function at any given \(x\).

• If \(\frac{d}{dx}\) is zero at some point, the function may have a minimum or a maximum there because the slope is flat (or horizontal).
• If the derivative is positive, the function is increasing, and if negative, it's decreasing.

In distance optimization problems, once we find \(x\) where the derivative is zero, further analysis (such as the second derivative test) helps confirm whether we found a minimum distance.

Distance Formula

The distance formula is key to solving problems involving the shortest distance between a point and a curve. In this case, the formula calculates how far a point \((x, \sqrt{x})\) on the curve \(y = \sqrt{x}\) is from another fixed point \((\frac{3}{2}, 0)\).

It is given by\(D(x) = \sqrt{(x - \frac{3}{2})^2 + (\sqrt{x} - 0)^2}\).This formula accounts for differences in both x-coordinates and y-coordinates. Using this, you find the actual distance between any two points, which is crucial for understanding distance minimization problems.

• The squared distance avoids square root complications and is easier to minimize because it leads to simpler expressions and calculations.
• Minimizing the squared distance yields the same results as minimizing the distance, requiring less computational effort.

The simplified version helps focus directly on finding where the shortest possible distance occurs between two points.

Squared Distance

The concept of squared distance is a helpful trick in calculus, especially for minimization problems. Instead of working directly with the distance formula, we use the squared version. This step sidesteps the complexity introduced by square roots.

Specifically, instead of minimizing\(D(x) = \sqrt{(x - \frac{3}{2})^2 + (\sqrt{x} - 0)^2}\),we minimize \(D^2(x) = (x - \frac{3}{2})^2 + x\).This function is cleaner to work with mathematically.

• It allows us to apply derivatives directly without worrying about square roots.
• It provides the same critical points and thus guides us to the same solution.

By rewriting the distance minimization problem in terms of squared distance, we find that \(x = 1\) gives our minimal squared distance of\(\frac{5}{4}\). This approach makes it easier to identify and verify the minimum point on the curve, leading to the shortest distance between the curve and the given external point.